Kaydet (Commit) 3fa41d5a authored tarafından Raymond Hettinger's avatar Raymond Hettinger

Docs for itertools.combinations(). Implementation in forthcoming checkin.

üst b1d70e22
......@@ -76,6 +76,45 @@ loops that truncate the stream.
yield element
.. function:: combinations(iterable, r)
Return successive *r* length combinations of elements in the *iterable*.
Combinations are emitted in a lexicographic sort order. So, if the
input *iterable* is sorted, the combination tuples will be produced
in sorted order.
Elements are treated as unique based on their position, not on their
value. So if the input elements are unique, there will be no repeat
values within a single combination.
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):
pool = tuple(iterable)
if pool:
n = len(pool)
vec = range(r)
yield tuple(pool[i] for i in vec)
while 1:
for i in reversed(range(r)):
if vec[i] == i + n-r:
continue
vec[i] += 1
for j in range(i+1, r):
vec[j] = vec[j-1] + 1
yield tuple(pool[i] for i in vec)
break
else:
return
.. versionadded:: 2.6
.. function:: count([n])
Make an iterator that returns consecutive integers starting with *n*. If not
......@@ -311,9 +350,12 @@ loops that truncate the stream.
The leftmost iterators are in the outermost for-loop, so the output tuples
cycle in a manner similar to an odometer (with the rightmost element
changing on every iteration).
changing on every iteration). This results in a lexicographic ordering
so that if the inputs iterables are sorted, the product tuples are emitted
in sorted order.
Equivalent to (but without building the entire result in memory)::
Equivalent to the following except that the actual implementation does not
build-up intermediate results in memory::
def product(*args):
pools = map(tuple, args)
......
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