Sorting arrays in NumPy by column

问题:

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]])

回答1:

@steve\'s is actually the most elegant way of doing it.

For the \"correct\" way see the order keyword argument of numpy.ndarray.sort

However, you\'ll need to view your array as an array with fields (a structured array).

The \"correct\" way is quite ugly if you didn\'t initially define your array with fields...

As a quick example, to sort it and return a copy:

In [1]: import numpy as np

In [2]: a = np.array([[1,2,3],[4,5,6],[0,0,1]])

In [3]: np.sort(a.view(\'i8,i8,i8\'), order=[\'f1\'], axis=0).view(np.int)
Out[3]: 
array([[0, 0, 1],
       [1, 2, 3],
       [4, 5, 6]])

To sort it in-place:

In [6]: a.view(\'i8,i8,i8\').sort(order=[\'f1\'], axis=0) #<-- returns None

In [7]: a
Out[7]: 
array([[0, 0, 1],
       [1, 2, 3],
       [4, 5, 6]])

@Steve\'s really is the most elegant way to do it, as far as I know...

The only advantage to this method is that the \"order\" argument is a list of the fields to order the search by. For example, you can sort by the second column, then the third column, then the first column by supplying order=[\'f1\',\'f2\',\'f0\'].

回答2:

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

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

回答3:

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.

回答4:

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]]]

回答5:

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...

回答6:

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]])

回答7:

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.

回答8:

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]])

回答9:

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]])