使用Python中的自定义顺序对列表进行排序 [英] Sort a list with a custom order in Python

问题描述

我有一个清单

mylist = [['123', 'BOOL', '234'], ['345', 'INT', '456'], ['567', 'DINT', '678']]

我想按1的顺序对其进行排序.DINT 2. INT 3. BOOL

I want to sort it with the order of 1. DINT 2. INT 3. BOOL

结果:

[['567', 'DINT', '678'], ['345', 'INT', '456'], ['123', 'BOOL', '234']]

我在stackoverflow中看到了其他类似的问题，但没有类似的问题或对我很容易适用.

I've seen other similar questions in stackoverflow but nothing similar or easily applicable to me.

推荐答案

SORT_ORDER = {"DINT": 0, "INT": 1, "BOOL": 2}

mylist.sort(key=lambda val: SORT_ORDER[val[1]])

我们在这里所做的全部工作是通过为列表中的每个元素而不是整个列表返回一个整数来提供一个新的元素进行排序.我们可以使用内联三元表达式，但这会有点笨拙.

All we are doing here is providing a new element to sort on by returning an integer for each element in the list rather than the whole list. We could use inline ternary expressions, but that would get a bit unwieldy.

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