R代写-SECTION 9

Data Mining 2018
SECTION 9 Classification (Cont’d)
drive - ”D:”
code.dir - paste(drive, ”DATA/Data Mining R-Code”, sep”/”)
data.dir - paste(drive, ”DATA/Data Mining Data”, sep”/”)
#
source(paste(code.dir, ”creategrid.r”, sep”/”))c
source (paste(code.dir, ”example_display.r”. sep”/”))
source(paste(code.dir, ”confusion_ex.r”, sep”/”))
source(paste(code.dir, ”ClassBoundary.r”, sep”/”))
Neural Networks
The basic concept in neural networks is the weighting of inputs to produce the desired output
result. Suppose we consider a modification of an example that is familiar to students - you
write 4 tests worth 10% each, do 2 assignments worth 5% each and a final exam worth 50%.
Your mark in the course will be found from
0.1 t1 0.1 t2 0.1 t3 0.1 t4 0.05 a1 0.05 a2 0.5 f1
If this mark exceeds a threshold (say 50%) you pass, otherwise you fail. In this case, you
know in advance what the weighting will be.
Now suppose that your professor does not tell you what the weights will be but instead
decides to set the weights so that the students that (s)he thinks should pass will pass and the
others will fail. The professor would then have to start off with weights (maybe at random)
and see what happens to each student. If a student that the professor feels should pass
exceeds the threshold, no change needs to be made in the weights. On the other hand, if the
expected result does not occur, the weights need to be altered. Of course, changing the
weights means that the professor then has to check on how these new weights affect the
outcome for students whose grades were calculated using the old weights. The process will
thus involve going through the student marks with various weights a number of times until
no changes occur. Note that there may be cases for which the result will not be that which
the professor desired - for example, if two students have the same marks and the professor
felt one should pass and the other fail, it is impossible to find weights to produce the desired
result.
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This type of process can be done with a simple neural net - a Threshold Logic Unit (TLU,
shown below):
Figure 1. Threshold Logic Unit
Here the n inputs (marks, signals) x1,x2, ..., xn with weights w1,w2, ..., wn are put into an
adder, so that the total input to a node is
a x1w1 x2w2 ... xnwn
i1
n
xiwi x w
Suppose the node has a threshold (call it ) such that if a the node will ‘fire’ so the
output is
y
1 if a
0 if a
In a simple case of a TLU having 2 inputs (with values of 0 or 1) with initial weights of 1
and a threshold of 1.5, we get a response table
x1 x2 w1 w2 a x w y
0 0 1 1 0 1.5 0
0 1 1 1 1 1.5 0
1 0 1 1 1 1.5 0
1 1 1 1 2 1.5 1
These inputs could be considered to be points at the corners of a rectangle in two dimensions
(see below). The TLU classifies them into 2 classes (0 - fail, 1 - pass).
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If we look at the situation in this way, we see that we will have a line
x w or x w 0
or x1w1 x2w2
so x2
w1
w2
x1

w2
i.e. x2 Ax1 B
that divides the plane. In our example the slope A 1 and the y-intercept B 1.5.
Figure 2. Logical AND
(We note that this is a logical AND). If this is what we wish to obtain, then we expect
x1 x2 t
0 0 0
0 1 0
1 0 0
1 1 1
The t value is the result that we want or the target value (if we know the results we want, we
are using ‘supervised’ learning).
Suppose that we do not know the weights but are required to find them by ‘training’. i.e. we
are given a set of weights (possibly random) and a threshold and have to adjust the weights
and threshold based on the results that they produce.
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For example, if we start with w1 0.2, w2 0.4, and 0.3, then
x2 0.20.4 x1
0.3
0.4
0.5x1 0.75
Figure 3. Linear non-separation
which does not do the desired classification (logical AND).
Because both are to be trained, we will consider a variation of the preceeding by letting the
threshold be treated as another “weight” attached to an input of 1. So the weight vector
now becomes
w w1,w2,
and the input vector becomes
x x1,x2,1 v,1
so now x w x1w1 x2w2 . This means that we have a new type of node that fires if the
“activation signal” a is 0. That is
y
1 if a x w 0
0 if a x w 0
What will happen if we adjust a weight? Let the first weight become w1 w1 so
x w x1w1 w1 x2w2
x1w1 x2w2 x1w1
x w x1w1
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Similarly we modify the second weight w2. From this we see that if we have an activation
level that is “too low”, we could ‘add’ [w1 w2 to the weight vector (similarly if the
activation level is too high).
If we look at
Response from
the neural net
y
Correct or desired
response
t
Change t y
0 0 no 0
0 1 1
1 0 1
1 1 no 0
we note that any change will involve t y. One possibility would be to add
t yx1, x2 t yv but this might overshoot or overcorrect so we use a learning rate (a
damping factor) and adjust the weights (including threshold) by
w t yv
To help illustrate, we need some functions. The following function sets
y 1 x w 0
0 x w 0
and returns w learning rate class y input (class is the t in the previous formula).
Note the use of formatC in the printing of the results. It allows us to display the results as
floating point values with 4 digits after the decimal point and a total width of 7.
comp.delta.w - function (input, w, class, rate){
activation - t(input)%*%w; # [x1,x2,-1].[w1,w2,thresh]
# if activation 0 y - 1 (fire)
y - (activation 0)*1
factor - rate*(class - y);
cat(formatC(c(w, input, activation, y, class, factor, factor*input,
1-(classy)), format”f”,
digit4, width7), ”\n”)
return(list(deltafactor*input, Err1-(class y)))
}
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The following plots the data with classes coloured, puts the separating line on it, and updates
the weights.
Note: pch sets the plotting character and cex sets the character size.
#
# Plot and update the weights
#
f.update.weights - function (inputs, w, target, rate, XR, YR){
Total.Error - 0
for (i in 1:(length(inputs[,1]))){
plot(XR, YR, type ”n”,
mainpaste(floor(w[1]*100.5)/100, floor(w[2]*100.5)/100,
floor(w[3]*100.5)/100),
xlab””, ylab””)
mtext(paste(”b”,floor(w[3]/w[2]*100.5)/100,
” m”,floor(w[1]/w[2]*100)/100),side1, line0.5, cex.8)
points (inputs[,1], inputs[,2], coltarget2, pch19, cex1.2)
plot.line (w)
points(inputs[i,1], inputs[i,2], col”blue”, pch”x”, cex2)
delta.w - comp.delta.w(inputs[i,], w, target[i], rate)
w - w delta.w\$delta
Total.Error - Total.Error delta.w\$Err
}
return (list(ww, ErrTotal.Error))
}
#
# Plot a coloured line (w w1, w2,
#
plot.line - function (w){
if (w[2] ! 0){
inter - w[3]/w[2]
slope - w[1]/w[2]
abline (inter,slope, col”red”)
}
else { # vertical line at xx
xx - w[3]/w[1]
abline (vxx, col”red”)
}
}
The function f.AND runs through the inputs (up to 200 times) and displays the results.
#
# Main function
#
f.AND - function(inputs, class, w, rate, XR, YR) {
# Now use up to 200 training cycles to get the weights
T.Err - 0
for (i in (1:200)) {
# Display the inputs (coloured by class)
cat(” w1 w2 b0 x1 x2 -1 act. y”,
” target delta dw1 dw2 db \n”)
old.w - w
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res - f.update.weights (inputs, w, class, rate, XR, YR)
w - res\$w
cat(”weights ”,w,” old ”,old.w, ”Err ”, res\$Err, ”\n”)
if (res\$Err 0) {
break
}
if (ret ”q”) break
}
}
Set the data for the problem:
rate - 0.25; # Learning rate
x1 - c(0,0,-1)
x2 - c(0,1,-1)
x3 - c(1,0,-1)
x4 - c(1,1,-1)
oldpar - par(mfrow c(5,4),marc(2,1,2,1),xaxt”n”,yaxt”n”)
f.AND(rbind(x1,x2,x3,x4), c(0,0,0,1), c(-.2, 0.2, -0.5), rate, -.5:1.5, -.5:1.5)
par(oldpar)
w1 w2 b0 x1 x2 -1 act. y target delta
-0.2000 0.2000 -0.5000 0.0000 0.0000 -1.0000 0.5000 1.0000 0.0000 -0.2500 0
-0.2000 0.2000 -0.2500 0.0000 1.0000 -1.0000 0.4500 1.0000 0.0000 -0.2500 0
-0.2000 -0.0500 0.0000 1.0000 0.0000 -1.0000 -0.2000 0.0000 0.0000 0.0000 0
-0.2000 -0.0500 0.0000 1.0000 1.0000 -1.0000 -0.2500 0.0000 1.0000 0.2500 0
weights 0.05 0.2 -0.25 old -0.2 0.2 -0.5 Err 3
e.g. [0.2000 0.2000 0.50000.0000 0.0000 1.0000 0.5 0 so the predicted class
is y 1. The true class is 0 so delta rate target y 0.25 and
delta x 0.0000 0.0000 0.2500. The last column indicates if y target.
