python - Counting same elements in an array and create dictionary -

this question might noob, still not able figure out how properly.

i have given array [0,0,0,0,0,0,1,1,2,1,0,0,0,0,1,0,1,2,1,0,2,3] (arbitrary elements 0-5) , want have counter occurence of zeros in row.

1 times 6 zeros in row 1 times 4 zeros in row 2 times 1 0  in row  => (2,0,0,1,0,1) 

so dictionary consists out of n*0 values index , counter value.

the final array consists of 500+ million values unsorted 1 above.

this should want:

import numpy np  = [0,0,0,0,0,0,1,1,2,1,0,0,0,0,1,0,1,2,1,0,2,3]  # find indexes of zeroes index_zeroes = np.where(np.array(a) == 0)[0]  # find discontinuities in indexes, denoting separated groups of zeroes # note: adding true @ end because otherwise last 0 ignored index_zeroes_disc = np.where(np.hstack((np.diff(index_zeroes) != 1, true)))[0]  # count number of zeroes in each group # note: adding 0 @ start first group of zeroes counted count_zeroes = np.diff(np.hstack((0, index_zeroes_disc + 1)))  # count number of groups same number of zeroes groups_of_n_zeroes = {} count in count_zeroes:     if groups_of_n_zeroes.has_key(count):         groups_of_n_zeroes[count] += 1     else:         groups_of_n_zeroes[count] = 1 

groups_of_n_zeroes holds:

{1: 2, 4: 1, 6: 1}