python 代写-ICS-33
时间:2020-12-16
Quiz #4: Iterators and Iterator Decorators via Generators ICS-33 Fall 2020
When working on this quiz, recall the rules stated on the Academic Integrity statement that you signed. You
can download the q4helper project folder (available for Friday, on the Weekly Schedule link) in which to
write/test/debug your code. Submit your completed q4solution module online by Wednesday, 11:30pm. I will
post my solutions to EEE reachable via the Solutions link on Wednesday right after 11:30 pm. In this way
you can see my solutions before the Thursday midterm.
Remember, if an argument is iterable, it means that you can call only iter on it, and then call next on the
value iter returns (for loops do this automatically). There is no guarantee you can call len on the iterable or
index/slice it. You may not copy all the values of an iterable into a list (or any other data structure) so that
you can perform these operations (that is not in the spirit of the assignment, and some iterables could produce
an infinite number of values, so such copying is impossible). You may create local data structures storing as
many values as the arguments or the result that the function returns, but not all the values in the iterable.
1. (20 pts) Write generator functions below (each one is worth 4 points) that satisfy the following
specifications. You may not import any of the generators in itertools or any other modules to write your
generators. You may use any of the standard functions like zip and enumerate.
a. The sequence generator takes any number of iterables as parameters: it produces every value from the first
iterable, followed by every value from the second iterable, etc.,. Hint: I used only for loops. For example
for i in sequence('abc', 'd', 'ef', 'ghi'):
print(i,end='')
prints abcdefghi: all characters from 'abc', followed by all in 'd', followed by all in 'ef' etc..
b. The group_when generator takes one iterable and one predicate as parameters: it produces lists that each
end in a value from the iterable where the predicate is True. If the iterable ends on a value for which the
predicate returns False, yield a final list containing all the values from the one after the previous end to the
last value produced by the iterable. Hint: I used a for loop . For example
for i in group_when('combustibles', lambda x : x in 'aeiou'):
print(i,end='')
prints the 5 lists ['c', 'o']['m', 'b', 'u']['s', 't', 'i']['b', 'l', 'e']['s'].
c. The drop_last generator takes one iterable and one int as a parameter (call it n): it produces every value
from the iterable except for the last n values (without being able to count how many values the iterable
produces) Hint: I used an explicit call to iter and a while loop and a comprehension that creates a list that
stores at most n values (so that data structure is allowed here). For example
for i in drop_last('combustible', 5):
print(i,end='')
prints combus; if the iterable produces < n values, it terminates immediately, on the first call to next.
d. The yield_and_skip generator takes one iterable and one function (which takes one argument and returns
an int) as parameters: it produces a value from the iterable but then it then skips the number of values
specified when the function argument is called on the just-produced value. Hint: I used an explicit call to iter
and a while and for loop. For example
for i in yield_and_skip('abbabxcabbcaccabb',lambda x : {'a':1,'b':2,'c':3}.get(x,0)):
print(i,end='')
prints abxccab; prints a then skips 1; prints b then skips 2; prints x then skips 0; prints c then skips 3; …
e. The alternate_all generator takes any number of iterables as parameters: it produces the first
value from the first parameter, then the first value from the second parameter, ..., then the first value
from the last parameter; then the second value from the first parameter, then the second value from the
second parameter, ..., then the second value from the last parameter; etc. If any iterable produces no
more values, it is ignored. Eventually, this generator produces every value in each iterable. Hint: I
used explicit calls to iter, and a while and for loop, and a try/except statement; you can create a
list whose length is the same as the number of parameters (I stored iter called on each parameter in
such a list). For example
for i in alternate_all('abcde','fg','hijk'):
print(i,end='')
prints afhbgicjdke.

2. (5 pts) Define a generator named min_key_order whose parameter is a dict that has keys that can all be
compared with each other (e.g., all numbers, all strings, etc; don’t check this property). The min_key_order
generator produces a 2-tuple of the dict’s keys and values, in increasing order of the keys. For example:
for i in min_key_order ({1:’a’, 2:’x’, 4:’m’, 8:’d’, 16:'f'}):
print(i,end='')
prints: (1,’a’)(2,’x’)(4,’m’)(8,’d’)(16,’f’)
It would be trivial to write this generator as
def min_key_order (adict):
for k,v in sorted(adict.items()): # iterate over a sorted list of keys/values
yield (k,v)
but the generator must have one more property: it must iterate over mutations made to the dict while the
dict is being iterated over. That is
d = ({1:’a’, 2:’x’, 4:’m’, 8: ’d’}
i = min_keys_order(d)
print(next(i))
print(next(i))
del d[8]
print(next(i))
d[16] = ’f’
d[32] = ’z’
print(next(i))
print(next(i))
Note that the process of iterating through the dict should not mutate the dict (although the dict can be
mutated directly by code during the iteration, as is shown above),
Hints: The first time next is called, find the minimum key in the dict (it is unique, if the dict isn’t
empty) and yield it and the key and its value first; later, repeatedly find the smallest key bigger than the
previously yielded key (so long as there is one) and yield the key and its value; you may create a local
list/tuple whose length is no bigger than the dict (although there is a way to solve this problem
without such a data structure, but it requires more code).

Prints:
(1, 'a')
(2, 'x')
(4, 'm')
(16, 'f')
(32, 'z')

























































































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