w1 w2 b0 x1 x2 -1 act. y target delta
0.0500 0.2000 -0.2500 0.0000 0.0000 -1.0000 0.2500 1.0000 0.0000 -0.2500 0
0.0500 0.2000 0.0000 0.0000 1.0000 -1.0000 0.2000 1.0000 0.0000 -0.2500 0
0.0500 -0.0500 0.2500 1.0000 0.0000 -1.0000 -0.2000 0.0000 0.0000 0.0000 0
0.0500 -0.0500 0.2500 1.0000 1.0000 -1.0000 -0.2500 0.0000 1.0000 0.2500 0
weights 0.3 0.2 0 old 0.05 0.2 -0.25 Err 3
w1 w2 b0 x1 x2 -1 act. y target delta
0.3000 0.2000 0.0000 0.0000 0.0000 -1.0000 0.0000 0.0000 0.0000 0.0000 0
0.3000 0.2000 0.0000 0.0000 1.0000 -1.0000 0.2000 1.0000 0.0000 -0.2500 0
0.3000 -0.0500 0.2500 1.0000 0.0000 -1.0000 0.0500 1.0000 0.0000 -0.2500 -0
0.0500 -0.0500 0.5000 1.0000 1.0000 -1.0000 -0.5000 0.0000 1.0000 0.2500 0
weights 0.3 0.2 0.25 old 0.3 0.2 0 Err 3
w1 w2 b0 x1 x2 -1 act. y target delta
0.3000 0.2000 0.2500 0.0000 0.0000 -1.0000 -0.2500 0.0000 0.0000 0.0000 0
0.3000 0.2000 0.2500 0.0000 1.0000 -1.0000 -0.0500 0.0000 0.0000 0.0000 0
0.3000 0.2000 0.2500 1.0000 0.0000 -1.0000 0.0500 1.0000 0.0000 -0.2500 -0
0.0500 0.2000 0.5000 1.0000 1.0000 -1.0000 -0.2500 0.0000 1.0000 0.2500 0
weights 0.3 0.45 0.25 old 0.3 0.2 0.25 Err 2
w1 w2 b0 x1 x2 -1 act. y target delta
0.3000 0.4500 0.2500 0.0000 0.0000 -1.0000 -0.2500 0.0000 0.0000 0.0000 0
0.3000 0.4500 0.2500 0.0000 1.0000 -1.0000 0.2000 1.0000 0.0000 -0.2500 0
0.3000 0.2000 0.5000 1.0000 0.0000 -1.0000 -0.2000 0.0000 0.0000 0.0000 0
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0.3000 0.2000 0.5000 1.0000 1.0000 -1.0000 0.0000 0.0000 1.0000 0.2500 0
weights 0.55 0.45 0.25 old 0.3 0.45 0.25 Err 2
w1 w2 b0 x1 x2 -1 act. y target delta
0.5500 0.4500 0.2500 0.0000 0.0000 -1.0000 -0.2500 0.0000 0.0000 0.0000 0
0.5500 0.4500 0.2500 0.0000 1.0000 -1.0000 0.2000 1.0000 0.0000 -0.2500 0
0.5500 0.2000 0.5000 1.0000 0.0000 -1.0000 0.0500 1.0000 0.0000 -0.2500 -0
0.3000 0.2000 0.7500 1.0000 1.0000 -1.0000 -0.2500 0.0000 1.0000 0.2500 0
weights 0.55 0.45 0.5 old 0.55 0.45 0.25 Err 3
w1 w2 b0 x1 x2 -1 act. y target delta
0.5500 0.4500 0.5000 0.0000 0.0000 -1.0000 -0.5000 0.0000 0.0000 0.0000 0
0.5500 0.4500 0.5000 0.0000 1.0000 -1.0000 -0.0500 0.0000 0.0000 0.0000 0
0.5500 0.4500 0.5000 1.0000 0.0000 -1.0000 0.0500 1.0000 0.0000 -0.2500 -0
0.3000 0.4500 0.7500 1.0000 1.0000 -1.0000 0.0000 0.0000 1.0000 0.2500 0
weights 0.55 0.7 0.5 old 0.55 0.45 0.5 Err 2
w1 w2 b0 x1 x2 -1 act. y target delta
0.5500 0.7000 0.5000 0.0000 0.0000 -1.0000 -0.5000 0.0000 0.0000 0.0000 0
0.5500 0.7000 0.5000 0.0000 1.0000 -1.0000 0.2000 1.0000 0.0000 -0.2500 0
0.5500 0.4500 0.7500 1.0000 0.0000 -1.0000 -0.2000 0.0000 0.0000 0.0000 0
0.5500 0.4500 0.7500 1.0000 1.0000 -1.0000 0.2500 1.0000 1.0000 0.0000 0
weights 0.55 0.45 0.75 old 0.55 0.7 0.5 Err 1
w1 w2 b0 x1 x2 -1 act. y target delta
0.5500 0.4500 0.7500 0.0000 0.0000 -1.0000 -0.7500 0.0000 0.0000 0.0000 0
0.5500 0.4500 0.7500 0.0000 1.0000 -1.0000 -0.3000 0.0000 0.0000 0.0000 0
0.5500 0.4500 0.7500 1.0000 0.0000 -1.0000 -0.2000 0.0000 0.0000 0.0000 0
0.5500 0.4500 0.7500 1.0000 1.0000 -1.0000 0.2500 1.0000 1.0000 0.0000 0
weights 0.55 0.45 0.75 old 0.55 0.45 0.75 Err 0
We see that, in the last set, the output y is identical to the target value (shown by the delta
values being 0) and so the TLU has been “trained”.
In the following figures, m is the slope and b is the y-intercept. In the figures, the blue
indicates which case is being ‘presented’ for training ( we use a similar process with
self-organizing maps (SOM)).
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Figure 4. First 5 passes Figure 5. Next 4 passes
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The linear separator has the equation
b0 w1 x w2 y 0
or y b0
w2
w1
w2 x
b mx
Note: This simple training algorithm is very restricted in its ability to be extended to deal
with more complicated partitioning using more nodes.
We have a value t y that is a measure of the error for a set of weights. Rather than
consider the output y (which is not continous), we will consider what happens with the
activation
a x w x1w1 ... xnwn .
Suppose we consider as our error for a single node
E i 12 t a
2
(the 12 is there to remove the 2 that appears on differentiation of E i).
We will try to find the set of weights w that gives the minimum error. To do this we note that
if a is considered to be a function of w (t is constant), then
E i
wj
t a a
wj
t a
x1w1 x2w2 ... xnwn
wj

t axj j 1,2, ..., n
t a1
The error curve is quadratic in w. What happens if we make a change in wj that is
proportional to the slope of the error curve?
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E
w
i.e. if we let
wj slope
E i
wj
t axj
learning rate then
E i E iwj
wj
t a2xj
2 0
so we move down towards the minimum.
When we did the original version, we used the fact that we had a 1 if the activation a 0 and
0 if a 0. We need to have our target activations take on the same behaviour. For this, we
will use t to be either 1 or 1.
We can replace the comp.delta.w function with a new version that uses the gradient.
comp.delta.w - function (input.i, w, target, rate){
activation - t(input.i)%*%w
# As before [x1,x2,x3].[w1,w2,w3] [x1,x2,-1].[w1,w2,b]
factor - rate*(target - activation)
cat(formatC(c(w, input.i, activation, target, factor,
factor*input.i, 0.5*(target-activation)^2),
format”f”, digit4,width7), ”\n”)
return(list(deltafactor*input.i, Err0.5*(target-activation)^2))
}
and make a slight change to the main function
f2.AND.G - function(inputs, class, w, rate, XR, YR) {
# Now use up to 200 training cycles to get the weights
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T.Err - 9999999
for (i in (1:200)) {
# Display the inputs (coloured by class)
cat(” w1 w2 b x1 x2 -1 act.”,
” target delta dw1 dw2 db \n”)
old.w - w
res - f.update.weights (inputs, w, class, rate, XR, YR)
w - res\$w
cat(”weights ”,w,” old ”,old.w,”\n”)
cat(”Err ”, res\$Err, ”\n”)
if (abs(res\$Err - T.Err) 0.01*res\$Err) {
T.Err - res\$Err
if (ret ”q”) break
} else {
break
}
}
}
x1 - c(0,0,-1)
x2 - c(0,1,-1)
x3 - c(1,0,-1)
x4 - c(1,1,-1)
oldpar - par(mfrow c(5,4), marc(2,1,2,1), xaxt”n”, yaxt”n”)
f2.AND.G(rbind(x1,x2,x3,x4), c(-1,-1,-1,1), c(-.2, 0.2, -0.5), rate,
-.5:1.5,-.5:1.5)
par(oldpar)
w1 w2 b x1 x2 -1 act. target delta dw1
-0.2000 0.2000 -0.5000 0.0000 0.0000 -1.0000 0.5000 -1.0000 -0.3750 0.0000 0
-0.2000 0.2000 -0.1250 0.0000 1.0000 -1.0000 0.3250 -1.0000 -0.3312 0.0000 -0
-0.2000 -0.1312 0.2062 1.0000 0.0000 -1.0000 -0.4063 -1.0000 -0.1484 -0.1484 0
-0.3484 -0.1312 0.3547 1.0000 1.0000 -1.0000 -0.8344 1.0000 0.4586 0.4586 0
weights 0.1101563 0.3273438 -0.1039063 old -0.2 0.2 -0.5
Err 3.861548
e.g. 0.2000 0.2000 0.5000 0.0000 0.0000 1.0000 0.5. The true class is 1
rate target activation 0.25 1.5 0.375 so
delta x 0.0000 0.0000 0.3750. The last column is
0.5 target activation2 1.52/2 1.125.
