根据另一个列表的顺序重新排列列表 [英] Rearranging list based on order of another list

问题描述

我确定以前可能已经问过这个问题，但是我似乎找不到正确的答案.如果我有两个列表

I am sure this question has possibly been asked before but I can't seem to find the correct answer. If I have two lists

_list1 = ["keyName", "test1", "test2"]
_list2 = ["keyName", "test2", "test1"]

我正在尝试使用_list1重新排列_list2中的元素，以便它们与顺序完全匹配.什么是最干净的方法?所需的输出:

I am trying to use _list1 to rearrange elements in _list2 so that they match the order exactly. What's the cleanest way to do that? Desired output:

_list1 = ["keyName", "test1", "test2"]
_list2 = ["keyName", "test1", "test2"]

很抱歉，如果这是重复的，但到目前为止，我只能使用压缩的sorted()方法找到数字列表的答案.

I am sorry if this is duplicate but so far I am only able to find answers for list of numbers and using zipped sorted() method.

如果_list2是列表列表怎么办?

What if the _list2 is a list of lists?

_list2 = [["test1", "test2", "keyName"], ["test2", "test1", "keyName"]]

所需的输出:

_list2 = [["keyName", "test1", "test2"], ["keyName", "test1", "test2"]]

如果发生以下情况，该怎么办:如果我想使用_list1作为键来对其他任何对象列表进行排序，该怎么办

One more what if: What if I wanted to sort any other list of objects using _list1 as a key

_list2 = [[object1, object2, object3], [object1, object2, object3]]

其中:

object1.Name = "keyName"
object3.Name = "test1"
object2.Name = "test2"

我希望如此有效地输出:

so effectively I would expect output of:

_list2 = [[object1, object3, objec1], [object1, object3, objec1]]

有可能吗?

推荐答案

In [84]: _list1 = ["keyName", "test1", "test2"]

In [85]: d = {k:v for v,k in enumerate(_list1)}

In [86]: _list2 = ["keyName", "test2", "test1"]

In [87]: _list2.sort(key=d.get)

In [88]: _list2
Out[88]: ['keyName', 'test1', 'test2']

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