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pythonnumpyscipypermutationshuffle
DotPi
DotPi

Reputation: 4127

shuffle vs permute numpy

What is the difference between numpy.random.shuffle(x) and numpy.random.permutation(x)?

I have read the doc pages but I could not understand if there was any difference between the two when I just want to randomly shuffle the elements of an array.

To be more precise suppose I have an array x=[1,4,2,8].

If I want to generate random permutations of x, then what is the difference between shuffle(x) and permutation(x)?

Upvotes: 106

Views: 75930

Answers (4)

Partha Sarathi Satapathy
Partha Sarathi Satapathy

Reputation: 31

The permutation() method returns a re-arranged array (and leaves the original array un-changed),this method will keep the original array intact and will return a shuffled array, for example x = [1,4,2,8] is the original array and the permutation method will return the rearranged array (lets say [8,4,1,2]).Now,you have two arrays, original array and the rearranged array.

In the other hand,

The shuffle() method makes changes to the original array,for example x = [1,4,2,8] is the original array and the shuffle method will return the shuffled array(lets say shuffled array is [8,4,1,2]). Now , the original array itself got changed to the Shuffled array, and you are only left with the shuffled array.

Reference :-https://www.w3schools.com/python/numpy_random_permutation.asp

Upvotes: 3

Mohanrac
Mohanrac

Reputation: 95

Adding on @ecatmur, Here is a brief explanation. To start with I have created an array which is of shape 3,3 and has numbers from 0 to 8

import numpy as np
x1 = np.array(np.arange(0,9)).reshape(3,3) #array with shape 3,3 and have numbers from 0 to 8

#step1: using np.random.permutation
x_per = np.random.permutation(x1)
print('x_per:', x_per)
print('x_1:', x_1)
#Inference: x1 is not changed and x_per has its rows randomly changed

#The outcome will be 
x1: [[0 1 2]
     [3 4 5]
     [6 7 8]]
x_per:[[3 4 5]
       [0 1 2]
       [6 7 8]]
#Lets apply shuffling
x2 = np.array(range(9)).reshape(3,3)
x2_shuffle = np.random.shuffle(x2)
print('x2_shuffle:', x2_shuffle)
print('x2', x2)

#Outcome: 
x2_shuffle: None
x2 [[3 4 5]
    [0 1 2]
    [6 7 8]]

Key inference is: When x is an array, both numpy.random.permutation(x) and numpy.random.shuffle(x) can permute the elements in x randomly along the first axis. numpy.random.permutation(x) actually returns a new variable and the original data is not changed. Where as numpy.random.shuffle(x) has changed original data and does not return a new variable. I just tried to show with an example so it can help others. Thanks!!

Upvotes: 1

hlin117
hlin117

Reputation: 22268

Adding on to what @ecatmur said, np.random.permutation is useful when you need to shuffle ordered pairs, especially for classification:

from np.random import permutation
from sklearn.datasets import load_iris
iris = load_iris()
X = iris.data
y = iris.target

# Data is currently unshuffled; we should shuffle 
# each X[i] with its corresponding y[i]
perm = permutation(len(X))
X = X[perm]
y = y[perm]

Upvotes: 37

ecatmur
ecatmur

Reputation: 157374

np.random.permutation has two differences from np.random.shuffle:

  • if passed an array, it will return a shuffled copy of the array; np.random.shuffle shuffles the array inplace
  • if passed an integer, it will return a shuffled range i.e. np.random.shuffle(np.arange(n))

If x is an integer, randomly permute np.arange(x). If x is an array, make a copy and shuffle the elements randomly.

The source code might help to understand this:

3280        def permutation(self, object x):
...
3307            if isinstance(x, (int, np.integer)):
3308                arr = np.arange(x)
3309            else:
3310                arr = np.array(x)
3311            self.shuffle(arr)
3312            return arr

Upvotes: 137

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