w1 w2 b x1 x2 -1 act. target delta dw1
0.1102 0.3273 -0.1039 0.0000 0.0000 -1.0000 0.1039 -1.0000 -0.2760 0.0000 0
0.1102 0.3273 0.1721 0.0000 1.0000 -1.0000 0.1553 -1.0000 -0.2888 0.0000 -0
0.1102 0.0385 0.4609 1.0000 0.0000 -1.0000 -0.3507 -1.0000 -0.1623 -0.1623 0
-0.0522 0.0385 0.6232 1.0000 1.0000 -1.0000 -0.6368 1.0000 0.4092 0.4092 0
weights 0.3570496 0.4477356 0.2139954 old 0.1101563 0.3273438 -0.1039063
Err 2.827031
w1 w2 b x1 x2 -1 act. target delta dw1
0.3570 0.4477 0.2140 0.0000 0.0000 -1.0000 -0.2140 -1.0000 -0.1965 0.0000 0
0.3570 0.4477 0.4105 0.0000 1.0000 -1.0000 0.0372 -1.0000 -0.2593 0.0000 -0
0.3570 0.1884 0.6698 1.0000 0.0000 -1.0000 -0.3128 -1.0000 -0.1718 -0.1718 0
0.1852 0.1884 0.8416 1.0000 1.0000 -1.0000 -0.4680 1.0000 0.3670 0.3670 0
weights 0.5522269 0.555414 0.474629 old 0.3570496 0.4477356 0.2139954
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Err 2.160428
w1 w2 b x1 x2 -1 act. target delta dw1
0.5522 0.5554 0.4746 0.0000 0.0000 -1.0000 -0.4746 -1.0000 -0.1313 0.0000 0
0.5522 0.5554 0.6060 0.0000 1.0000 -1.0000 -0.0506 -1.0000 -0.2374 0.0000 -0
0.5522 0.3181 0.8433 1.0000 0.0000 -1.0000 -0.2911 -1.0000 -0.1772 -0.1772 0
0.3750 0.3181 1.0206 1.0000 1.0000 -1.0000 -0.3275 1.0000 0.3319 0.3319 0
weights 0.706878 0.6499282 0.6886811 old 0.5522269 0.555414 0.474629
Err 1.721121
w1 w2 b x1 x2 -1 act. target delta dw1
0.7069 0.6499 0.6887 0.0000 0.0000 -1.0000 -0.6887 -1.0000 -0.0778 0.0000 0
0.7069 0.6499 0.7665 0.0000 1.0000 -1.0000 -0.1166 -1.0000 -0.2209 0.0000 -0
0.7069 0.4291 0.9874 1.0000 0.0000 -1.0000 -0.2805 -1.0000 -0.1799 -0.1799 0
0.5270 0.4291 1.1672 1.0000 1.0000 -1.0000 -0.2112 1.0000 0.3028 0.3028 0
weights 0.8297923 0.7318663 0.864451 old 0.706878 0.6499282 0.6886811
Err 1.430988
w1 w2 b x1 x2 -1 act. target delta dw1
0.8298 0.7319 0.8645 0.0000 0.0000 -1.0000 -0.8645 -1.0000 -0.0339 0.0000 0
0.8298 0.7319 0.8983 0.0000 1.0000 -1.0000 -0.1665 -1.0000 -0.2084 0.0000 -0
0.8298 0.5235 1.1067 1.0000 0.0000 -1.0000 -0.2769 -1.0000 -0.1808 -0.1808 0
0.6490 0.5235 1.2875 1.0000 1.0000 -1.0000 -0.1150 1.0000 0.2787 0.2787 0
weights 0.9277692 0.8022292 1.008743 old 0.8297923 0.7318663 0.864451
Err 1.239578
w1 w2 b x1 x2 -1 act. target delta dw1
0.9278 0.8022 1.0087 0.0000 0.0000 -1.0000 -1.0087 -1.0000 0.0022 0.0000 0
0.9278 0.8022 1.0066 0.0000 1.0000 -1.0000 -0.2043 -1.0000 -0.1989 0.0000 -0
0.9278 0.6033 1.2055 1.0000 0.0000 -1.0000 -0.2777 -1.0000 -0.1806 -0.1806 0
0.7472 0.6033 1.3860 1.0000 1.0000 -1.0000 -0.0355 1.0000 0.2589 0.2589 0
weights 1.006081 0.8621967 1.127163 old 0.9277692 0.8022292 1.008743
Err 1.113613
w1 w2 b x1 x2 -1 act. target delta dw1
1.0061 0.8622 1.1272 0.0000 0.0000 -1.0000 -1.1272 -1.0000 0.0318 0.0000 0
1.0061 0.8622 1.0954 0.0000 1.0000 -1.0000 -0.2332 -1.0000 -0.1917 0.0000 -0
1.0061 0.6705 1.2871 1.0000 0.0000 -1.0000 -0.2810 -1.0000 -0.1798 -0.1798 0
0.8263 0.6705 1.4668 1.0000 1.0000 -1.0000 0.0300 1.0000 0.2425 0.2425 0
weights 1.068833 0.9129927 1.224327 old 1.006081 0.8621967 1.127163
Err 1.031035
w1 w2 b x1 x2 -1 act. target delta dw1
1.0688 0.9130 1.2243 0.0000 0.0000 -1.0000 -1.2243 -1.0000 0.0561 0.0000 0
1.0688 0.9130 1.1682 0.0000 1.0000 -1.0000 -0.2553 -1.0000 -0.1862 0.0000 -0
1.0688 0.7268 1.3544 1.0000 0.0000 -1.0000 -0.2856 -1.0000 -0.1786 -0.1786 0
0.8902 0.7268 1.5330 1.0000 1.0000 -1.0000 0.0840 1.0000 0.2290 0.2290 0
weights 1.119231 0.9558044 1.304034 old 1.068833 0.9129927 1.224327
Err 0.977192
w1 w2 b x1 x2 -1 act. target delta dw1
1.1192 0.9558 1.3040 0.0000 0.0000 -1.0000 -1.3040 -1.0000 0.0760 0.0000 0
1.1192 0.9558 1.2280 0.0000 1.0000 -1.0000 -0.2722 -1.0000 -0.1819 0.0000 -0
1.1192 0.7739 1.4100 1.0000 0.0000 -1.0000 -0.2907 -1.0000 -0.1773 -0.1773 0
0.9419 0.7739 1.5873 1.0000 1.0000 -1.0000 0.1285 1.0000 0.2179 0.2179 0
weights 1.159793 0.9917371 1.369408 old 1.119231 0.9558044 1.304034
Err 0.9423397
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Figure 7. First 5 passes Figure 8. Second 5 passes
w1 w2 b x1 x2 -1 act. target delta dw1
1.1598 0.9917 1.3694 0.0000 0.0000 -1.0000 -1.3694 -1.0000 0.0924 0.0000 0
1.1598 0.9917 1.2771 0.0000 1.0000 -1.0000 -0.2853 -1.0000 -0.1787 0.0000 -0
1.1598 0.8131 1.4557 1.0000 0.0000 -1.0000 -0.2959 -1.0000 -0.1760 -0.1760 0
0.9838 0.8131 1.6317 1.0000 1.0000 -1.0000 0.1651 1.0000 0.2087 0.2087 0
weights 1.192501 1.021792 1.423018 old 1.159793 0.9917371 1.369408
Err 0.9199996
Press Enter...
w1 w2 b x1 x2 -1 act. target delta dw1
1.1925 1.0218 1.4230 0.0000 0.0000 -1.0000 -1.4230 -1.0000 0.1058 0.0000 0
1.1925 1.0218 1.3173 0.0000 1.0000 -1.0000 -0.2955 -1.0000 -0.1761 0.0000 -0
1.1925 0.8457 1.4934 1.0000 0.0000 -1.0000 -0.3009 -1.0000 -0.1748 -0.1748 0
1.0177 0.8457 1.6682 1.0000 1.0000 -1.0000 0.1952 1.0000 0.2012 0.2012 0
weights 1.218922 1.046857 1.466975 old 1.192501 1.021792 1.423018
Err 0.905868
Press Enter...
w1 w2 b x1 x2 -1 act. target delta dw1
1.2189 1.0469 1.4670 0.0000 0.0000 -1.0000 -1.4670 -1.0000 0.1167 0.0000 0
1.2189 1.0469 1.3502 0.0000 1.0000 -1.0000 -0.3034 -1.0000 -0.1742 0.0000 -0
1.2189 0.8727 1.5244 1.0000 0.0000 -1.0000 -0.3055 -1.0000 -0.1736 -0.1736 0
1.0453 0.8727 1.6980 1.0000 1.0000 -1.0000 0.2200 1.0000 0.1950 0.1950 0
weights 1.240296 1.067708 1.503013 old 1.218922 1.046857 1.466975
Err 0.8970905
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Figure 9. Final passes
We see that we have a correct pass through the training set (all the activations a and targets
have the same sign). Further iterations may reduce the error.
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The type of problems described above are problems in which we can do a linear separation.
A simple example in which this will not work is the logical XOR function
x1 x2 t
0 0 0
0 1 1
1 0 1
1 1 0
Figure 10. Logical XOR
It is not possible to separate the classes with a straight line
In order to solve problems that are not linearly separable we need to introduce a more
complicated structure for the neural network. This will come in the form of one (or more)
additional layers of nodes. Because they are not visible as input or output layers, they are
referred to as hidden layers.
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A network with one hidden layer can be pictured as:
Figure 11. Feed-forward Neural Network
As before, we take xi as inputs and wij as weights.
The hidden nodes will have activation functions (as do the output nodes) but, in order to
allow us to update the weights, we will need an activation function that behaves in a manner
similar to a step function but is continuous (so that we can differentiate it).
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A sigmoidal function such as
0
0.2
0.4
0.6
0.8
1
-10 -5 5 10
x
Figure 12. f x ex/1 ex
or
0
0.2
0.4
0.6
0.8
1
-10 -5 5 10
x
Figure 13. f x tanh x
is typically used (recall the Logistic Regression).
In our network above, we can take a common hidden node activation function fh and a
common output activation function fo to be
1/1 ex
(Note: We could have different functions at different nodes).
We will ‘present’ the N cases to the neural network where, for case i, we have
XiT 1 xi1 xi2 xip
(Note that the 1 is for the bias unit and is similar to the 1 in regression that corresponds to the
presence of an intercept term).
