Dec 24, 2022 · Dec 24, 2022 · Dec 24, 2022
diff --git a/Doc/library/itertools.rst b/Doc/library/itertools.rst

 .. 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))
       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)

    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]

    >>> 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']
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 DownExpand 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 DownExpand 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 DownExpand 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