Question

我有n个子列表的列表x。我想制作长度为new_list的{{1}}，其中包含每个子列表的最大索引。我该怎么办呢？

例如：

x = [[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]

（为了便于使用，我在每个子列表中采用了相同的最大值）

因此输出必须是：

new_list = [4,1,0,3,2]

Answer 1

如果每个子列表中没有重复项，请尝试以下操作：

new_list = [sub.index(max(sub)) for sub in x]

>>> x = [[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]
>>> new_list = [sub.index(max(sub)) for sub in x]
>>> new_list
[4, 1, 0, 3, 2]
>>>

Answer 2

您可以使用lambda表达式并映射到l，以获取x中最大元素的索引

l=[[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]
print map(lambda x:x.index(max(x)),l)

输出：

[4, 1, 0, 3, 2]

如果你正在使用numpy数组，

>>> print map(lambda x:int(np.where(x==max(x))[0]),l)
[4, 1, 0, 3, 2]

Answer 3

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numpy

Answer 4

通过循环并获取每个子列表的最大值来创建一个新列表：

像这样：

x = [[1,4,2,3,5],[3,5,2,1,3],[5,2,1,4,3],[1,2,3,5,4],[4,1,5,2,3]]

max_index = [i.index(max(i)) for i in x]

找出python中子列表的最大值索引

4 个答案: