How do you sort a list of lists by the second element using Python?
To sort a list of lists by the second element in each list use the
key parameter to either the
.sort() list function (if modifying the original list) or the
sorted() function (if returning a new list).
Here’s an example demonstrating how you can do this with code:
>>> list_2d = [['2nd', 2], ['3rd', 3], ['1st', 1]] >>> list_2d.sort(key=lambda x: x) >>> print(list_2d) [['1st', 1], ['2nd', 2], ['3rd', 3]]
As you can see from the above code the
key parameter in the
.sort() list function uses a
lambda expression which takes each list and extracts the second element. This is used as the key to determine how each list will be sorted.
This approach mutates the original list.
If you need to return a new list then use the
sorted() built-in function, like so:
>>> list_2d = [['6th', 6], ['4th', 4], ['5th', 5]] >>> sorted(list_2d, key=lambda x: x) [['4th', 4], ['5th', 5], ['6th', 6]] >>> print(list_2d) [['6th', 6], ['4th', 4], ['5th', 5]]
The result from the
sorted function produces a new 2d list, as can be seen when printing the original
list_2d – this has not changed.
Sort List By Value
With this same technique, you should be able to sort a list if lists by any index within the lists. Whether it be the second or third or… nth index within the list, the same process occurs: use the
key parameter referencing a
lambda function that extracts the element you want to compare across all lists.
To sort a list of lists by the second element use the
key property in either the
.sort() list method or the
sorted() function which is set to
lambda x: x where
1 in the
lambda function represents the nth index you would want to sort the lists by.