Swapping columns in a numpy array?

Question

from numpy import *
def swap_columns(my_array, col1, col2):
    temp = my_array[:,col1]
    my_array[:,col1] = my_array[:,col2]
    my_array[:,col2] = temp

Then

swap_columns(data, 0, 1)

Doesn't work. However, calling the code directly

temp = my_array[:,0]
my_array[:,0] = my_array[:,1]
my_array[:,1] = temp

Does. Why is this happening and how can I fix it? The Error says "IndexError: 0-d arrays can only use a single () or a list of newaxes (and a single ...) as an index", which implies the arguments aren't ints? I already tried converting the cols to int but that didn't solve it.

Sven Marnach · Accepted Answer · 2022-10-07 11:27:33Z

133

There are two issues here. The first is that the data you pass to your function apparently isn't a two-dimensional NumPy array -- at least this is what the error message says.

The second issue is that the code does not do what you expect:

my_array = numpy.arange(9).reshape(3, 3)
# array([[0, 1, 2],
#        [3, 4, 5],
#        [6, 7, 8]])
temp = my_array[:, 0]
my_array[:, 0] = my_array[:, 1]
my_array[:, 1] = temp
# array([[1, 1, 2],
#        [4, 4, 5],
#        [7, 7, 8]])

The problem is that Numpy basic slicing does not create copies of the actual data, but rather a view to the same data. To make this work, you either have to copy explicitly

temp = numpy.copy(my_array[:, 0])
my_array[:, 0] = my_array[:, 1]
my_array[:, 1] = temp

or use advanced slicing

my_array[:, [0, 1]] = my_array[:, [1, 0]]

edited Oct 7, 2022 at 11:27

answered Feb 1, 2011 at 1:21

Sven Marnach

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5 Comments

Mateen Ulhaq Over a year ago

Advanced slicing does not appear to copy the data either.

Sven Marnach Over a year ago

@MateenUlhaq Be assured, it does. See the linked documentation, or try it yourself.

AI0867 Over a year ago

It certainly copies the data, but that means that the assignment assigns to a temporary object that is immediately discarded.

Sven Marnach Over a year ago

@AI0867 Only the advanced slicing on the right-hand side creates a copy. The advanced slicing in assignment-target position does not. You can try the code – it works just fine. (The reason for the different behaviour is that the [] operator in value context is handled by __getitem__(), while assignments with [] are handled by __setitem__(). It never makes sense to assign to a temporary copy, so the __setitem__() implementation for a NumPy array doesn't.)

AI0867 Over a year ago

I stand corrected.

blaz · Accepted Answer · 2015-10-27 07:41:25Z

47

I find the following the fastest:

my_array[:, 0], my_array[:, 1] = my_array[:, 1], my_array[:, 0].copy()

Time analysis of:

import numpy as np
my_array = np.arange(900).reshape(30, 30)

is as follows:

%timeit my_array[:, 0], my_array[:, 1] = my_array[:, 1], my_array[:, 0].copy()
The slowest run took 15.05 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 1.72 µs per loop

The advanced slicing times are:

%timeit my_array[:,[0, 1]] = my_array[:,[1, 0]]
The slowest run took 7.38 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 3: 6.9 µs per loop

answered Oct 27, 2015 at 7:41

blaz

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1 Comment

magu_ Over a year ago

Upvote for the copy() part, I was missing this in my code.

Jensun Ravichandran · Accepted Answer · 2022-09-20 22:41:27Z

17

An elegant way to swap the columns in NumPy is analogous to swapping two variables in Python like so: x, y = y, x.

i, j = 0, 1
A.T[[i, j]] = A.T[[j, i]]  # swap the columns i and j

Suppose you have a numpy array A like this:

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

A.T[[0, 1]] = A.T[[1, 0]] will swap the first two columns:

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

edited Sep 20, 2022 at 22:41

answered May 17, 2018 at 0:03

Jensun Ravichandran

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Comments

Renaud · Accepted Answer · 2014-11-27 17:26:23Z

15

Building up on @Sven's answer:

import numpy as np
my_array = np.arange(9).reshape(3, 3)
print my_array

[[0 1 2]
 [3 4 5]
 [6 7 8]]

def swap_cols(arr, frm, to):
    arr[:,[frm, to]] = arr[:,[to, frm]]

swap_cols(my_array, 0, 1)
print my_array

[[1 0 2]
 [4 3 5]
 [7 6 8]]

def swap_rows(arr, frm, to):
    arr[[frm, to],:] = arr[[to, frm],:]

my_array = np.arange(9).reshape(3, 3)
swap_rows(my_array, 0, 2)
print my_array

[[6 7 8]
 [3 4 5]
 [0 1 2]]

answered Nov 27, 2014 at 17:26

Renaud

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Comments

Andrew Holmgren · Accepted Answer · 2021-11-10 19:36:03Z

1

If you want to simultaneously swap columns and assign to a new variable, the most clear and concise way I could figure out how to do it was this:

import numpy as np
test_arr = np.arange(12).reshape(4,3)
swapped = np.concatenate((test_arr[:, [1,0]], test_arr[:, 2:]), axis=1)

Note, if the further columns don't exist it will concatenate an on an empty array, meaning it will just swap the first two columns.

answered Nov 10, 2021 at 19:36

Andrew Holmgren

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