Replace values in numpy array equal to list of values

Question

I have a numpy array - image with various values: example image = [1,2,2, 3, 4, 4, 4, 4, 5, 6, 6,7,8,8,8,8] I want to replace only those numbers occurring less than twice - by a specific number, let's say 0. I managed to create the list of those numbers like this:

(unique, counts) = np.unique(image, return_counts=True)
frequencies = np.asarray((unique, counts)).T
freq = frequencies[frequencies[:,1] < 2,0]
print(freq)
array([1, 3, 5, 7], dtype=int64)

How do I replace those numbers by zero?

the outcome should look like: [0,2,2, 0, 4, 4, 4, 4, 0, 6, 6,0,8,8,8,8]

Thanks in advance!

Answer 1

If both image and freq are numpy arrays:

freq = np.array([1, 3, 5, 7])
image = np.array([1 ,2 ,2, 3, 4, 4, 4, 4, 5, 6, 6 ,7 ,8 ,8 ,8 ,8])

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Solution 1

You can then find the indices of image entries appearing in freq , then set them to zero:

image[np.argwhere(np.isin(image, freq)).ravel()] = 0

Based on: Get indices of items in numpy array, where values is in list .

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Solution 2

Use np.in1d :

image = np.where(np.in1d(image,freq),0,image)

More info: Numpy - check if elements of a array belong to another array

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Solution 3

You can also use a list comprehension:

image = [each if each not in freq else 0 for each in image]

Can find more info here: if/else in a list comprehension .

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The last one will result in a list, not a numpy array, but other than that, all of these yield the same result.

Answer 2

You could compare each item to the rest of the array to form a 2D matrix and sum each count. Then assign the items meeting the frequency condition with the desired value:

import numpy as np

img = np.array([1 ,2 ,2, 3, 4, 4, 4, 4, 5, 6, 6 ,7 ,8 ,8 ,8 ,8])

img[np.sum(img==img[:,None],axis=1)<2] = 0

array([0, 2, 2, 0, 4, 4, 4, 4, 0, 6, 6, 0, 8, 8, 8, 8])

Probably not very efficient but it should work.