将一个列表分成长度不同的块

Question

给定一个项目序列和另一个块长度序列，如何将序列拆分为所需长度的块？

a = range(10)
l = [3, 5, 2]
split_lengths(a, l) == [[0, 1, 2], [3, 4, 5, 6, 7], [8, 9]]

理想情况下，解决方案应该适用于通用可迭代对象和，而不仅仅是列表。

- ecatmur

Answer 1

使用itertools.islice来切片一个列表的迭代器。

In [12]: a = range(10)

In [13]: b = iter(a)

In [14]: from itertools import islice

In [15]: l = [3, 5, 2]

In [16]: [list(islice(b, x)) for x in l]
Out[16]: [[0, 1, 2], [3, 4, 5, 6, 7], [8, 9]]

或：

In [17]: b = iter(a)

In [18]: [[next(b) for _ in range(x)] for x in l]
Out[18]: [[0, 1, 2], [3, 4, 5, 6, 7], [8, 9]]

- Ashwini Chaudhary

Answer 2

def split_lengths(a,l):
    resultList = []
    index=0

    for length in l:
        resultList.append(a[index : index + length])
        index = index + length

    retun resultList

- Manish Mulani