如何按内部列表的特定索引对列表列表进行排序？

>>> from operator import itemgetter
>>> L=[[0, 1, 'f'], [4, 2, 't'], [9, 4, 'afsd']]
>>> sorted(L, key=itemgetter(2))
[[9, 4, 'afsd'], [0, 1, 'f'], [4, 2, 't']]

也可以在此处使用lambda函数，但是在这种简单情况下，lambda函数的运行速度较慢

This is a job for itemgetter

>>> from operator import itemgetter
>>> L=[[0, 1, 'f'], [4, 2, 't'], [9, 4, 'afsd']]
>>> sorted(L, key=itemgetter(2))
[[9, 4, 'afsd'], [0, 1, 'f'], [4, 2, 't']]

It is also possible to use a lambda function here, however the lambda function is slower in this simple case

回答 1

到位

>>> l = [[0, 1, 'f'], [4, 2, 't'], [9, 4, 'afsd']]
>>> l.sort(key=lambda x: x[2])

使用排序不到位：

>>> sorted(l, key=lambda x: x[2])

in place

>>> l = [[0, 1, 'f'], [4, 2, 't'], [9, 4, 'afsd']]
>>> l.sort(key=lambda x: x[2])

not in place using sorted:

>>> sorted(l, key=lambda x: x[2])

回答 2

Itemgetter使您可以按多个条件/列进行排序：

sorted_list = sorted(list_to_sort, key=itemgetter(2,0,1))

Itemgetter lets you to sort by multiple criteria / columns:

sorted_list = sorted(list_to_sort, key=itemgetter(2,0,1))

回答 3

也可以通过lambda函数实现多个条件

sorted_list = sorted(list_to_sort, key=lambda x: (x[1], x[0]))

multiple criteria can also be implemented through lambda function

sorted_list = sorted(list_to_sort, key=lambda x: (x[1], x[0]))

回答 4

array.sort(key = lambda x:x[1])

您可以使用此代码段轻松进行排序，其中1是元素的索引。

array.sort(key = lambda x:x[1])

You can easily sort using this snippet, where 1 is the index of the element.

回答 5

像这样：

import operator
l = [...]
sorted_list = sorted(l, key=operator.itemgetter(desired_item_index))

Like this:

import operator
l = [...]
sorted_list = sorted(l, key=operator.itemgetter(desired_item_index))

回答 6

我认为lambda函数可以解决您的问题。

old_list = [[0,1,'f'], [4,2,'t'],[9,4,'afsd']]

#let's assume we want to sort lists by last value ( old_list[2] )
new_list = sorted(old_list, key=lambda x: x[2])

#Resulst of new_list will be:

[[9, 4, 'afsd'], [0, 1, 'f'], [4, 2, 't']]

I think lambda function can solve your problem.

old_list = [[0,1,'f'], [4,2,'t'],[9,4,'afsd']]

#let's assume we want to sort lists by last value ( old_list[2] )
new_list = sorted(old_list, key=lambda x: x[2])

#Resulst of new_list will be:

[[9, 4, 'afsd'], [0, 1, 'f'], [4, 2, 't']]

回答 7

**old_list = [[0,1,'f'], [4,2,'t'],[9,4,'afsd']]
    #let's assume we want to sort lists by last value ( old_list[2] )
    new_list = sorted(old_list, key=lambda x: x[2])**

如果我错了，请纠正我，但不是’x [2]’调用列表中的第三项，而不是嵌套列表中的第三项吗？应该是x [2] [2]吗？

**old_list = [[0,1,'f'], [4,2,'t'],[9,4,'afsd']]
    #let's assume we want to sort lists by last value ( old_list[2] )
    new_list = sorted(old_list, key=lambda x: x[2])**

correct me if i’m wrong but isnt the ‘x[2]’ calling the 3rd item in the list, not the 3rd item in the nested list? should it be x[2][2]?

回答 8

更容易理解（Lambda实际在做什么）：

ls2=[[0,1,'f'],[4,2,'t'],[9,4,'afsd']]
def thirdItem(ls):
    #return the third item of the list
    return ls[2]
#Sort according to what the thirdItem function return 
ls2.sort(key=thirdItem)

More easy to understand (What is Lambda actually doing):

ls2=[[0,1,'f'],[4,2,'t'],[9,4,'afsd']]
def thirdItem(ls):
    #return the third item of the list
    return ls[2]
#Sort according to what the thirdItem function return 
ls2.sort(key=thirdItem)

回答 9

排序多维数组在这里执行

arr=[[2,1],[1,2],[3,5],[4,5],[3,1],[5,2],[3,8],[1,9],[1,3]]



arr.sort(key=lambda x:x[0])
la=set([i[0] for i in Points])

for i in la:
    tempres=list()
    for j in arr:
        if j[0]==i:
            tempres.append(j[1])

    for j in sorted(tempres,reverse=True):
        print(i,j)

Sorting a Multidimensional Array execute here

arr=[[2,1],[1,2],[3,5],[4,5],[3,1],[5,2],[3,8],[1,9],[1,3]]



arr.sort(key=lambda x:x[0])
la=set([i[0] for i in Points])

for i in la:
    tempres=list()
    for j in arr:
        if j[0]==i:
            tempres.append(j[1])

    for j in sorted(tempres,reverse=True):
        print(i,j)

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