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Misc Itertools recipe tweaks by rhettinger · Pull Request #100493 · python/cpython · GitHub

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54 changes: 49 additions & 5 deletions Doc/library/itertools.rst
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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

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