Efficient way to implement "change to other number" in a Numpy array

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Question

Suppose I now have an array that looks like arr = np.random.randint(1, 4, 100), where the unique numbers are 1, 2, and 3.

Now I would like to model the data corruption, where some of the numbers are likely to become others. For example, if arr[k] is 1, then it is likely to remain the same, but it is also possible to become 2 or 3 (all with equal probabilities).

I could implement this using following code

import numpy as np
arr = np.random.randint(1, 4, 100)

mask = np.random.choice([0, 1], size=100, p=[0.8, 0.2])
for idx in range(100):
    if mask[idx] != 0:
        arr[idx] = np.random.choice([1, 2, 3)

This works fine but I really do not like the loop. Is there some way I could eliminate the (ugly) loop?

Frank Yellin · Accepted Answer · 2020-09-17 16:34:51Z

1

import numpy as np
arr = np.random.randint(1, 4, 100)
arr2 = np.random.randint(1, 4, 100)
mask = np.random.choice([0, 1], size=100, p=[0.8, 0.2]) != 0
arr[mask] = arr2[mask]

Alternatively, you can count how many Trues there are in mask and just make arr2 precisely that many elements long.

answered Sep 17, 2020 at 16:34

Frank Yellin

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1 Comment

Mad Physicist Over a year ago

Your suggestion is the more efficient option

Michael Szczesny · Accepted Answer · 2020-09-17 16:28:34Z

0

I think this works fine

np.where(np.random.choice([1,0],100, p=[0.8,0.2]), arr, np.random.randint(1,4,100))

To compute the final proportion of corruption

y = np.where(np.random.choice([1,0],100, p=[0.8,0.2]), arr, np.random.randint(1,4,100))
(arr != y).mean()

edited Sep 17, 2020 at 16:28

answered Sep 17, 2020 at 16:23

Michael Szczesny

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