How to count the particuler range of elements in an numpy array

Question

I have an array like:

import numpy as np

data=np.array([[0,0,0,1,1,1,0,0,0,0,1,1,1,1,1],
              [0,0,0,1,0,0,0,0,1,0,0,0,0,0,0],
              [0,0,0,0,0,0,1,0,0,0,0,0,0,0,0]])

Requirement:

Needs to count the number of ones in an array, I can count by using this function.

print(np.count_nonzero(data==1))

the output I got:

But, one special requirement is needed, like if consecutive ones are more than 3 times then only count the ones, in that case, the excepted count of number ones more than 3 consecutive are 5

expected output:

Why is the expected output 5 (I see only 4)? Why are you working with a 2D array? What's the full list of "requirements"? — ShlomiF
– ShlomiF, Commented Sep 4, 2022 at 20:56
I just need to count the element 1's if there are more than 3 times(for ex:) repeated consecutively. otherwise, ignore the count — Bhar Jay
– Bhar Jay, Commented Sep 4, 2022 at 20:59
in another case: if element value 1 is not repeated more than 3 times we can consider it as 0 — Bhar Jay
– Bhar Jay, Commented Sep 4, 2022 at 21:00
In pure numpy: length = np.diff(np.flatnonzero(np.diff(data, append=0, prepend=0, axis=1)))[::2]; length[length > 3].sum() — Michael Szczesny
– Michael Szczesny, Commented Sep 4, 2022 at 21:27
@Michael smart one (however I think this might require some explanations to the non aficionados :p) — mozway
– mozway, Commented Sep 4, 2022 at 21:37

mozway · Accepted Answer · 2022-09-04 21:19:13Z

1

You can erode/dilate your data to remove the stretches of less than N consecutive 1s.

from scipy.ndimage import binary_erosion, binary_dilation

N = 3
k = np.ones((1, N+1))

binary_dilation(binary_erosion(data, k), k).sum()

Output: 5

Output on data=np.array([[0,0,0,1,1,1,0,0,0,0,1,1,1,1,1], [0,0,0,1,1,1,1,1,1,1,1,0,0,0,0], [0,0,0,0,0,0,1,0,0,0,0,0,0,0,0]]): 13

answered Sep 4, 2022 at 21:19

mozway

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ShlomiF · Accepted Answer · 2022-09-04 21:04:24Z

1

A simple method would be to use a "running average" of window-size 3 and then compare to 1.
I still don't understand why the OP is using a 2d array, but instead of changing the example I'll just flatten it out:

import numpy as np

data=np.array([[0,0,0,1,1,1,0,0,0,0,1,1,1,1,1],
               [0,0,0,1,0,0,0,0,1,0,0,0,0,0,0],
               [0,0,0,0,0,0,1,0,0,0,0,0,0,0,0]])    

def running_avg(x, N):
    return np.convolve(x, np.ones(N)/N, mode='valid')

print(sum(running_avg(data.flatten(), 3) == 1))
# 4, which is actually the correct answer for the given example data as far as I can tell

answered Sep 4, 2022 at 21:04

ShlomiF

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3 Comments

Bhar Jay Over a year ago

for this example: data=np.array([[0,0,0,1,1,1,0,0,0,0,1,1,1,1,1], [0,0,0,1,1,1,1,1,1,1,1,0,0,0,0], [0,0,0,0,0,0,1,0,0,0,0,0,0,0,0]]) def running_avg(x, N): return np.convolve(x, np.ones(N)/N, mode='valid') print(sum(running_avg(data.flatten(), 3) == 1)) output it gives:10 instead of count :12

mozway Over a year ago

Did you mean 13?

ShlomiF Over a year ago

Looking at your example, I'd want an answer of 10. Or are we counting rows and columns separately? the op was not very clear despite being asked about the 2d nature explicitly...

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