The weights from p inputs (and the bias unit) to the hidden node m (Hm) are
wm
T w0m w1m w2m ... wpm
so the input to Hm for case i is
zm w0m1 w1mxi1 w2mxi2 ... wpmxip
w0m
i1
p
wimxi
wm
T Xi
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Figure 14 Inputs to hidden node
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This input is ‘evaluated’ by the activation function fH and the output from a hidden node due
to case i from Hm is
T im fH w0m w1mxi1 w2mxi2 ... wpmxip
fH w0m
i1
p
wimxi fHwmT Xi
The weights from all M hidden nodes to the output node Ok are given by
WkT W0k W1k W2k ... WM k
so that the input to Ok is for case i is
Zk W0k
m1
M
Wmk fH w0m
i1
p
wimxi
W0k
m1
M
Wmk T im
or
Zk W0k W1kT i1 W2kT i2 ....WM kT iM
WkT Ti
Figure 15. Inputs to output node
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The output from Ok for case i is
fOW0k W1kT i1 W2kT i2 ....WM kT iM fO WkT Ti
To find the “error” at Ok we take the difference between this output and the “true” value at
the node - t ik - and square it
E ik t ik fO WkT Ti 2
The error from all output nodes for case i is
E i
k1
K
t ik fO WkT Ti 2
so the total error of the network for all N input values is
E
i1
N
E i
i1
N

k1
K
t ik fO WkT Ti 2
We wish to choose weights which minimize E so we start by differentiating E with respect to
the weight from Hm to Ok :
E i
Wmk

Wmk

k1
K
t ik fO WkT Ti 2

Wmk

k1
K
t ik fOW0k W1kT i1 W2kT i2 WM kT iM 2
2 t ik fO WkT Ti fOW0k W1kT i1 W2kT i2 WM kT iM Wmk
2 t ik fO WkT Ti fO WkT Ti T im
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and also differentiate with respect to the weight from each input node I l to Hm :
E i
wlm

wlm

k1
K
t ik fO WkT Ti 2

k1
K
2 t ik fO WkT Ti
fO WkT Ti
wlm

k1
K
2 t ik fO WkT Ti fO WkT Ti W0k W1kT i1 W2kT i2 WM kT iM wlm .
Note that
W0k W1kT i1 W2kT i2 ....WM kT iM W0k W1k fHw01 w11xi1 w21xi2 wp1xip
W2k fHw02 w12xi1 w22xi2 wp2xip
.... WM k fHw0M w1Mxi1 w2Mxi2 wpMxip
so
E
wlm

k1
K
2 t ik fO WkT Ti fO WkT Ti Wmk fH wmT Xi xil
With these derivatives, it is possible to do a gradient descent update (with a learning rate s
so that at step s 1 :
Wmk
s1 Wmk
s s
i1
N
E i
wlm
s
and wlm
s1 wlm
s s
i1
N
E
wlm
s
This process is referred to as back-propagation.
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In order to illustrate the neural network consider the following:
Example: Neural Network on Flea Beetles
We know the correct classes for the flea beetles (given by flea.species). We will use
that as our target values for the neural net.
nnet expects the target information to be given in the form of a class.ind (that is a
matrix with the the columns indicating classes for the cases - the transpose is used to make
the dispaly more compact).
t(class.ind(flea.species))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
C 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Hk 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Hp 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
C 1 1 1 1 1 1 1 0 0 0 0 0
Hk 0 0 0 0 0 0 0 0 0 0 0 0
Hp 0 0 0 0 0 0 0 1 1 1 1 1
[,27] [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35] [,36] [,37] [,38]
C 0 0 0 0 0 0 0 0 0 0 0 0
Hk 0 0 0 0 0 0 0 0 0 0 0 0
Hp 1 1 1 1 1 1 1 1 1 1 1 1
[,39] [,40] [,41] [,42] [,43] [,44] [,45] [,46] [,47] [,48] [,49] [,50]
C 0 0 0 0 0 0 0 0 0 0 0 0
Hk 0 0 0 0 0 1 1 1 1 1 1 1
Hp 1 1 1 1 1 0 0 0 0 0 0 0
[,51] [,52] [,53] [,54] [,55] [,56] [,57] [,58] [,59] [,60] [,61] [,62]
C 0 0 0 0 0 0 0 0 0 0 0 0
Hk 1 1 1 1 1 1 1 1 1 1 1 1
Hp 0 0 0 0 0 0 0 0 0 0 0 0
[,63] [,64] [,65] [,66] [,67] [,68] [,69] [,70] [,71] [,72] [,73] [,74]
C 0 0 0 0 0 0 0 0 0 0 0 0
Hk 1 1 1 1 1 1 1 1 1 1 1 1
Hp 0 0 0 0 0 0 0 0 0 0 0 0
We can find how many of each species are present:
apply(class.ind(flea.species),2,sum)
C Hk Hp
21 31 22
We try for the simplest possible solution (i.e. the smallest hidden layer) consistent with a
good result (low misclassification). The size parameter indicates the number of nodes in
the hidden layer. If we omit the hidden layer size0 we must indicate skipT. (See
?nnet.)
In order to enable us to look at how well the classifier works we will divide the data into a
training and a test set using the functions from previous lectures.
”%w/o%” - function(x,y) x[!x %in% y]
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#
# Set the indices for the training/test sets
#
get.train - function (data.sz, train.sz) {
# Take subsets of data for training/test samples
# Return the indices
train.ind - sample(data.sz, train.sz)
test.ind - (1:data.sz) %w/o% train.ind
list(traintrain.ind, testtest.ind)
}
Train.sz - 52
(tt.ind - get.train(dim(df.flea)[1], Train.sz))
\$train
[1] 52 61 13 12 7 8 33 40 16 4 22 60 58 23 6 9 66 21 48 65 38 5 3 41 27 42
[28] 71 32 43 74 50 51 1 34 47 69 45 56 28 10 67 24 37 68 18 2 62 54 64 57 19
\$test
[1] 11 14 15 17 20 25 26 29 30 31 35 36 39 44 46 49 53 55 59 63 70 72
(For reproducibility we will use these values rather than the random ones.)
tt.ind\$train - c(52,61,13,12,7,8,33,40,16,4,22,60,58,23,6,9,66,21,48,65,38,
5,3,41,27,42,73,71,32,43,74,50,51,1,34,47,69,45,56,28,10,
67,24,37,68,18,2,62,54,64,57,19)
tt.ind\$test -
c(11,14,15,17,20,25,26,29,30,31,35,36,39,44,46,49,53,55,59,63,70,72)
apply(class.ind(flea.species[tt.ind\$train]),2,sum)
C Hk Hp
16 22 14
apply(class.ind(flea.species[tt.ind\$test]),2,sum)
C Hk Hp
5 9 8
We see that we have 16/21 C, 22/31 Hk, and 14/22 Hp in the training set and 5/21 C, 9/31
Hk, and 8/22 Hp in the test set.
NOTE: There is always a danger that the classes may be represented in an inappropriate
fashion in the training set. For example, if there are no representatives of class C in the
training set then the neural network will never classify anything as class C.
We will start with no hidden layer (size0), connections from the input to the output layer
(skipT), range for initial random weights 0.1,0.1 (range0.1), weight decay
parameter 0.0005 (decay5e-4), and a maximum of 500 iterations (maxit500):
flea.nn - nnet(df.flea[tt.ind\$train,], class.ind(flea.species[tt.ind\$train]),
size0, skipT, rang0.1, decay5e-4, maxit500)
# weights: 21
initial value 38.003178
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iter 10 value 25.673878
iter 20 value 17.542510
iter 30 value 4.471108
iter 40 value 3.329062
iter 50 value 3.298340
iter 60 value 3.212452
iter 70 value 3.146040
iter 80 value 3.129967
iter 90 value 3.114434
iter 100 value 3.105363
iter 110 value 3.098298
iter 120 value 3.095578
iter 130 value 3.095131
iter 140 value 3.094971
iter 150 value 3.094963
final value 3.094956
converged
The neural net has 21 weights (each of the 6 inputs is connected to each output as is the bias,
for a total of 7 3).
summary(flea.nn)
a 6-0-3 network with 21 weights
options were - skip-layer connections decay5e-04
b-o1 i1-o1 i2-o1 i3-o1 i4-o1 i5-o1 i6-o1
-3.16 0.65 -0.03 -9.88 2.80 -3.83 0.53
b-o2 i1-o2 i2-o2 i3-o2 i4-o2 i5-o2 i6-o2
0.02 0.79 -0.26 0.19 -0.49 0.57 -0.75
b-o3 i1-o3 i2-o3 i3-o3 i4-o3 i5-o3 i6-o3
0.03 -0.57 0.24 0.24 0.20 -0.14 0.21
The above output shows the connections and the corresponding weights.
We can investigate the results of this (and many other functions) by use of expressions of the
form flea.nn[i].
flea.nn[11]
\$wts
[1] -3.16234631 0.64955583 -0.02654692 -9.87994802 2.79725402 -3.83345049
[7] 0.52646205 0.02468363 0.79217231 -0.25674016 0.19259577 -0.48754806
[13] 0.57420624 -0.75152200 0.02583289 -0.57245962 0.23516685 0.23850064
[19] 0.19531903 -0.14429595 0.21283729
(We could also use flea.nn\$wts.)
t(flea.nn\$fitted.values)
52 61 13 12 7 8 33
C 0 1.0000000 1.000000e00 1.0000000000 8.993480e-01 9.999995e-01 0.000000
Hk 1 0.9999208 8.611791e-06 0.0000000000 6.284825e-07 0.000000e00 0.000000
Hp 0 0.0000000 0.000000e00 0.0006301502 3.469448e-07 4.432988e-07 0.999971
40 16 4 22 60 58
C 0.0000000 1.086797e-05 9.876069e-01 0.0000000 0.06561292 1.087303e-05
Hk 0.0000000 2.812531e-04 0.000000e00 0.0000000 0.99999945 1.000000e00
Hp 0.9975123 0.000000e00 2.131585e-05 0.9985126 0.00000000 0.000000e00
23 6 9 66 21 48
C 0.0005646545 1.000000000 0.944343721 0.0000000 0.000000e00 0.0001635885
Hk 0.0000000000 0.000000000 0.000000000 0.9999484 0.000000e00 1.0000000000
Hp 0.9974711616 0.001730916 0.003764649 0.0000000 2.489253e-05 0.0000000000
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65 38 5 3 41 27 42 73
C 0.06260702 5.306533e-05 1.0000000000 0.999994919 0 0.07486592 0.0000000 0
Hk 0.99442958 0.000000e00 0.0000000000 0.003979733 0 0.00000000 0.0000000 1
Hp 0.00000000 9.999784e-01 0.0003857919 0.000000000 1 0.99994208 0.9983447 0
71 32 43 74 50 51 1 34
C 0 0 0.000000 0.0000000 1.540079e-06 1.462929e-06 9.999757e-01 9.045183e-06
Hk 1 0 0.000000 0.9916207 1.000000e00 1.000000e00 3.837811e-07 0.000000e00
Hp 0 1 0.999999 0.0000000 0.000000e00 0.000000e00 0.000000e00 1.000000e00
47 69 45 56 28 10 67 24 37 68 18
C 0 0 0 0 0.0000000 1.000000e00 0 0.03017668 0.0000000 0 1.000000000
Hk 1 1 1 1 0.0000000 0.000000e00 1 0.00000000 0.0000000 1 0.000000000
Hp 0 0 0 0 0.9999996 9.697477e-05 0 0.99782310 0.9991032 0 0.001895541
2 62 54 64 57 19
C 1.000000e00 0 0 1.095646e-05 0.0000000 1.000000000
Hk 0.000000e00 1 1 1.000000e00 0.9999972 0.008531558
Hp 1.746592e-06 0 0 0.000000e00 0.0000000 0.000000000
(We get the same information from t(predict(flea.nn,
df.flea[tt.ind\$train,])).)
