[Python-checkins] Misc Itertools recipe tweaks (GH-100493)

Sat Dec 24 03:21:35 EST 2022

https://github.com/python/cpython/commit/0769f957514300a75be51fc6d1b963c8e359208b
commit: 0769f957514300a75be51fc6d1b963c8e359208b
branch: main
author: Raymond Hettinger <rhettinger at users.noreply.github.com>
committer: rhettinger <rhettinger at users.noreply.github.com>
date: 2022年12月24日T00:21:30-08:00
summary:
Misc Itertools recipe tweaks (GH-100493)
files:
M Doc/library/itertools.rst

diff --git a/Doc/library/itertools.rst b/Doc/library/itertools.rst
index 9688aae68e6c..b3634aecd10d 100644
--- a/Doc/library/itertools.rst
+++ b/Doc/library/itertools.rst
@@ -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))
@@ -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)
@@ -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]
 
@@ -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']
 
