如何按列对多维数组进行排序？

Is there a way to use the sort() method or any other method to sort a list by column? Lets say I have the list:

有没有办法使用sort（）方法或任何其他方法按列对列表进行排序？让我们说清单：

[
[John,2],
[Jim,9],
[Jason,1]
]

And I wanted to sort it so that it would look like this:

我想对它进行排序，使它看起来像这样：

[
[Jason,1],
[John,2],
[Jim,9],
]

What would be the best approach to do this?

这样做的最佳方法是什么？

Edit:

编辑：

Right now I am running into an index out of range error. I have a 2 dimensional array that is lets say 1000 rows b 3 columns. I want to sort it based on the third column. Is this the right code for that?

现在我遇到索引超出范围错误。我有一个二维数组，可以说1000行b 3列。我想根据第三列对其进行排序。这是正确的代码吗？

sorted_list = sorted(list_not_sorted, key=lambda x:x[2])

5 个解决方案

#1

Yes. The sorted built-in accepts a key argument:

是。已排序的内置接受一个关键参数：

sorted(li,key=lambda x: x[1])
Out[31]: [['Jason', 1], ['John', 2], ['Jim', 9]]

note that sorted returns a new list. If you want to sort in-place, use the .sort method of your list (which also, conveniently, accepts a key argument).

请注意，sorted返回一个新列表。如果要在就地排序，请使用列表的.sort方法（也可以方便地接受键参数）。

or alternatively,

或者，

from operator import itemgetter
sorted(li,key=itemgetter(1))
Out[33]: [['Jason', 1], ['John', 2], ['Jim', 9]]

#2

You can use the sorted method with a key.

您可以使用带键的排序方法。

sorted(a, key=lambda x : x[1])

#3

The optional key parameter to sort/sorted is a function. The function is called for each item and the return values determine the ordering of the sort

排序/排序的可选键参数是一个函数。为每个项调用该函数，返回值确定排序的顺序

>>> lst = [['John', 2], ['Jim', 9], ['Jason', 1]]
>>> def my_key_func(item):
...     print("The key for {} is {}".format(item, item[1]))
...     return item[1]
... 
>>> sorted(lst, key=my_key_func)
The key for ['John', 2] is 2
The key for ['Jim', 9] is 9
The key for ['Jason', 1] is 1
[['Jason', 1], ['John', 2], ['Jim', 9]]

taking the print out of the function leaves

从功能中取出打印件

>>> def my_key_func(item):
...     return item[1]

This function is simple enough to write "inline" as a lambda function

这个函数很简单，可以将“inline”写成lambda函数

>>> sorted(lst, key=lambda item: item[1])
[['Jason', 1], ['John', 2], ['Jim', 9]]

#4

You can use list.sort with its optional key parameter and a lambda expression:

您可以将list.sort与其可选的键参数和lambda表达式一起使用：

>>> lst = [
...     ['John',2],
...     ['Jim',9],
...     ['Jason',1]
... ]
>>> lst.sort(key=lambda x:x[1])
>>> lst
[['Jason', 1], ['John', 2], ['Jim', 9]]
>>>

This will sort the list in-place.

这将对列表进行就地排序。

Note that for large lists, it will be faster to use operator.itemgetter instead of a lambda:

请注意，对于大型列表，使用operator.itemgetter而不是lambda会更快：

>>> from operator import itemgetter
>>> lst = [
...     ['John',2],
...     ['Jim',9],
...     ['Jason',1]
... ]
>>> lst.sort(key=itemgetter(1))
>>> lst
[['Jason', 1], ['John', 2], ['Jim', 9]]
>>>

#5

sorted(list, key=lambda x: x[1])

Note: this works on time variable too.

注意：这也适用于时间变量。

#1