从数组中随机翻转m个值

Question

I have an array with length n , I want to randomly choose m elements from it and flip their value. What is the most efficient way?

there are two cases, m=1 case is a special case. It can be discussed separately, and m=/=1 .

My attempt is:

import numpy as np
n = 20
m = 5
#generate an array a
a = np.random.randint(0,2,n)*2-1
#random choose `m` element and flip it.
for i in np.random.randint(0,n,m):
    a[m]=-a[m]

Suppose m are tens and n are hundreds.

Answer 1

To make sure we are not flipping the same element twice or even more times, we can create unique indices in that length range with np.random.choice using its optional replace argument set as False. Then, simply indexing into the input array and flipping in one go should give us the desired output. Thus, we would have an implementation like so -

idx = np.random.choice(n,m,replace=False)
a[idx] = -a[idx]

Faster version : For a faster version of np.random_choice , I would suggest reading up on this post that explores using np.argpartition to simulate identical behavior.

Answer 2

You can make a random permutation of the array indices, take the first m of them and flip their values:

a[np.random.permutation(range(len(a)))[:m]]*=-1

Using permutation validate you don't choose the same index twice.

Answer 3

You need to change the index of the array from m to i to actually change the value. Result:

import numpy as np
n = 20
m = 5
#generate an array a
a = np.random.randint(0,2,n)*2-1
print(a)
#random choose `i` element and flip it.
for i in np.random.randint(0,n,m):
    a[i] = -a[i]

print(a)

My output:

[ 1  1 -1 -1  1 -1 -1  1  1 -1 -1  1 -1  1  1  1  1 -1  1 -1]
[ 1  1 -1 -1 -1 -1  1  1  1 -1 -1  1 -1 -1  1 -1  1 -1 -1 -1]