Python 使用列表理解的嵌套 For 循环
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Nested For Loops Using List Comprehension
提问by John Howard
If I had two strings, 'abc'and 'def', I could get all combinations of them using two for loops:
如果我有两个字符串'abc'和'def'，我可以使用两个 for 循环获得它们的所有组合：
for j in s1:
for k in s2:
print(j, k)
However, I would like to be able to do this using list comprehension. I've tried many ways, but have never managed to get it. Does anyone know how to do this?
但是，我希望能够使用列表理解来做到这一点。我尝试了很多方法，但从未设法得到它。有谁知道如何做到这一点？
采纳答案by aaronasterling
lst = [j + k for j in s1 for k in s2]
or
或者
lst = [(j, k) for j in s1 for k in s2]
if you want tuples.
如果你想要元组。
Like in the question, for j...is the outer loop, for k...is the inner loop.
就像在问题中一样，for j...是外循环，for k...是内循环。
Essentially, you can have as many independent 'for x in y' clauses as you want in a list comprehension just by sticking one after the other.
从本质上讲，只需一个接一个地粘贴，您就可以在列表理解中拥有任意数量的独立“for x in y”子句。
回答by miles82
Since this is essentially a Cartesian product, you can also use itertools.product. I think it's clearer, especially when you have more input iterables.
由于这本质上是笛卡尔积，因此您也可以使用itertools.product。我认为它更清晰，尤其是当你有更多的输入迭代时。
itertools.product('abc', 'def', 'ghi')
回答by Stefan Gruenwald
Try recursion too:
也尝试递归：
s=""
s1="abc"
s2="def"
def combinations(s,l):
if l==0:
print s
else:
combinations(s+s1[len(s1)-l],l-1)
combinations(s+s2[len(s2)-l],l-1)
combinations(s,len(s1))
Gives you the 8 combinations:
给你8种组合：
abc
abf
aec
aef
dbc
dbf
dec
def