The fitted class for each case is given by the largest value for the case. To determine which
class has been predicted for which case, we can use:
max.col(flea.nn\$fitted.values)
[1] 2 2 1 1 1 1 3 3 1 1 3 2 2 3 1 1 2 1 2 2 3 1 1 3 3 3 2 2 3 3 2 2 2 1 3 2 2 2
[39] 2 3 1 2 3 3 2 1 1 2 2 2 2 1
It is difficult to see what the classes are in this form so a useful method for testing the
accuracy of classification is the confusion matrix.
confusion.expand(max.col(flea.nn\$fitted.values), flea.species[tt.ind\$train])
true
object C Hk Hp | Row Sum
1 14 1 0 | 15
2 1 21 0 | 22
3 1 0 14 | 15
------- ---- ---- ---- | ----
Col Sum 16 22 14 | 52
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
(Watch out! The confusion matrix displays the classes in alphabetical order - not the given
order!)
Note the use of max.col in the confusion matrix. It does as the name suggests - determines
which column has the maximum value (note that ties are broken at random).
Here 14 of class C have been classified as class 1, 1 as class 2, and 1 as class 3, 21 of the Hk
were classified as being in class 2, and 1 as class 1, while all the Hp are classified as class 3.
Thus we have an error rate of 3/52 or 5.8%. The problem is that we are obtaining this error
rate by using the same data that went into creating the model. Because we have reserved
some data for testing, we can use it to get a more accurate representation of the error rate.
We first need to determine how the test data can be classified. It is typical of classifiers that
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they have a prediction method so that we can see how the cases can be classified.
predict(flea.nn, df.flea[tt.ind\$test,])
C Hk Hp
11 0.3513851292 1.61017e-05 0.000000e00
14 1.0000000000 0.00000e00 4.722636e-07
15 1.0000000000 0.00000e00 4.124901e-06
17 0.0433338860 0.00000e00 1.977635e-06
20 1.0000000000 0.00000e00 1.260786e-06
25 0.9999701704 0.00000e00 9.999987e-01
26 0.0000000000 0.00000e00 9.999587e-01
29 0.0000000000 0.00000e00 9.999708e-01
30 0.0000000000 0.00000e00 1.000000e00
31 0.9997923461 0.00000e00 9.999988e-01
35 0.0000000000 0.00000e00 9.976731e-01
36 0.0000000000 0.00000e00 9.999655e-01
39 0.0626155629 0.00000e00 9.999328e-01
44 0.0000000000 1.00000e00 0.000000e00
46 0.9998812554 1.00000e00 0.000000e00
49 0.0000000000 1.00000e00 0.000000e00
53 0.0004629659 1.00000e00 0.000000e00
55 0.0000000000 1.00000e00 0.000000e00
59 0.0000922833 1.00000e00 0.000000e00
63 0.0000000000 1.00000e00 0.000000e00
70 0.0000000000 9.99988e-01 0.000000e00
72 0.0000000000 1.00000e00 0.000000e00
confusion.expand(max.col(predict(flea.nn,
df.flea[tt.ind\$test,])),flea.species[tt.ind\$test])
true
object C Hk Hp | Row Sum
1 5 0 0 | 5
2 0 9 0 | 9
3 0 0 8 | 8
------- ---- ---- ---- | ----
Col Sum 5 9 8 | 22
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
Here we have a perfect classification, but remember we are dealing with a small data set that
we have seen can be separated quite well.
We might also try a neural net that has one node (size1) in the hidden layer:
flea.nn - nnet(df.flea[tt.ind\$train,], class.ind(flea.species[tt.ind\$train]),
size1, rang0.1, decay5e-4, maxit500)
# weights: 13
initial value 39.061500
iter 10 value 34.003834
iter 20 value 27.969509
iter 30 value 18.598327
iter 40 value 17.403684
iter 50 value 17.302171
iter 60 value 16.063792
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iter 70 value 13.520698
iter 80 value 11.554444
iter 90 value 11.269187
iter 100 value 11.108513
iter 110 value 11.006646
iter 120 value 11.005451
iter 130 value 11.001329
final value 11.000691
converged
(Notice that we have fewer weights.)
summary(flea.nn)
a 6-1-3 network with 13 weights
options were - decay5e-04
b-h1 i1-h1 i2-h1 i3-h1 i4-h1 i5-h1 i6-h1
3.24 0.07 -0.10 0.03 -0.06 0.32 -0.01
b-o1 h1-o1
-0.02 -1.75
b-o2 h1-o2
-7.13 14.50
b-o3 h1-o3
2.39 -34.26
This describes the network in terms of what nodes it contains, how they are connected, and
the weights.
i1
i2
i3
i4
i5
i6
h1
o1
o2
o3
0.07
-0.1
0.03
-0.06
0.32
-0.01
-1.75
14.5
-34.26
+1Bias
3.24 2.39
-7.13
-0.02
Figure 16. Neural net for flea beetles
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confusion.expand(max.col(flea.nn\$fitted.values), flea.species[tt.ind\$train])
true
object C Hk Hp | Row Sum
1 16 0 0 | 16
2 0 22 0 | 22
3 0 0 14 | 14
------- ---- ---- ---- | ----
Col Sum 16 22 14 | 52
Here all the cases are correctly classified.
Suppose we look at how this neural network will predict the class by looking at the first case:
df.flea[1,]
tars1 tars2 head aede1 aede2 aede3
1 191 131 53 150 15 104
To see how we get our result, look at the input vector (the 1 is the bias component)
(i.case.1 - c(1, as.matrix(df.flea[1,])))
[1] 1 191 131 53 150 15 104
and the weights to the hidden node:
(w.i.h - flea.nn\$wts[1:7])
[1] 3.239477309 0.069190234 -0.098314207 0.026150991 -0.064498570
[6] 0.318486158 -0.009023526
The ‘signal’ to the hidden node is thus:
(to.h - sum(i.case.1*w.i.h))
[1] -0.8742865
which is then passed through the sigmoidal function of the hidden layer to give the hidden
layer output:
(from.h - 1/(1exp(-to.h)))
[1] 0.2943632
The input to the output nodes comes from the bias and the hidden node i.e. c(1,
from.h)and the weights are:
(w.h.o1 - flea.nn\$wts[8:9])
[1] -0.01884662 -1.74534267
(w.h.o2 - flea.nn\$wts[10:11])
[1] -7.132326 14.497100
(w.h.o3 - flea.nn\$wts[12:13])
[1] 2.385936 -34.257359
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so the inputs to the output nodes are:
(to.o1 - sum(c(1, from.h)*w.h.o1))
[1] -0.5326112
(to.o2 - sum(c(1, from.h)*w.h.o2))
[1] -2.864914
(to.o3 - sum(c(1, from.h)*w.h.o3))
[1] -7.698168
and the outputs are:
from.o1 - 1/(1exp(-to.o1))
from.o2 - 1/(1exp(-to.o2))
from.o3 - 1/(1exp(-to.o3))
cat(from.o1, from.o2, from.o3, ”\n”)
0.3699081 0.05391547 0.0004534519
Using
predict(flea.nn, df.flea[1,])
C Hk Hp
1 0.3699081 0.05391547 0.0004534519
191
131
53
150
15
104
0.07 (13.215)
-0.1 (-12.88)
0.03 (1.386)
-0.06 (-9.675)
0.32 (4.777)
-0.01 (-0.939)
Input to
hidden node
-0.875
Ouput from
hidden node
0.294
-1.75
14.5
-34.26
(-0.533)
(-2.865)
(-7.699)
+1Bias
3.24
(3.24)
2.39
(2.39)
-7.13
(-7.13)
-0.02
(-0.02)
-0.533
-2.865
-7.699
0.369
0.053
4e-04
Figure 17. Results for first case
In the Figure 17 above, the black numbers represent the weights, the blue numbers represent
the output from a neuron and the red numbers represent the inputs.
Classificationn-Neural Nets 422 © Mills 2018
Data Mining 2018
Now look at the test cases.
confusion.expand(max.col(predict(flea.nn,
df.flea[tt.ind\$test,])),flea.species[tt.ind\$test])
true
object C Hk Hp | Row Sum
1 5 0 0 | 5
2 0 9 0 | 9
3 0 0 8 | 8
------- ---- ---- ---- | ----
Col Sum 5 9 8 | 22
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
It would appear that this neural network is a good classifier.
Suppose that we rerun with the same values (omitting some output):
flea.nn - nnet(df.flea[tt.ind\$train,], class.ind(flea.species[tt.ind\$train]),
size1, rang0.1, decay5e-4, maxit500)
confusion(max.col(flea.nn\$fitted.values), flea.species[tt.ind\$train])
© Mills 2018 Classification-Neural Nets 423
Data Mining 2018
# weights: 13
initial value 38.378513
iter 10 value 34.004917
...
iter 160 value 18.370829
final value 18.370750
converged
true
object C Hk Hp
1 2 0 0
2 14 22 0
3 0 0 14
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
Different
final value and
# of iterations.
# weights: 13
initial value 39.608202
...
final value 18.549964
converged
true
object C Hk Hp
1 1 0 0
2 15 22 0
3 0 0 14
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
Different
final value and
# of iterations.
# weights: 13
initial value 38.641683
...
iter 90 value 15.034509
final value 15.034442
converged
true
object C Hk Hp
1 16 0 14
2 0 22 0
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
Different
classification.
