Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Misc Itertools recipe tweaks #100493

Merged
merged 2 commits into from
Dec 24, 2022
Merged
Changes from all commits
Commits
File filter

Filter by extension

Filter by extension

Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
54 changes: 49 additions & 5 deletions Doc/library/itertools.rst
Original file line number Diff line number Diff line change
Expand Up @@ -788,6 +788,11 @@ which incur interpreter overhead.

.. testcode::

import collections
import math
import operator
import random

def take(n, iterable):
"Return first n items of the iterable as a list"
return list(islice(iterable, n))
Expand Down Expand Up @@ -892,6 +897,21 @@ which incur interpreter overhead.
data[2] = 1
return iter_index(data, 1) if n > 2 else iter([])

def factor(n):
"Prime factors of n."
# factor(97) --> 97
# factor(98) --> 2 7 7
# factor(99) --> 3 3 11
for prime in sieve(n+1):
while True:
quotient, remainder = divmod(n, prime)
if remainder:
break
yield prime
n = quotient
if n == 1:
return

def flatten(list_of_lists):
"Flatten one level of nesting"
return chain.from_iterable(list_of_lists)
Expand Down Expand Up @@ -1134,11 +1154,6 @@ which incur interpreter overhead.

Now, we test all of the itertool recipes

>>> import operator
>>> import collections
>>> import math
>>> import random

>>> take(10, count())
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Expand Down Expand Up @@ -1251,6 +1266,35 @@ which incur interpreter overhead.
>>> set(sieve(10_000)).isdisjoint(carmichael)
True

list(factor(0))
[]
list(factor(1))
[]
list(factor(2))
[2]
list(factor(3))
[3]
list(factor(4))
[2, 2]
list(factor(5))
[5]
list(factor(6))
[2, 3]
list(factor(7))
[7]
list(factor(8))
[2, 2, 2]
list(factor(9))
[3, 3]
list(factor(10))
[2, 5]
all(math.prod(factor(n)) == n for n in range(1, 1000))
True
all(set(factor(n)) <= set(sieve(n+1)) for n in range(1, 1000))
True
all(list(factor(n)) == sorted(factor(n)) for n in range(1, 1000))
True

>>> list(flatten([('a', 'b'), (), ('c', 'd', 'e'), ('f',), ('g', 'h', 'i')]))
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']

Expand Down