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Finish-up docs for combinations() and permutations() in itertools.

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Raymond Hettinger 2008-03-02 10:59:31 +00:00
parent 47dff40a40
commit f287f17779
1 changed files with 30 additions and 10 deletions
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40
Doc/library/itertools.rst
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@ -104,26 +104,24 @@ loops that truncate the stream.
Each result tuple is ordered to match the input order. So, every Each result tuple is ordered to match the input order. So, every
combination is a subsequence of the input *iterable*. combination is a subsequence of the input *iterable*.
Example: ``combinations(range(4), 3) --> (0,1,2), (0,1,3), (0,2,3), (1,2,3)``
Equivalent to:: Equivalent to::
def combinations(iterable, r): def combinations(iterable, r):
'combinations(range(4), 3) --> (0,1,2) (0,1,3) (0,2,3) (1,2,3)'
pool = tuple(iterable) pool = tuple(iterable)
n = len(pool) n = len(pool)
assert 0 <= r <= n indices = range(r)
vec = range(r) yield tuple(pool[i] for i in indices)
yield tuple(pool[i] for i in vec)
while 1: while 1:
for i in reversed(range(r)): for i in reversed(range(r)):
if vec[i] != i + n - r: if indices[i] != i + n - r:
break break
else: else:
return return
vec[i] += 1 indices[i] += 1
for j in range(i+1, r): for j in range(i+1, r):
vec[j] = vec[j-1] + 1 indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in vec) yield tuple(pool[i] for i in indices)
.. versionadded:: 2.6 .. versionadded:: 2.6
@ -369,7 +367,29 @@ loops that truncate the stream.
value. So if the input elements are unique, there will be no repeat value. So if the input elements are unique, there will be no repeat
values in each permutation. values in each permutation.
Example: ``permutations(range(3),2) --> (1,2) (1,3) (2,1) (2,3) (3,1) (3,2)`` Equivalent to::
def permutations(iterable, r=None):
'permutations(range(3), 2) --> (0,1) (0,2) (1,0) (1,2) (2,0) (2,1)'
pool = tuple(iterable)
n = len(pool)
r = n if r is None else r
indices = range(n)
cycles = range(n-r+1, n+1)[::-1]
yield tuple(pool[i] for i in indices[:r])
while n:
for i in reversed(range(r)):
cycles[i] -= 1
if cycles[i] == 0:
indices[:] = indices[:i] + indices[i+1:] + indices[i:i+1]
cycles[i] = n - i
else:
j = cycles[i]
indices[i], indices[-j] = indices[-j], indices[i]
yield tuple(pool[i] for i in indices[:r])
break
else:
return
.. versionadded:: 2.6 .. versionadded:: 2.6