How to sort a list of lists by a specific index of the inner list?

Question

I have a list of lists. For example,

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

If I wanted to sort the outer list by the string field o

Answer 1

Make sure that you do not have any null or NaN values in the list you want to sort. If there are NaN values, then your sort will be off, impacting the sorting of the non-null values.

Check out Python: sort function breaks in the presence of nan

Answer 2

multiple criteria can also be implemented through lambda function

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

Answer 3

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)

Answer 4

Like this:

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

Answer 5

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)

Answer 6

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