Replace values in numpy array equal to list of values-CodePudding

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!

CodePudding user response：

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])

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.

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

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.

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

CodePudding user response：

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.