How to sort a list within an array?

Question

I have a array in Python given by

matrix = [[10, -9, 8], [1, 15, 5]]

I want to sort the elements of the matrix[0] list so that matrix[1] is organized according to the ascending order of matrix[0]. The expected result would be:

matrix = [[-9, 8, 10], [15, 5, 1]]

My matrix represents values of X and Y. I want to sort X in ascending order so that Y is sorted according to the positions of X, in order to not lose the pairs formed by (X[i], Y[i]).

How can i do that?

Number of rows and columns of matrix will be same or the shape will change? — shaik moeed, Aug 29 '23 at 20:07
Do you want to modify the existing data structure or create a new one? And does the result need to be a list of lists or can it be a list of tuples? — Kelly Bundy, Aug 29 '23 at 20:10
My matrix represents values of ```X``` and ```Y```. I want to sort ```X``` in ascending order so that Y is sorted according to the positions of ```X```, in order to not lose the pairs formed by ```(X[i], Y[i])```. — Projetos Programação, Aug 29 '23 at 20:15
@ProjetosProgramação Are you using or fine to use `numpy`? — shaik moeed, Aug 29 '23 at 20:20
How about: `[list(x) for x in zip(*sorted(zip(matrix[0], matrix[1]), key=lambda y:y[1], reverse=True))]` just for fun. Obviously a bit complex so wouldn't typically do this in a single line of code, but just providing for interest's sake. — jarmod, Aug 29 '23 at 20:22
@shaikmoeed correct, as the OP requested afaik (see the OP's expected result). I assume the goal is to retain the entire column. — jarmod, Aug 29 '23 at 20:47
Important terminology note, that is a *list* that contains other *lists* — juanpa.arrivillaga, Aug 29 '23 at 20:49
@jarmod In OP, *matrix[1] is organized according to the ascending order of matrix[0]*, I assume, OP wants to sort the first row and arrange values of the second row based on sorted indices of the first row. — shaik moeed, Aug 29 '23 at 20:52
@shaikmoeed thanks, I think you're right and I need to swap. Let me check and will fix comments shortly. — jarmod, Aug 29 '23 at 20:56
Fix above one-liner: `[list(x) for x in zip(*sorted(zip(matrix[0], matrix[1]), key=lambda y:y[0]))]` — jarmod, Aug 29 '23 at 21:01

shaik moeed · Answer 1 · 2023-08-29T20:21:48.757

1

You can do it easily with numpy, it make your job easy, if you have more matrix related operations, check @Andrej Kesely's answer for plain python version

matrix = np.array([[10, -9, 8], [1, 15, 5]])
matrix[:, matrix[0, :].argsort()]

Output:

array([[-9,  8, 10],
       [15,  5,  1]])

edited Aug 29 '23 at 20:21

answered Aug 29 '23 at 20:16

shaik moeed

5,300
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score 0 · Accepted Answer · answered Aug 29 '23 at 20:17

0

Try:

matrix = [[10, -9, 8], [1, 15, 5]]

matrix[0] = sorted(enumerate(matrix[0]), key=lambda k: k[::-1])
matrix[0], matrix[1] = [v for _, v in matrix[0]], [matrix[1][i] for i, _ in matrix[0]]

print(matrix)

Prints:

[[-9, 8, 10], [15, 5, 1]]

answered Aug 29 '23 at 20:17

Andrej Kesely

168,389
15
48
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score -1 · Answer 3 · edited Aug 30 '23 at 07:45

Here's a working example:

import copy

# Here's your original data
matrix = [[10, -9, 8], [1, 15, 5]]

# Sorting will modify the container, so making a shallow copy (rather than a 
# reference) will keep you for mixing up your output
m_0_orig = copy.copy(matrix[0])

# Now, sort the first list
m_0 = matrix[0]
m_0.sort()

# Using a list comprehension, loop over the sorted elements and get the 
# indexes of the elements prior to the sort, now in the sorted order, from
# matrix[1]
m_1 = [matrix[1][m_0_orig.index(x)] for x in m_0]

# store the result
matrix_result = [m_0,m_1]

# show the result
print(matrix)
print(matrix_result)

## [[-9, 8, 10], [1, 15, 5]]
## [[-9, 8, 10], [15, 5, 1]]

How to sort a list within an array?

3 Answers3