将列表的子列表拆分为其他子列表

Question

I am having problems with 'splitting' a larger list into several of it's combinations. Here is an example:

Let's say I have this list:

  x = [['a','b'],['c','f'],['q','w','t']]

and I want to end up with

  x = [['a','b'],['c','f'],['q','w'],['q','t'],['w','t']]

so essentially

 ['q','w','t']

becomes

 ['q','w'],['q','t'],['w','t']

I see how I can convert

 ['q','w','t']

to

 [['q','w'],['q','t'],['w','t']] #notice the extra brackets

with itertools combinations, but then I am stuck with

 x = [['a','b'],['c','f'],[['q','w'],['q','t'],['w','t']]] #notice the extra brackets

Which is not what I want.

How should I do this?

EDIT:

Here is the "solution", that does not give me the result that I want:

from itertools import combinations

x = [['a','b'],['c','f'],['q','w','t']]

new_x = []

for sublist in x:
    if len(sublist) == 2:
        new_x.append(sublist)
    if len(sublist) > 2:
        new_x.append([list(ele) for ele in (combinations(sublist,2))])

Thank You

Answer 1

I generally use a nested list comprehension to flatten a list like this:

>>> x = [['a','b'],['c','f'],['q','w','t']]
>>> [c for s in x for c in itertools.combinations(s, 2)]
[('a', 'b'), ('c', 'f'), ('q', 'w'), ('q', 't'), ('w', 't')]

Answer 2

Not the best way to do it but pretty clear for understanding:

from itertools import combinations

a = [['a','b'],['c','f'],['q','w','t']]

def get_new_list(x):

    newList = []
    for l in x:
        if len(l) > 2:
            newList.extend([list(sl) for sl  in combinations(l, 2)])
        else:
            newList.append(l)
    return newList

print get_new_list(a)
>>> [['a','b'],['c','f'],['q','w'],['q','t'],['w','t']]