a = [0,1,2,1,1,1,1]

 b = {1:3, 4:1}

a = [a,a,a,b,b,b,a,a,a] 
b = {3:3}

a = [1,1,1,1,0,1,0,1,1,1,1]
b = {1:3, 4:2}

a = [1,0,0,1,1,1,0,0,0,0,1,1,1,1,1]
b = {1:1, 2:1, 3:1, 4:1, 5:1}

a = [0,1,2,1,1,1,1,2,3,0] # исходный массив
b = {} #результат
for row in a:
    try:
        b[row] += 1
    except KeyError:
        b[row] = 1
        
print(b)

a = [1,1,1,1,0,1,0,1,1,1,1]

b = {0: 2, 1: 9}

b = {1:3, 4:2}

a = [1,1,1,1,0,0,1,0]
#построение частотного словаря
dictionary = []
for x1 in range(len(a)):
    for y in range(1,len(a) - x1 + 1):
        if dictionary.count(a[x1:(x1+y)]) == 0: dictionary.append(a[x1:(x1+y)])
#оставляем только моно резульаты
mono_dictionary = []
for ent in dictionary:
    if ent.count(ent[0]) == len(ent) : mono_dictionary.append(ent)

str('1,1,1,1,0,0,1,0')

#!/usr/bin/python
test_data=[['a','a','a','b','b','b','a','a','a'],
           [1,1,1,1,0,1,0,1,1,1,1],
           [1,0,0,1,1,1,0,0,0,0,1,1,1,1,1]]
          
def get_consecutive_stat(in_list):
    resDict = dict()
    counter = 1
    for i in xrange(len(in_list)-1):
        if in_list[i] == in_list[i+1]:
            counter += 1
        else:
            if counter in resDict:
                resDict[counter] += 1
            else:
                resDict[counter] = 1
            counter = 1
    if counter in resDict:
        resDict[counter] += 1
    else:
        resDict[counter] = 1
    return resDict
        
for test in test_data:
    print(get_consecutive_stat(test))

from itertools import groupby
from collections import Counter
 
a = [1,0,0,1,1,1,0,0,0,0,1,1,1,1,1]
dictionary = {}
 
occurrences_list = [len(list(group)) for key, group in groupby(a)]
for key1, value in Counter(occurrences_list).items():
    dictionary[key1] = value
 
print dictionary

{1: 1, 2: 1, 3: 1, 4: 1, 5: 1}

occurrences_list = [len(list(group)) for key, group in groupby(a)]