1

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!

Improve this question

2 Answers 2

Reset to default
0

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.

Improve this answer
Sign up to request clarification or add additional context in comments.

5 Comments

Thanks a lot for the quick response works like a charm with the example, but with my image I get: "\ipykernel_launcher.py:4: DeprecationWarning: elementwise comparison failed; this will raise an error in the future." I checked the image and the values stayed as in the original
Rewritten my answer, removing the confusing thing about res.
Thanks for the explanation! I tried all of them. But none of them works as I expected. I guess it worked for the first row of the image though. The image is about 500 rows by 500 cols, sorry for not clarifying this sooner
Oh, your image array is multidimensional? Sadly this means that these solutions aren't going to work.
I recommend asking a new question emphasizing that arrays are multidimensional, with arrays similar to your real arrays (ie they should have the same number of dimensions) - other people are probably going to assume what I did unless they are reading these comments.
0

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.

Improve this answer

3 Comments

Thanks for the reply! Very efficient, but if I wanted to set the number to "0" or "nan", would that be possible?
Changed it to a masked assignment, you can use any value you want in that form.
This worked like a charm, even in 2D array, which I did not mention, Thanks a lot!

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.