count occurrence of numbers in a sequence and return corresponding value

[michalmonday]

X = [2,2,2,2,2,3,4,5,6,6,6,6,6,7,8]
Y = [4,4,4,4,4,0,0,0,5,5,5,5,5,0,0]

Aren't 4's in the Y supposed to be 5's because there are 5 occurences of 2's in the X?

so in this case it should return 4 and 6 and I would also need to sum up the occurrences of 4 and 6 if this kind of pattern repeats again throughout the time series. so the final output would be as below
Y Count
4 1
6 1

I fail to grasp what is the origin of 4 and 6 in this final output and how their Count is 1.

If the origin of 4 and 6 in the final output is based on the Y then I just don't get it. In the example you posted the Y consists of 5 occurences of 4, and 5 occurences of 5. No 6's to be seen.

If the origin of 4 and 6 in the final output is based on the X then it is kinda understandable what 6 is taken from but then why 4 is also there instead of 2?

______________________________

You mentioned "consecutively", should the function check whether the repetitions are consecutive?
Or X will never actually contain this kind of sequence:
X = [1,1,1,1,1,1,2,1,1,3] # where 1's are separated by "2" or other numbers

If X will actually contain such sequence then is the output below correct?
Y = [6,6,6,6,6,6,0,0,0,0]

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In case if the X values don't have to be checked for consecutiveness then you could use something like this:

import numpy as np

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

counts = np.array([(X==i).sum() for i in X])

# np.where 2nd and 3rd arguments can be single values or arrays
# the return values sometimes are taken from 2nd and sometimes are taken
# from 3rd array/value, depending on whether the condition from 1st argument
# is met
Y = np.where(counts < 5, 0, counts)
print('Y =\n', Y)

print('np.unique =\n', np.unique(X, return_counts=True))

Output:
Y =
 [5 5 5 5 5 0 0 0 5 5 5 5 5 0 0]
np.unique =
 (array([2, 3, 4, 5, 6, 7, 8]), array([5, 1, 1, 1, 5, 1, 1]))