[英]All combinations of list wIthout itertools
I'm trying to make a recursive function that finds all the combinations of a python list. 我正在尝试创建一个递归函数,以查找python列表的所有组合。
I want to input ['a','b','c'] in my function and as the function runs I want the trace to look like this: 我想在函数中输入['a','b','c'],并且在函数运行时,我希望跟踪显示如下:
['a','b','c']
['['a','a'],['b','a'],['c','a']]
['['a','a','b'],['b','a','b'],['c','a','b']]
['['a','a','b','c'],['b','a','b','c'],['c','a','b','c']]
My recursive function looks like this: 我的递归函数如下所示:
def combo(lst,new_lst = []):
for item in lst:
new_lst.append([lst[0],item])
print([lst[0],item])
return combo(new_lst,lst[1:])
The right answer is that you should use itertools.combinations
. 正确的答案是您应该使用
itertools.combinations
。 But if for some reason you don't want to, and want to write a recursive function, you can use the following piece of code. 但是,如果由于某种原因您不想这样做,并且想编写一个递归函数,则可以使用以下代码。 It is an adaptation of the erlang way of generating combinations, so it may seem a bit weird at first:
它是对生成组合的erlang方法的一种改编,因此乍一看似乎有点奇怪:
def combinations(N, iterable):
if not N:
return [[]]
if not iterable:
return []
head = [iterable[0]]
tail = iterable[1:]
new_comb = [ head + list_ for list_ in combinations(N - 1, tail) ]
return new_comb + combinations(N, tail)
This a very elegant way of thinking of combinations of size N
: you take the first element of an iterable ( head ) and combine it with smaller ( N-1
) combinations of the rest of the iterable ( tail ). 这是思考大小为
N
的组合的一种非常优雅的方式:您将可迭代( 头 )的第一个元素与其他可迭代( 尾 )的较小的( N-1
)组合在一起。 Then you add same size ( N
) combinations of the tail to that. 然后,向其添加相同大小的尾巴组合(
N
)。 That's how you get all possible combinations. 这就是您获得所有可能组合的方式。
If you need all combinations, of all lengths you would do: 如果需要所有长度的所有组合,则可以执行以下操作:
for n in range(1, len(iterable) + 1):
print(combinations(n, iterable))
Seems that you want all the product of a list, you can use itertools.product
within the following function to return a list of generators: 似乎需要列表的所有乘积,可以在以下函数中使用
itertools.product
返回生成器列表:
>>> from itertools import product
>>> def pro(li):
... return [product(l,repeat=i) for i in range(2,len(l)+1)]
...
>>> for i in pro(l):
... print list(i)
...
[('a', 'a'), ('a', 'b'), ('a', 'c'), ('b', 'a'), ('b', 'b'), ('b', 'c'), ('c', 'a'), ('c', 'b'), ('c', 'c')]
[('a', 'a', 'a'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'b', 'a'), ('a', 'b', 'b'), ('a', 'b', 'c'), ('a', 'c', 'a'), ('a', 'c', 'b'), ('a', 'c', 'c'), ('b', 'a', 'a'), ('b', 'a', 'b'), ('b', 'a', 'c'), ('b', 'b', 'a'), ('b', 'b', 'b'), ('b', 'b', 'c'), ('b', 'c', 'a'), ('b', 'c', 'b'), ('b', 'c', 'c'), ('c', 'a', 'a'), ('c', 'a', 'b'), ('c', 'a', 'c'), ('c', 'b', 'a'), ('c', 'b', 'b'), ('c', 'b', 'c'), ('c', 'c', 'a'), ('c', 'c', 'b'), ('c', 'c', 'c')]
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