获取列表中嵌套元素的所有排列 - python 2

Question

So its not the permutations Im having trouble with, rather that the outcome lists are not equivalent to the original list I started with. The list is this,

B=[[m, b], [c, g], [d, f]]

and the code Ive used to get all the permutation is this, along with the outcome,

C=list(itr.permutations(B))
C
 [([m, b], [c, g], [d, f]),
 ([m, b], [d, f], [c, g]),
 ([c, g], [m, b], [d, f]),
 ([c, g], [d, f], [m, b]),
 ([d, f], [m, b], [c, g]),
 ([d, f], [c, g], [m, b])]

Is there any way to not have the inner lists in parentheses, rather instead in square bracket, because as of now C[0] is not the same as B, when in fact they should be equal.

C[0]==B
False 

Ive defined the letters (m, b, d, etc) as symbols in sympy rather then as strings. Any advice is appreciated.

Answer 1

You can use map() to turn the tuples into lists:

>>> import itertools as itr
>>> B=[[m, b], [c, g], [d, f]]
>>> C = map(list, itr.permutations(B))
>>> C
[[['m', 'b'], ['c', 'g'], ['d', 'f']], [['m', 'b'], ['d', 'f'], ['c', 'g']], [['c', 'g'], ['m', 'b'], ['d', 'f']], [['c', 'g'], ['d', 'f'], ['m', 'b']], [['d', 'f'], ['m', 'b'], ['c', 'g']], [['d', 'f'], ['c', 'g'], ['m', 'b']]]
>>> C[0] == B
True

Also, you can use itertoolsimap() if you want to keep it as an iterator.

Answer 2

Just use C[0][0] == B[0]

>>> print(C)
[(['m', 'b'], ['c', 'g'], ['d', 'f']), (['m', 'b'], ['d', 'f'], ['c', 'g']), (['c', 'g'], ['m', 'b'], ['d', 'f']), (['c', 'g'], ['d', 'f'], ['m', 'b']), (['d', 'f'], ['m', 'b'], ['c', 'g']), (['d', 'f'], ['c', 'g'], ['m', 'b'])]
>>> print(B)
[['m', 'b'], ['c', 'g'], ['d', 'f']]
>>> C[0][0]
['m', 'b']
>>> B[0]
['m', 'b']
>>> C[0][0] == B[0]
True
>>> C[0][1] == B[1]
True