# weights: 13
initial value 37.963702
...
final value 14.778896
converged
true
object C Hk Hp
1 16 0 14
2 0 22 0
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
Different
classification.
# weights: 13
initial value 38.419771
iter 10 value 34.000557
final value 34.000455
converged
true
object C Hk Hp
2 16 22 14
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
Different
classification.
Classificationn-Neural Nets 424 © Mills 2018
Data Mining 2018
We find that, even with the same parameters, we get different classifications, different
numbers of iterations, and, even for the same classification, different minima
final value. Why?
To help us understand what has happened, suppose we run the neural network 10 times on
the flea beetle data using one hidden node, the 6 inputs variables and 2 outputs (the number
of classes). If we use random initialization we find that we get the following minima at
termination:
[ 1] 48.51438 48.51438 48.51446 15.00773 15.42307 15.00780 48.51439 48.51446
[ 9] 48.51443 15.41893
The answer is that the nnet routine initializes the weights at random, so we start at
different locations - and then tries to find the absolute minimum in a 13-dimensional weight
space.
To try to get an idea of what is happening, we can take the data and change it so that there
are only two input vectors, 2 classes (and hence 2 output nodes). With reduction of the data,
we can get down to 7 weights:
targets - class.ind(c(rep(”C”, 5), rep(”Hb”, 4)))
plot(df.flea[1:9,1],df.flea[1:9,2], pchtargets1)
Figure 18.
Now run the following several times:
flea.nn - nnet(df.flea[1:9,1:2], targets, size1, rang0.1, decay5e-4, maxit200
# weights: 7
initial value 4.529719
iter 10 value 4.444539
© Mills 2018 Classification-Neural Nets 425
Data Mining 2018
...
final value 4.444498
converged
# weights: 7
initial value 4.456285
...
final value 2.982019 converged
# weights: 7
initial value 4.496130
...
final value 2.765461
converged
# weights: 7
initial value 4.474008
...
final value 2.780494
Again we see different final values.
It is interesting to look at what happens if we fix the initialization for 5 of the weights and let
the other 2 vary over a rectangle.
The following shows a perspective plot and a contour plot of the minima attained for the
various ‘free’ weights:
library(rgl)
z - matrix(0, 81, 81)
x-1:81
y-1:81
for (i in x) {
for (j in y) {
for (i in x) {
for (j in y) {
mywts - c((i-41)/200, (j-41)/200, 0, 0, 0, 0, 0)
flea.nn - nnet(df.flea[1:9,1:2], targets, Wtsmywts, size1,
rang0.1, decay5e-4, maxit200,traceF)
z[i,j] - flea.nn\$value
cat(”x ”,i,” y ”,j,” min ”, flea.nn\$value,” wts
”,floor(1000*flea.nn\$wts)/1000,”\n”)
}
}
x 1 y 1 min 2.780487 wts -1.549 -5.527 8.013 3.711 -4.403 -3.712 4.
x 1 y 2 min 2.780484 wts -1.619 -5.505 7.982 3.715 -4.407 -3.716 4.
x 1 y 3 min 2.967719 wts 1.678 -5.951 8.06 0.916 -5.114 -0.917 5.113
x 1 y 4 min 2.780537 wts -1.331 -5.538 8.028 3.69 -4.385 -3.691 4.384
x 1 y 5 min 2.967729 wts 1.544 -5.965 8.08 0.917 -5.126 -0.918 5.125
x 1 y 6 min 2.780551 wts -1.286 -5.557 8.054 3.699 -4.389 -3.7 4.388
x 1 y 7 min 2.780498 wts -1.534 -5.556 8.055 3.712 -4.406 -3.713 4.
x 1 y 8 min 2.780496 wts -1.511 -5.503 7.978 3.698 -4.391 -3.699 4.
x 1 y 9 min 2.967752 wts 1.443 -5.987 8.111 0.914 -5.12 -0.915 5.119
x 1 y 10 min 2.967774 wts 1.39 -5.949 8.06 0.919 -5.115 -0.92 5.114
x 1 y 11 min 2.967721 wts 1.642 -5.943 8.05 0.915 -5.117 -0.916 5.116
...
x 1 y 40 min 2.780484 wts -1.608 -5.513 7.993 3.719 -4.412 -3.72 4.
x 1 y 41 min 4.444497 wts -0.004 -0.061 -0.044 0.222 0.001 -0.223 -
x 1 y 42 min 2.765456 wts 1.582 5.408 -7.846 -0.691 4.946 0.69 -4.947
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x 1 y 43 min 2.780488 wts -1.541 -5.517 7.999 3.712 -4.404 -3.713 4
...
x 2 y 38 min 2.780486 wts -1.57 -5.52 8.004 3.714 -4.407 -3.715 4.406
x 2 y 39 min 2.780486 wts -1.567 -5.509 7.988 3.707 -4.401 -3.708 4
x 2 y 40 min 2.780543 wts -1.536 -5.606 8.127 3.691 -4.381 -3.692 4
x 2 y 41 min 4.444497 wts -0.005 -0.065 -0.046 0.222 0.001 -0.223 -
x 2 y 42 min 2.765472 wts 1.464 5.343 -7.75 -0.691 4.938 0.69 -4.939
x 2 y 43 min 2.780489 wts -1.553 -5.531 8.019 3.718 -4.412 -3.719 4
...
x 3 y 24 min 2.781137 wts -0.748 -5.418 7.849 3.827 -4.516 -3.828 4
x 3 y 25 min 2.780488 wts -1.532 -5.512 7.992 3.71 -4.403 -3.711 4.
x 3 y 26 min 2.780491 wts -1.584 -5.54 8.033 3.698 -4.391 -3.699 4.
x 3 y 27 min 4.444514 wts -0.19 -0.07 0 0.222 -0.001 -0.223 0
x 3 y 28 min 4.444514 wts -0.19 -0.065 0 0.222 -0.001 -0.223 0
x 3 y 29 min 4.444514 wts -0.19 -0.06 0 0.222 -0.001 -0.223 0
x 3 y 30 min 4.444514 wts -0.19 -0.055 0 0.222 -0.001 -0.223 0
x 3 y 31 min 2.780486 wts -1.569 -5.523 8.008 3.717 -4.409 -3.718 4
x 3 y 32 min 2.780491 wts -1.507 -5.525 8.011 3.716 -4.409 -3.717 4
x 3 y 33 min 2.780548 wts -1.297 -5.54 8.03 3.699 -4.388 -3.7 4.387
x 3 y 34 min 2.780499 wts -1.468 -5.531 8.018 3.721 -4.412 -3.722 4
...
x 4 y 20 min 2.780486 wts -1.566 -5.506 7.983 3.709 -4.402 -3.71 4.
x 4 y 21 min 2.780506 wts -1.429 -5.519 8.001 3.703 -4.397 -3.704 4
x 4 y 22 min 2.967788 wts 1.367 -6.024 8.162 0.917 -5.118 -0.918 5.
x 4 y 23 min 2.780491 wts -1.505 -5.518 8.001 3.713 -4.406 -3.714 4
x 4 y 24 min 4.444515 wts -0.185 -0.085 0 0.222 -0.001 -0.223 0
x 4 y 25 min 4.444514 wts -0.185 -0.08 0 0.222 -0.001 -0.223 0
x 4 y 26 min 4.444514 wts -0.185 -0.075 0 0.222 -0.001 -0.223 0
x 4 y 27 min 4.444514 wts -0.185 -0.07 0 0.222 -0.001 -0.223 0
x 4 y 28 min 4.444513 wts -0.185 -0.065 0 0.222 -0.001 -0.223 0
x 4 y 29 min 4.444513 wts -0.185 -0.06 0 0.222 -0.001 -0.223 0
x 4 y 30 min 4.444513 wts -0.185 -0.055 0 0.222 -0.001 -0.223 0
x 4 y 31 min 2.780522 wts -1.423 -5.552 8.049 3.741 -4.432 -3.742 4
x 4 y 32 min 2.781296 wts -0.37 -5.571 8.068 3.719 -4.412 -3.72 4.411
x 4 y 33 min 2.780542 wts -1.313 -5.505 7.98 3.736 -4.429 -3.737 4.
x 4 y 34 min 2.780495 wts -1.584 -5.544 8.038 3.688 -4.381 -3.689 4
x 4 y 35 min 2.780493 wts -1.521 -5.512 7.992 3.701 -4.392 -3.702 4
x 4 y 36 min 2.780527 wts -1.353 -5.54 8.03 3.716 -4.406 -3.717 4.405
x 4 y 37 min 2.780487 wts -1.551 -5.522 8.006 3.709 -4.402 -3.71 4.
x 4 y 38 min 2.967742 wts 1.478 -5.972 8.09 0.917 -5.13 -0.918 5.129
x 4 y 39 min 4.444513 wts -0.185 -0.045 -0.041 0.222 0.011 -0.223 -
x 4 y 40 min 2.780504 wts -1.438 -5.526 8.011 3.706 -4.397 -3.707 4
x 4 y 41 min 4.444498 wts -0.007 -0.071 -0.049 0.222 0.003 -0.223 -
x 4 y 42 min 2.982028 wts -1.526 6.122 -8.298 -3.638 4.557 3.637 -4
x 4 y 43 min 2.780488 wts -1.531 -5.515 7.996 3.709 -4.402 -3.71 4.
x 4 y 44 min 2.765450 wts 1.521 5.395 -7.826 -0.691 4.924 0.69 -4.925
...
x 5 y 20 min 2.780489 wts -1.528 -5.516 7.997 3.712 -4.404 -3.713 4
x 5 y 21 min 2.780484 wts -1.625 -5.515 7.997 3.715 -4.408 -3.716 4
x 5 y 22 min 4.444515 wts -0.18 -0.095 0 0.222 0 -0.223 0
x 5 y 23 min 4.444514 wts -0.18 -0.09 0 0.222 -0.001 -0.223 0
x 5 y 24 min 4.444514 wts -0.18 -0.085 0 0.222 -0.001 -0.223 0
x 5 y 25 min 4.444513 wts -0.18 -0.08 0 0.222 -0.001 -0.223 0
x 5 y 26 min 4.444513 wts -0.18 -0.075 0 0.222 -0.001 -0.223 0
x 5 y 27 min 4.444513 wts -0.18 -0.07 0 0.222 -0.001 -0.223 0
x 5 y 28 min 4.444512 wts -0.18 -0.065 0 0.222 -0.001 -0.223 0
x 5 y 29 min 4.444512 wts -0.18 -0.06 0 0.222 -0.001 -0.223 0
x 5 y 30 min 4.444512 wts -0.18 -0.055 0 0.222 -0.001 -0.223 0
x 5 y 31 min 4.444511 wts -0.18 -0.05 0 0.222 -0.001 -0.223 0
x 5 y 32 min 2.967738 wts 1.498 -5.964 8.079 0.918 -5.125 -0.919 5.
x 5 y 33 min 2.780486 wts -1.557 -5.516 7.998 3.71 -4.402 -3.711 4.
x 5 y 34 min 2.780535 wts -1.336 -5.533 8.021 3.694 -4.385 -3.695 4
...
