Question

我有以下内容：

a_list = [A,B,C]
r1 = range(1,5)
r2 = range(1,5)
r3 = range(1,5)

我希望能够根据范围找到此列表中元素的各种组合。例如：

combi1 = [A, B, C, C, C]
combi2 = [A, A, B, C, C]
combi3 = [A, A, A, B, C]
combi4 = [A, B, B, C, C]

等

如果只有range，我可以这样做，但我不确定如何使3 range适合。

inc = range(1, 5)
desc = range(5, 1, -1)
combis = [list(itertools.chain(*(itertools.repeat(elem, n) for elem, n in zip(list, [i,j])))) for i,j in zip(inc,desc)]

SOLUTION：

def all_exist(avalue, bvalue):
    return all(any(x in y for y in bvalue) for x in avalue)

combins = itertools.combinations_with_replacement(a_list, 5)
combins_list = [list(i) for i in combins]
for c in combins_list:
    if all_exist(a_list, c) == True:
        print c

输出：

['A', 'A', 'A', 'B', 'C']
['A', 'A', 'B', 'B', 'C']
['A', 'A', 'B', 'C', 'C']
['A', 'B', 'B', 'B', 'C']
['A', 'B', 'B', 'C', 'C']
['A', 'B', 'C', 'C', 'C']

Answer 1

@doyz. I think this is may be what you are looking for :

From a list abc = ['A','B','C'], you want to obtain its various combinations with replacement. Python has built-in itertools to do this.

import itertools

abc = ['A', 'B', 'C'];

combins = itertools.combinations_with_replacement(abc, 5);
combins_list = [list(i) for i in combins];

print(combins_list[0:10]);

This is the first 10 combinations with replacement :

[['A', 'A', 'A', 'A', 'A'], ['A', 'A', 'A', 'A', 'B'], ['A', 'A', 'A', 'A', 'C'], \
 ['A', 'A', 'A', 'B', 'B'], ['A', 'A', 'A', 'B', 'C'], ['A', 'A', 'A', 'C', 'C'], \
['A', 'A', 'B', 'B', 'B'], ['A', 'A', 'B', 'B', 'C'], ['A', 'A', 'B', 'C', 'C'], ['A', 'A', 'C', 'C', 'C']]

If you want to include all elements in abc, here is one way, that also includes the permutations :

import itertools
abc = ['A', 'B', 'C']; 
combins = itertools.combinations_with_replacement(abc, 5); 
combins_list = list(combins);
combins_all =[];
for i in combins_list:
    if len(set(i))==len(abc):
        combins_all.append(i);
print(combins_all);

include_permutations=[];
for i in combins_all:
    permut = list(itertools.permutations(i));
    include_permutations.append(permut);
print(include_permutations);

Is this okay?

*Note : itertools.combinations_woth_replacement and itertools.permutations do not result in a list, or tuple, but a different object itself, so you can't treat it as those.

使用itertools从列表和3个范围创建组合

1 个答案: