我有

listName = [[0,1,2,15,16,17,2,3,4,6,8,9]]


我的代码行

[list(g) for k, g in groupby(listName, key=lambda i,j=count(): i-next(j))]


listName拆分为[[0,1,2],[15,16,17],[2,3,4],[6,8,9]]
我希望只有下一个数字小于前一个数字时才进行拆分。即
我希望我的listName分成

[[0,1,2,15,16,17],[2,3,4,6,8,9]]


谢谢! :)

最佳答案

使用生成器函数(使用itertools.chain创建迭代器并展平列表)要容易得多:

listName = [[0, 1, 2, 15, 16, 17, 2, 3, 4, 6, 8, 9]]

from itertools import chain
def split(l):
    it = chain(*l)
    prev = next(it)
    tmp = [prev]
    for ele in it:
        if ele < prev:
            yield tmp
            tmp = [ele]
        else:
            tmp.append(ele)
        prev = ele
    yield tmp


print(list(split(listName)))

07-24 22:21