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x 6 y 18 min 2.780489 wts -1.533 -5.518 8 3.711 -4.405 -3.712 4.404
x 6 y 19 min 2.967737 wts 1.526 -5.988 8.111 0.915 -5.113 -0.916 5.
x 6 y 20 min 4.444515 wts -0.175 -0.105 0 0.222 0 -0.223 0
x 6 y 21 min 4.444514 wts -0.175 -0.1 0 0.222 0 -0.223 0
x 6 y 22 min 4.444514 wts -0.175 -0.095 0 0.222 0 -0.223 0
x 6 y 23 min 4.444513 wts -0.175 -0.09 0 0.222 -0.001 -0.223 0
x 6 y 24 min 4.444513 wts -0.175 -0.085 0 0.222 -0.001 -0.223 0
x 6 y 25 min 4.444513 wts -0.175 -0.08 0 0.222 -0.001 -0.223 0
x 6 y 26 min 4.444512 wts -0.175 -0.075 0 0.222 -0.001 -0.223 0
x 6 y 27 min 4.444512 wts -0.175 -0.07 0 0.222 -0.001 -0.223 0
x 6 y 28 min 4.444511 wts -0.175 -0.065 0 0.222 -0.001 -0.223 0
x 6 y 29 min 4.444511 wts -0.175 -0.06 0 0.222 -0.001 -0.223 0
x 6 y 30 min 4.444511 wts -0.175 -0.055 0 0.222 -0.001 -0.223 0
x 6 y 31 min 4.444511 wts -0.175 -0.05 0 0.222 -0.001 -0.223 0
x 6 y 32 min 2.967729 wts 1.608 -5.988 8.111 0.915 -5.118 -0.916 5.
x 6 y 33 min 2.780487 wts -1.543 -5.52 8.003 3.716 -4.409 -3.717 4.
It can be seen that, in some cases, the weights for a similar minima are quite similar:
x 6 y 21 min 4.444514 wts -0.175 -0.1 0 0.222 0 -0.223 0
x 6 y 22 min 4.444514 wts -0.175 -0.095 0 0.222 0 -0.223 0
x 6 y 23 min 4.444513 wts -0.175 -0.09 0 0.222 -0.001 -0.223 0
x 6 y 24 min 4.444513 wts -0.175 -0.085 0 0.222 -0.001 -0.223 0
while in other cases there are considerable differences:
x 4 y 38 min 2.967742 wts 1.478 -5.972 8.09 0.917 -5.13 -0.918 5.129
x 4 y 42 min 2.982028 wts -1.526 6.122 -8.298 -3.638 4.557 3.637 -4
persp3d(x,y,z, col ”lightblue”)
contour((x-41)/200,(y-41)/200,z, nlevels10,colc(1,2,3,4,5,6,7,8,9,10))
Figure 19 (a) Perspective plot Figure 19 (b) Contour plot
We see from the images above that slight changes in the initial weights (only two of the
seven changed - the other five started at 0) resulted in convergence to quite different minima.
Keep in mind that the images represent the minima to which the process converges based on
the initial values of the two “free” weights.
Classificationn-Neural Nets 428 © Mills 2018
Data Mining 2018
The minima have values in the following range:
range(z)
[1] 2.765447 4.444516
This had the weights initialized as c((i-41)/200, (j-41)/200, 0, 0, 0, 0,
0).
The following use the same code with different fixed weights - 0.5 rather than 0:
mywts - c((i-41)/200, (j-41)/200, 0.5, 0.5, 0.5, 0.5, 0.5)
Figure 20 (a) Perspective plot Figure 20 (b) Contour plot
range(z)
[1] 2.765447 4.444488
Again with the fixed weights at 1.0:
mywts - c((i-41)/200, (j-41)/200, 1, 1, 1, 1, 1)
Figure 21 (a) Perspective plot Figure 21 (b) Contour plot
range(z)
[1] 2.765447 4.444475
© Mills 2018 Classification-Neural Nets 429
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The following sets the first weight and the last four to 0:
mywts - c(0, (i-41)/200, (j-41)/200, 0, 0, 0, 0)
Figure 22 (a) Perspective plot Figure 22 (b) Contour plot
range(z)
[1] 2.765447 4.444515
and then to 1.0:
mywts - c(1, (i-41)/200, (j-41)/200, 1, 1, 1, 1)
Figure 23 (a) Perspective plot Figure 23 (b) Contour plot
range(z)
[1] 2.765447 4.444516
Classificationn-Neural Nets 430 © Mills 2018
Data Mining 2018
and then the first three to 0 and the last two to 1.0:
mywts - c(0, 0, 0, (i-41)/200, (j-41)/200, 1, 1)
Figure 24 (a) Perspective plot Figure 24 (b) Contour plot
range(z)[1] 2.765447 4.444507
We see that there are several different minima for this data. Depending on the initial
weights, we may converge to a relative minimum rather than the absolute minimum. Care
must be taken to avoid such suboptimal solutions.
© Mills 2018 Classification-Neural Nets 431
Data Mining 2018
As before, we will try neural networks on the synthetic data intoduced in the previous
lecture.
We read in the function for plotting the examples, the data, and the training/test set indices:
source(paste(code.dir, ”example_display.r”, sep”/”))
#
syn.train.class - res\$tp[tt.ind[[1]],]
syn.train.data - res\$data[tt.ind[[1]],]
#
syn.test.class - res\$tp[tt.ind[[2]],]
syn.test.data - res\$data[tt.ind[[2]],]
The following is the usual menu function to allow us to run neural nets on the different data
sets with different numbers of hidden nodes:
while (TRUE) {
cat(”Enter the number of the data set”,”\n”)
cat (”0 to quit\n”)
resN - menu(c(”1:2”, ”1/3:2/4”, ”1:2/3:4”, ”1:2 rotated”, ”1/3:2/4 rotated”
”1:2/3:4 rotated”, ”Hole”, ”y^2”))
if (resN 0) break
size - readline(”Number of nodes - 0 to 20: ”)
example.display(syn.train.data, syn.train.class[,resN], syn.test.data,
syn.test.class[,resN], c(100,100), paste(”NN -”, size), model.exp, predict
as.integer(size))
}
}
The following illustrates the nature of the distributions of data that can be classified using
neural networks. Note that, if the class boundary is a straight line, no hidden layer is
required. A hidden layer with at least one node is required if the class boundary is curved.
Note that for the circle, several examples are given with the same number of nodes in the
hidden layer with different results, indicating that the results are dependent on the initial
weights.
In the confusion matrices, the results are given as trainingtest where there is a difference.
Classificationn-Neural Nets 432 © Mills 2018
Data Mining 2018
true
object 0 1
0 22059 0
1 0 18041
true
object 0 1
0 11825
110
26
1 8518
87
31
true
object 0 1
0 15938
18
9
1 445
179
48
true
object 0 1
0 20143
3
0
1 20
196
57
true
object 0 1 2
0 21858
0
1
0
0
1 10
85
17
2
0
2 11
0
0
93
23
true
object 0 1
0 20254 0
1 0 19846
true
object 0 1
0 10219
80
23
1 9727
121
31
true
object 0 1
0 13635
51
9
1 6311
150
45
Figure 25. Set 1 Set 2
Set 2 Set 2
Figure 26. Set 3 Set 4
Set 5 Set 5
© Mills 2018 Classification-Neural Nets 433
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true
object 0 1
0 16335
53
17
1 3610
148
37
true
object 0 1
0 19645
3
0
1 31
198
54
true
object 0 1
0 16438
49
14
1 358
152
40
true
object 0 1
0 19846
0
2
1 10
201
52
true
object 0 1 2
0 20153
1
0
1
0
1 01
104
19 0
2 10
1
0
91
27
true
object 1 0
1 31784
8
1
true
object 0 1
0 4810
39
9
1 356
278
75
true
object 0 1
0 8214
2
1
1 12
31
83
Figure 27. Set 5 Set 5
Set 5 Set 5
Figure 28. Set 6 Set 7
Set 7 Set 7
Classificationn-Neural Nets 434 © Mills 2018
Data Mining 2018
true
object 0 1
0 27875
32
2
1 267
64
16
true
object 0 1
0 30482
0
0
1 0 9918
Figure 29. Set 8 Set 8
© Mills 2018 Classification-Neural Nets 435
Data Mining 2018
Weight decay
In all of our examples we use a value of 0.0005 for the weight decay without an
understanding of what it does or if this is an appropriate value. We will now look at what
happens if we use different values of the weight decay in one of our synthetic examples and
then look at the role of this parameter.
One of the problems with examining the effects of changing a parameter is that, unless we
specify the initial weights, the nnet function will assign them for us.
