Question

我有一个清单，说

listx = [1, 2, 0, 5, 2, 3, 0, 3, 4, 2, 3, 5, 6, 0, 4]

如何将列表拆分为不包含0的连续部分？例如，此列表将拆分为

separated = [[1, 2], [5, 2, 3], [3, 4, 2, 3, 5, 6], [4]]

Answer 1

import itertools

In [69]: L = [1, 2, 0, 5, 2, 3, 0, 3, 4, 2, 3, 5, 6, 0, 4]

In [70]: [list(i[1]) for i in itertools.groupby(L, bool) if i[0]]
Out[70]: [[1, 2], [5, 2, 3], [3, 4, 2, 3, 5, 6], [4]]

Answer 2

如果没有重复0's：

listx = [1, 2, 0, 5, 2, 3, 0, 3, 4, 2, 3, 5, 6, 0, 4]

final = [[]]
for ele in listx:
    final[-1].append(ele) if ele else final.append([])

print(final)
[[1, 2], [5, 2, 3], [3, 4, 2, 3, 5, 6], [4]]

如果您重复0's：

def split_l(l):
    final = [[]]
    for ele in l:
        final[-1].append(ele) if ele else final.append([])
    return filter(None, final)

In [146]: timeit split_l(L)
100 loops, best of 3: 19.6 ms per loop

In [147]: timeit  [list(i[1]) for i in itertools.groupby(L, bool) if i[0]]
10 loops, best of 3: 33.8 ms per loop

将列表拆分为单独的列表[Python]

2 个答案: