Efficiently permute array in row wise using Numpy

Question

Given a 2D array, I would like to permute this array row-wise.

Currently, I will create a for loop to permute the 2D array row by row as below:

for i in range(npart):
    pr=np.random.permutation(range(m))
    # arr_rand3 is the same as arr, but with each row permuted
    arr_rand3[i,:]=arr[i,pr]

But, I wonder whether there is some setting within Numpy that can perform this in single line (without the for-loop).

The full code is

import numpy as np

arr=np.array([[0,0,0,0,0],[0,4,1,1,1],[0,1,1,2,2],[0,3,2,2,2]])
npart=len(arr[:,0])
m=len(arr[0,:])
# Permuted version of arr
arr_rand3=np.zeros(shape=np.shape(arr),dtype=np.int)
# Nodal association matrix for C
X=np.zeros(shape=(m,m),dtype=np.double)
# Random nodal association matrix for C_rand3
X_rand3=np.zeros(shape=(m,m),dtype=np.double)
  
for i in range(npart):
    pr=np.random.permutation(range(m))
    # arr_rand3 is the same as arr, but with each row permuted
    arr_rand3[i,:]=arr[i,pr]

What numpy version are you using?

Dani Mesejo
– Dani Mesejo

2021-09-29 15:26:09 +00:00
Commented Sep 29, 2021 at 15:26 — Dani Mesejo
– Dani Mesejo, Commented Sep 29, 2021 at 15:26
Thanks for interest in this OP, Im using numpy==1.19.5

rpb
– rpb

2021-09-29 15:27:32 +00:00
Commented Sep 29, 2021 at 15:27 — rpb
– rpb, Commented Sep 29, 2021 at 15:27

Dani Mesejo · Accepted Answer · 2021-09-29 15:33:44Z

3

In Numpy 1.19+ you should be able to do:

import numpy as np

arr = np.array([[0, 0, 0, 0, 0], [0, 4, 1, 1, 1], [0, 1, 1, 2, 2], [0, 3, 2, 2, 2]])

rng = np.random.default_rng()
arr_rand3 = rng.permutation(arr, axis=1)

print(arr_rand3)

Output

[[0 0 0 0 0]
 [4 0 1 1 1]
 [1 0 1 2 2]
 [3 0 2 2 2]]

According to the documentation, the method random.Generator.permutation receives a new parameter axis:

axis int, optional
The axis which x is shuffled along. Default is 0.

edited Sep 29, 2021 at 15:33

answered Sep 29, 2021 at 15:28

Dani Mesejo

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