如何将列表分成较小的列表？

Question

I have a list:

lists = (['1','2','3','S','3','4','S','4','6','7'])

And I want to split the list into s smaller list everytime 'S' appears and eliminate 'S' into something like:

([['1','2','3'],['3','4],['4','6','7']])

My code:

def x (lists):
    empty = ''
    list = []

    for x in lists:
        if x == empty:
            list[-1:].append(x)
        else:
            list.append([x])

    return (list)

I tried something like this, but I am quite new to python, and Im getting nowhere. Nothing fancy please, how would I fix what I have?

Answer 1

Try itertools.groupby() :

>>> from itertools import groupby
>>> lists = ['1','2','3','S','3','4','S','4','6','7']
>>> [list(g[1]) for g in groupby(lists, lambda i:i!='S') if g[0]]
[['1', '2', '3'], ['3', '4'], ['4', '6', '7']]

Answer 2

Maybe something like map(list,''.join(lists).split('S'))

Alternately, [list(s) for s in ''.join(lists).split('S'))

Answer 3

Well, may be funny, but this should work:

[s.split('#') for s in '#'.join(lists).split('#S#')]

Instead of the '#' any character can be used if it's unlikely to appear in lists .