从列表中获取所有最小元素及其索引 [英] Getting all the min elements and its indices from a list

问题描述

我有一个列表，其中包含一个最小元素，该元素多次出现，例如

I have a list that has a minimum element that is present multiple times like

a = [1,2,1,1,4,5,6]

我想让Python返回元素1和存在1的列表中的所有索引.我尝试使用

And I want Python to return the element 1 and all the indices in the list where 1 is present. I tried using

min_index, min_value = min(enumerate(a), key=operator.itemgetter(1))

仅给出首先出现1的索引.

which gives only the index where 1 occurs first.

推荐答案

确定最小元素，然后将其与列表中的其他元素进行比较.

determine the minimum element, and then check it against other elements in the list.

def locate_min(a):
    smallest = min(a)
    return smallest, [index for index, element in enumerate(a) 
                      if smallest == element]

将返回一个元组(min_element，[位置，位置，...]).如果我理解正确，这就是我想要的.以您为例:

which will return a tuple (min_element, [location, location, ...]). If I understand you correctly, this is what I think you want. For you example:

>>> locate_min([1, 2, 1, 1, 4, 5, 6])
(1, [0, 2, 3])

此示例使用列表理解.如果您不熟悉此功能，则它大致等效于以下for循环版本. (使用第一个版本，这只是为了帮助您了解其工作原理)

This example uses a list comprehension. If you're not familiar with this, it's roughly equivalent to the following for-loop version. (use the first version, this is just to help your understanding of how it works)

def locate_min(a):
    min_indicies = []
    smallest = min(a)
    for index, element in enumerate(a):
            if smallest == element: # check if this element is the minimum_value
                    min_indicies.append(index) # add the index to the list if it is

    return smallest, min_indicies

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