问题描述
当我使用 itertools.product ?还是说，是否在迭代中不去查看它们?因为如果列表数量太多，跳过可能会很耗时.
How can I skip the tuples which has duplicate elements in the iteration when I use itertools.product? Or let's say, is there anyway not to look at them in the iteration? Because skipping may be time consuming if the number of lists are too much.
Example,
lis1 = [1,2]
lis2 = [2,4]
lis3 = [5,6]
[i for i in product(lis1,lis2,lis3)] should be [(1,2,5), (1,2,6), (1,4,5), (1,4,6), (2,4,5), (2,4,6)]
它将没有(2,2,5)和(2,2,6)，因为此处2是重复的.我该怎么办?
It will not have (2,2,5) and (2,2,6) since 2 is duplicate in here. How can I do that?
推荐答案
itertools通常适用于输入中唯一的位置，而不适用于唯一的值.因此，当您要删除重复的值时，通常必须对itertools结果序列进行后处理，或者自行滚动".因为在这种情况下后期处理效率可能非常低，所以请自己动手:
itertools generally works on unique positions within inputs, not on unique values. So when you want to remove duplicate values, you generally have to either post-process the itertools result sequence, or "roll your own". Because post-processing can be very inefficient in this case, roll your own:
def uprod(*seqs):
def inner(i):
if i == n:
yield tuple(result)
return
for elt in sets[i] - seen:
seen.add(elt)
result[i] = elt
for t in inner(i+1):
yield t
seen.remove(elt)
sets = [set(seq) for seq in seqs]
n = len(sets)
seen = set()
result = [None] * n
for t in inner(0):
yield t
然后，例如
>>> print list(uprod([1, 2, 1], [2, 4, 4], [5, 6, 5]))
[(1, 2, 5), (1, 2, 6), (1, 4, 5), (1, 4, 6), (2, 4, 5), (2, 4, 6)]
>>> print list(uprod([1], [1, 2], [1, 2, 4], [1, 5, 6]))
[(1, 2, 4, 5), (1, 2, 4, 6)]
>>> print list(uprod([1], [1, 2, 4], [1, 5, 6], [1]))
[]
>>> print list(uprod([1, 2], [3, 4]))
[(1, 3), (1, 4), (2, 3), (2, 4)]
这可能会更加高效，因为甚至都不会考虑重复的值(既不可在输入内迭代，也不能跨输入).
This can be much more efficient, since a duplicate value is never even considered (neither within an input iterable, nor across them).
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