[python] Sorting arrays in NumPy by column

How can I sort an array in NumPy by the nth column?

For example,

a = array([[9, 2, 3],
           [4, 5, 6],
           [7, 0, 5]])

I'd like to sort rows by the second column, such that I get back:

array([[7, 0, 5],
       [9, 2, 3],
       [4, 5, 6]])

def sort_np_array(x, column=None, flip=False):
    x = x[np.argsort(x[:, column])]
    if flip:
        x = np.flip(x, axis=0)
    return x

Array in the original question:

a = np.array([[9, 2, 3],
              [4, 5, 6],
              [7, 0, 5]])

The result of the sort_np_array function as expected by the author of the question:

sort_np_array(a, column=1, flip=False)

[2]: array([[7, 0, 5],
            [9, 2, 3],
            [4, 5, 6]])

You can sort on multiple columns as per Steve Tjoa's method by using a stable sort like mergesort and sorting the indices from the least significant to the most significant columns:

a = a[a[:,2].argsort()] # First sort doesn't need to be stable.
a = a[a[:,1].argsort(kind='mergesort')]
a = a[a[:,0].argsort(kind='mergesort')]

This sorts by column 0, then 1, then 2.

Simply using sort, use coloumn number based on which you want to sort.

a = np.array([1,1], [1,-1], [-1,1], [-1,-1]])
print (a)
a=a.tolist() 
a = np.array(sorted(a, key=lambda a_entry: a_entry[0]))
print (a)

import numpy as np
a=np.array([[21,20,19,18,17],[16,15,14,13,12],[11,10,9,8,7],[6,5,4,3,2]])
y=np.argsort(a[:,2],kind='mergesort')# a[:,2]=[19,14,9,4]
a=a[y]
print(a)

Desired output is [[6,5,4,3,2],[11,10,9,8,7],[16,15,14,13,12],[21,20,19,18,17]]

note that argsort(numArray) returns the indices of an numArray as it was supposed to be arranged in a sorted manner.

example

x=np.array([8,1,5]) 
z=np.argsort(x) #[1,3,0] are the **indices of the predicted sorted array**
print(x[z]) #boolean indexing which sorts the array on basis of indices saved in z

answer would be [1,5,8]

#for sorting along column 1

indexofsort=np.argsort(dataset[:,0],axis=-1,kind='stable') 
dataset   = dataset[indexofsort,:]

From the NumPy mailing list, here's another solution:

>>> a
array([[1, 2],
       [0, 0],
       [1, 0],
       [0, 2],
       [2, 1],
       [1, 0],
       [1, 0],
       [0, 0],
       [1, 0],
      [2, 2]])
>>> a[np.lexsort(np.fliplr(a).T)]
array([[0, 0],
       [0, 0],
       [0, 2],
       [1, 0],
       [1, 0],
       [1, 0],
       [1, 0],
       [1, 2],
       [2, 1],
       [2, 2]])

Here is another solution considering all columns (more compact way of J.J's answer);

ar=np.array([[0, 0, 0, 1],
             [1, 0, 1, 0],
             [0, 1, 0, 0],
             [1, 0, 0, 1],
             [0, 0, 1, 0],
             [1, 1, 0, 0]])

Sort with lexsort,

ar[np.lexsort(([ar[:, i] for i in range(ar.shape[1]-1, -1, -1)]))]

Output:

array([[0, 0, 0, 1],
       [0, 0, 1, 0],
       [0, 1, 0, 0],
       [1, 0, 0, 1],
       [1, 0, 1, 0],
       [1, 1, 0, 0]])

A little more complicated lexsort example - descending on the 1st column, secondarily ascending on the 2nd. The tricks with lexsort are that it sorts on rows (hence the .T), and gives priority to the last.

In [120]: b=np.array([[1,2,1],[3,1,2],[1,1,3],[2,3,4],[3,2,5],[2,1,6]])
In [121]: b
Out[121]: 
array([[1, 2, 1],
       [3, 1, 2],
       [1, 1, 3],
       [2, 3, 4],
       [3, 2, 5],
       [2, 1, 6]])
In [122]: b[np.lexsort(([1,-1]*b[:,[1,0]]).T)]
Out[122]: 
array([[3, 1, 2],
       [3, 2, 5],
       [2, 1, 6],
       [2, 3, 4],
       [1, 1, 3],
       [1, 2, 1]])

I suppose this works: a[a[:,1].argsort()]

This indicates the second column of a and sort it based on it accordingly.

It is an old question but if you need to generalize this to a higher than 2 dimension arrays, here is the solution than can be easily generalized:

np.einsum('ij->ij', a[a[:,1].argsort(),:])

This is an overkill for two dimensions and a[a[:,1].argsort()] would be enough per @steve's answer, however that answer cannot be generalized to higher dimensions. You can find an example of 3D array in this question.

Output:

[[7 0 5]
 [9 2 3]
 [4 5 6]]

From the Python documentation wiki, I think you can do:

a = ([[1, 2, 3], [4, 5, 6], [0, 0, 1]]); 
a = sorted(a, key=lambda a_entry: a_entry[1]) 
print a

The output is:

[[[0, 0, 1], [1, 2, 3], [4, 5, 6]]]

In case someone wants to make use of sorting at a critical part of their programs here's a performance comparison for the different proposals:

import numpy as np
table = np.random.rand(5000, 10)

%timeit table.view('f8,f8,f8,f8,f8,f8,f8,f8,f8,f8').sort(order=['f9'], axis=0)
1000 loops, best of 3: 1.88 ms per loop

%timeit table[table[:,9].argsort()]
10000 loops, best of 3: 180 µs per loop

import pandas as pd
df = pd.DataFrame(table)
%timeit df.sort_values(9, ascending=True)
1000 loops, best of 3: 400 µs per loop

So, it looks like indexing with argsort is the quickest method so far...

I had a similar problem.

My Problem:

I want to calculate an SVD and need to sort my eigenvalues in descending order. But I want to keep the mapping between eigenvalues and eigenvectors. My eigenvalues were in the first row and the corresponding eigenvector below it in the same column.

So I want to sort a two-dimensional array column-wise by the first row in descending order.

My Solution

a = a[::, a[0,].argsort()[::-1]]

So how does this work?

a[0,] is just the first row I want to sort by.

Now I use argsort to get the order of indices.

I use [::-1] because I need descending order.

Lastly I use a[::, ...] to get a view with the columns in the right order.