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I have searched around and tried to find a solution to what seems to be a simple problem, but have come up with nothing. The problem is to sort a matrix based on its columns, progressively. So, if I have a numpy matrix like:

import numpy as np
X=np.matrix([[0,0,1,2],[0,0,1,1],[0,0,0,4],[0,0,0,3],[0,1,2,5]])
print(X)
[[0 0 1 2]
 [0 0 1 1]
 [0 0 0 4]
 [0 0 0 3]
 [0 1 2 5]]

I would like to sort it based on the first column, then the second, the third, and so on, to get a result like:

Xsorted=np.matrix([[0,0,0,3],[0,0,0,4],[0,0,1,1],[0,0,1,2],[0,1,2,5]])
print(Xsorted)
[[0,0,0,3]
 [0,0,0,4]
 [0,0,1,1]
 [0,0,1,2]
 [0,1,2,5]]

While I think it is possible to sort a matrix like this by naming the columns and all that, I would prefer to have a method for sorting that doesn't depend so much on how big the matrix is. I am using Python 3.4, if that is important.

Any help would be greatly appreciated!

Matthew Baker
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2 Answers2

2

It's not going to be particularly fast, but you can always convert your rows to tuples, then use Python's sort:

np.matrix(sorted(map(tuple, X.A)))

You can also use np.lexsort, as suggested in this answer to a somewhat related question:

X[np.lexsort(X.T[::-1])]

The lexsort approach appears to be faster, though you should test with your actual data to make sure:

In [20]: X = np.matrix(np.random.randint(10, size=(100,100)))

In [21]: %timeit np.matrix(sorted(map(tuple, X.A)))
100 loops, best of 3: 2.23 ms per loop

In [22]: %timeit X[np.lexsort(X.T[::-1])]
1000 loops, best of 3: 1.22 ms per loop
Community
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perimosocordiae
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1

Here:

data = [[0,0,1,2],[0,0,1,1],[0,0,0,4],[0,0,0,3],[0,1,2,5]]
x  = pandas.DataFrame(data)
# order of columns to sort
z = x.sort([0,1,2,3])
output = z.as_matrix()

output:

array([[0, 0, 0, 3],
   [0, 0, 0, 4],
   [0, 0, 1, 1],
   [0, 0, 1, 2],
   [0, 1, 2, 5]])
agold
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rohitkulky
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