python list元素展开

xiaoxiao2021-02-28  126

单层嵌套

import itertools a = [[1,2,3],[4,5,6], [7], [8,9]] out = list(itertools.chain.from_iterable(a))#如果有嵌套的化需要多次调用此函数

输出:[1, 2, 3, 4, 5, 6, 7, 8, 9] 当遇到多层嵌套(要求嵌套的结构必须相同)时只要重复调用上述函数即可

import itertools a = [[[1,1],[2,2],[3,3]],[[4,4],[5,5],[6,6]], [[7,7]], [[8,8],[9,9]]] out = list(itertools.chain.from_iterable(a)) #输出:[[1, 1], [2, 2], [3, 3], [4, 4], [5, 5], [6, 6], [7, 7], [8, 8], [9, 9]] out_new= list(itertools.chain.from_iterable(out)) #输出:[1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9]
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