Find same values in numpy array in a row

April 5, 2022

I’d like to find equal values in array(and their indexes) if they accure in a row more then 2 times.

[0, 3, 0, 1, 0, 1, 2, 1, 2, 2, 2, 2, 1, 3, 4]

so in this example I would find value "2" accured 4 times from position "8". Is there any build in function to do that?

I found the way with collections.Counter

collections.Counter(a)
# Counter({0: 3, 1: 4, 3: 2, 5: 1, 4: 1})

but this is not what I am looking for.
Of course I can write a loop and compare two values and then count them, but may be there is a more elegant solution?

>Solution :

Find consecutive runs and length of runs with condition

import numpy as np

arr = np.array([0, 3, 0, 1, 0, 1, 2, 1, 2, 2, 2, 2, 1, 3, 4])

res = np.ones_like(arr)
np.bitwise_xor(arr[:-1], arr[1:], out=res[1:])  # set equal, consecutive elements to 0
#np.not_equal(arr[:-1], arr[1:], out=res[1:])   # us this instead for np.floats
idxs = np.flatnonzero(res)                      # get indices of non zero elements
values = arr[idxs]
counts = np.diff(idxs, append=len(arr))         # difference between consecutive indices are the length

cond = counts > 2
values[cond], counts[cond], idxs[cond]

Output