Menu
I am trying to find a nice way to sort a 2d list , first by the 1st value , and then by the 2nd value.
I think an example will be the best If I have a list
[[1,4],
[2,7],
[10,1],
[1,2],
[10,6]
[2,1]]
I want that is will be sorted like this
[[1,2],
[1,4],
[2,1],
[2,7],
[10,1],
[10,6]]
861
To sort a two-dimensional list in Python use the sort() list method, which mutates the list, or the sorted() function, which does not. Set the key parameter for both types using a lambda function and return a tuple of the columns to sort according to the required sort order.
Use sorted() with a lambda function to sort a multidimensional list by column. Call sorted(iterable, key=None) with key set to a lambda function of syntax lambda x: x[i] to sort a multidimensional list iterable by the i th element in each inner list x .
Python sort list of tuples by first and second element. To sort the first tuple element we can use an index() method like list[0]. Similar to the second element we can use the list[1] index. By using the sort method we need to sort the lists in place and order.
Use the key argument of the sorted() function to sort a list of tuples by the second element, e.g. sorted_list = sorted(list_of_tuples, key=lambda t: t[1]) . The function will return a new list, sorted by the second tuple element.
l=[[1,4],
[2,7],
[10,1],
[1,2],
[10,6],
[2,1]]
print(sorted(l,key=lambda x: (x[0],x[1]))) # use lambda to sort by "x[0]"-> first element of the sublists or x[1] -> second element, if its a tie
[[1, 2], [1, 4], [2, 1], [2, 7], [10, 1], [10, 6]]
Or simply sorted(l) of l.sort() as your elements sort naturally.
A better example would be to sort by the second value only:
print(sorted(l,key=lambda x: (x[1])))
[[10, 1], [2, 1], [1, 2], [1, 4], [10, 6], [2, 7]]
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