We will use the following weight set for our illustration:
Wts - c(-0.076,0.034,-0.046,0.076,-0.056,0.024,0.029,-0.002,-0.013,0.096,
0.02,-0.1,-0.03,0.094,0.079,-0.077,0.092,0.021,-0.1,-0.029,0.067,
-0.005,-0.066,-0.036,0.07,-0.045,-0.073,-0.05,-0.002,0.006,0.026,
-0.027,-0.004,-0.01,-0.067,0.073,-0.028,0.016,-0.002,0.048,-0.004,
-0.06,0,0.085,-0.084,0.012,-0.076,-0.048,-0.088,-0.029,0.017,0.076)
Because the example.display and plot.class.boundary functions are not
flexible enough to use for this purpose, a revised version is used:
wt.decay.demo - function(train.data, train.class, size, decay, Wts) {
plot.class(train.data, train.class, {}, {},
paste(”Set ”,7, ” Size ”, size,” Decay ”, decay))
#
model -nnet(data.frame(train.data), class.ind(train.class),
skipF, softmax T, sizesize, decaydecay, maxit 1000,
trace F, WtsWts)
res - f.create.grid(train.data, c(100,100))
xp - res\$xp
yp - res\$yp
zp - res\$zp
dimnames(zp)[[2]]- dimnames(train.data)[[2]]
Z - as.integer(predict(model, zp, type”class”))
points(zp[,1], zp[,2], pch”.”, col (matrix(Z, length(xp), length(yp))2))
for (c in 1:2) {
contour(xp, yp, matrix(as.numeric(Z(c-1)), length(xp), length(yp)),
}
model\$wts
}
This combines parts of both functions and allows us to alter both the decay and initial
weights.
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As can be seen in the function, we will use synthetic set 7 for this example. First, we will
have a weight decay of 0 (the numbers that follow are the final weights):
wt.decay.demo(syn.train.data, syn.train.class[,7], 10, 0, Wts)
[1] 46.952944 -72.321578 -99.543309 -67.505512 -128.530383 -112.573710 24.
[8] 2.406699 -86.910964 4.655850 -5.063192 9.925622 165.129026 -52.
[15] -111.858845 73.052638 20.531977 -147.603460 -17.720219 -20.236981 -131.
[22] -63.040967 7.908969 56.876267 30.960780 -98.432307 39.383626 -19.
[29] -42.131693 9.172620 -207.027488 78.284520 -197.114643 41.715497 168.
[36] -142.902458 80.409175 93.197333 48.933154 106.898010 -161.048555 206.
[43] -78.311520 197.195643 -41.809497 -168.752360 142.899458 -80.485175 -93.
[50] -48.964154 -106.833010 161.120555
Figure 30. Figure 31.
then our default value of 0.0005:
wt.decay.demo(syn.train.data, syn.train.class[,7], 10, 0.0005, Wts)
[1] -5.87928080 5.03279020 12.19698210 -4.85386908 11.01464585 2.96233865
[7] 2.15728880 -0.18055929 0.07299016 -6.39541033 -15.08979694 1.96283858
[13] 2.17110174 -0.18986848 0.06882179 -2.99942496 5.81146042 4.75170420
[19] -5.80564095 -12.47637429 -8.74936714 8.63444402 -17.02970141 12.67236964
[25] 7.85763930 -0.30151434 18.40550548 -6.86009708 -7.98902089 13.30337368
[31] -3.64575415 -11.07458910 -9.05655467 -3.29344431 -12.86662984 -3.29296441
[37] -6.43938568 -10.88162586 13.14176398 11.87692163 -13.07925292 3.64575466
[43] 11.07458950 9.05655347 3.29344570 12.86663065 3.29296445 6.43938681
[49] 10.88162693 -13.14176352 -11.87692260 13.07925186
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and finally 0.02:
wt.decay.demo(syn.train.data, syn.train.class[,7], 10, 0.02, Wts)
[1] 3.104485804 -1.455452680 -7.090912984 -2.877788537 -5.136496553 5.016078302
[7] 0.119925527 -0.008408174 0.029998843 0.120244244 -0.009551926 0.032034089
[13] -2.423589572 -3.290603562 -4.717679548 -2.305383046 3.733994023 -4.991637432
[19] -2.883649417 6.541906313 2.752506359 0.120419263 -0.009514419 0.030042892
[25] 1.829243424 -3.362771870 3.387582234 -2.401569808 -6.188614088 -1.550652177
[31] -0.672603942 5.595841465 -6.059654330 -0.357572600 -0.357828448 -4.605786190
[37] -4.879244836 -5.863651265 -0.357939658 3.040065787 -5.058816013 0.672605305
[43] -5.595840383 6.059651082 0.357576369 0.357830653 4.605786310 4.879247883
[49] 5.863654152 0.357940901 -3.040068393 5.058813126
Figure 32.
When we look at the final weights, we see that the magnitudes are decreasing as the weight
decay value increases. It is also obvious from the figures that increasing the weight decay
has resulted in a smoother curve. In data smoothing, we found that wider windows resulted
in smoother curves that did not fit the training data as well but might be better at predicting
for test data (i.e.we might be less likely to overfit the data).
When we were deriving the equations for the neural network, we tried to minimize the error
term
E
i1
N

k1
K
t ik fO WkT Ti 2
We also know that the absolute minimum would come when the network fits every training
case exactly (overfitting).
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To avoid this, we can add a penalty term to the error to give an error term in the form of
E 1
l0
p
wlm
2 2
k0
M
Wmk2
(for simplicity, the tuning parameters 1 and 2 have been replaced by a single value in
nnet). The effect of such a penalty term is to force (shrink) the coefficients towards 0 for
increasing values of 1 and 2.
Recall that it was mentioned that we might split our data into three sets. Here is a situation in
which that might be required. The first set could be used to train the data for various values
of 1 and 2; the second could be used to determine the best values of 1 and 2 to use; and
the third set to test the model.
As before, it might be instructive to look at the separating surfaces in three dimensions.
We will look at the flea beetles in three dimensions using variables 2, 3, and 5. The classes
are not as well separated in these directions so the surfaces may be a bit more complex.
We use the function disp.3d from earlier in order to find the points at which the classes
change:
library(MASS)
We create the three dimensional dataset:
first - 5
second - 3
third - 2
data.for.3d - data.frame(d.flea[,first],d.flea[,second],d.flea[,third])
colnames(data.for.3d) - colnames(d.flea[,c(first,second,third)])
Because neural networks can converge to different minima depending on the initial weights,
we will set:
Wts -
c(-0.99,10.71,-4.93,0.95,-15.76,-1.8,0.37,0.16,-15.34,13.5,11.89,10.6,-8.59,
-13.63,-9.84,-6.11,12.75)
as our initial values.
Because we would like surfaces other than planes, we will use two (2) nodes in the hidden
layer for one example and 10 nodes in the hidden layer for the second example:
flea.nn.3d.2 - nnet(data.for.3d, class.ind(flea.species), size2, skipF,
rang0.1,
decay5e-4, maxit500, WtsWts)
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# weights: 17
initial value 17.170432
iter 10 value 15.382686
iter 20 value 15.381768
final value 15.381755
converged
summary(flea.nn.3d.2)
a 3-2-3 network with 17 weights
options were - decay5e-04
b-h1 i1-h1 i2-h1 i3-h1
-0.94 10.74 -4.94 0.95
b-h2 i1-h2 i2-h2 i3-h2
-15.73 -1.79 0.38 0.16
b-o1 h1-o1 h2-o1
-15.33 13.50 11.88
b-o2 h1-o2 h2-o2
10.62 -8.59 -13.63
b-o3 h1-o3 h2-o3
-9.83 -6.12 12.74
confusion.expand(max.col(flea.nn.3d.2\$fitted.values), flea.species)
true
object C Hk Hp | Row Sum
1 16 1 0 | 17
2 5 29 0 | 34
3 0 1 22 | 23
------- ---- ---- ---- | ----
Col Sum 21 31 22 | 74
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
flea.nn.3d.2[11] # or flea.nn.3d.2\$wts
\$wts
[1] -0.9429123 10.7362345 -4.9361434 0.9533044 -15.7269710 -1.7932020
[7] 0.3789444 0.1601158 -15.3255563 13.4954494 11.8781092 10.6156708
[13] -8.5906243 -13.6286737 -9.8275448 -6.1223647 12.7414013
We get the points at which the classes change:
data.3d.2 - disp.3d(flea.nn.3d.2, data.for.3d, ”net”, 60)
and display them:
library(rgl)
plot3d(d.flea[,c(first, second, third)], type”s”, col species1, size0.5)
points3d(data.3d.2, size2)
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Figure 33. 2 nodes in hidden layer Figure 34. 10 nodes in hidden layer
Now repeat with 10 nodes in the hidden layer:
Wts - c(16.54,2.17,-0.53,-0.17,-0.01,0.05,0.07,0.18,7.36,2.86,2.32,-1.3,17.5,
-0.87,1.29,-0.54,-1.03,5.59,-0.82,-0.2,0,0.01,0.05,0.14,1.78,-3.16,
-10.6,4.61,-5.24,5.15,6.69,-3.25,-6.26,0.58,8.4,-3.23,-2.99,8.87,
-4.52,0.98,-4.44,-12.29,-4.47,-3.67,-15.14,1.44,-1.77,7.25,11.69,
13.3,8.87,0.71,12.26,0.74,4.37,14.72,8.72,-0.75,-7.52,-11.99,-13.7,
-8.6,0.76,-0.38,0.76,-0.65,1.88,-7.29,0.84,1.77,-0.55,0.73,-4.73)
flea.nn.3d.10 - nnet(data.for.3d, class.ind(flea.species), size10, skipF,
rang0.1,
decay5e-4, maxit500, WtsWts)
confusion.expand(max.col(flea.nn.3d.10\$fitted.values), flea.species)
true
object C Hk Hp | Row Sum
1 20 0 0 | 20
2 1 31 0 | 32
3 0 0 22 | 22
------- ---- ---- ---- | ----
Col Sum 21 31 22 | 74
attr(,”error”)
[1] NaN
attr(,”mismatch”)
[1] 1
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