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

This commit is contained in:
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
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::
def combinations(iterable, r):
'combinations(range(4), 3) --> (0,1,2) (0,1,3) (0,2,3) (1,2,3)'
pool = tuple(iterable)
n = len(pool)
assert 0 <= r <= n
vec = range(r)
yield tuple(pool[i] for i in vec)
indices = range(r)
yield tuple(pool[i] for i in indices)
while 1:
for i in reversed(range(r)):
if vec[i] != i + n - r:
if indices[i] != i + n - r:
break
else:
return
vec[i] += 1
indices[i] += 1
for j in range(i+1, r):
vec[j] = vec[j-1] + 1
yield tuple(pool[i] for i in vec)
indices[j] = indices[j-1] + 1
yield tuple(pool[i] for i in indices)
.. 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
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