How to count number-sequences - random

Question

I want to do the following: I want to enter a "random" squence of 1 and 0.
Then I want to analyze this squence for sequences of 1's, so that I get the length and number of those squences and that I get in the end something like: there are x chains of n 1's in the squence for all lengths n that appear. For example:
11011100011
-> there are 2 chains of n=2 ('11')
and there is 1 chain of n=3 ('111').
Can I realize that with the count function? And if so, how? And if not, how else?
I hope my english is good enough for you to get my question : )

Answer 1

Sorry, I´m using Python.
I´ve got this so far, and it´s working:
from collections import Counter
a = [1,0,0,1,1,0,1,0,0,1,1,1,0,0,1,0,0,1,0]
x=0
h=0
n_list=[]
for x in range (0,len(a)):
if (a[x]==1):
h=h+1
else:
if (h>0):
n_list.append(h)
h=0
x=x+1
print (n_list)
counts = Counter (n_list)
print (counts)
But now I would like to be able to write this:
10101010101010101010100 as a string and convert to a list like above.
(I want to show some people, that even if you are just "randomly" hammer on the keys 1 and 0 you will never create a "real" random number, and because of that it would be bad if they had to typ a comma between every two digits)
Another Problem: there hast to be a zero at the end of the List, how can I change that?

Answer 2

Here's short code that requires long comments.
import collections
import itertools
def count(seq):
# count the number of items in a sequence.
return sum(1 for _ in seq)
def count_things(input_string, what):
grouped = itertools.groupby(input_string)
counts = [count(subseq) for (key, subseq) in grouped if key == what]
return collections.Counter(counts)
Now try it:
>>> one_counts = count_things('11100111101110110011', '1')
>>> print(one_counts)
Counter({2: 2, 3: 2, 4: 1})
How many sequences of length 3?
>>> print one_counts[3]
2
We of course can count anything with it if we need:
>>> print(count_things('aaabbccaaabaa', 'a'))
Counter({3: 2, 2: 1})
How does all this work?
The strangest thing is groupby. It takes a sequence and cuts it into a sequence of pairs. Each pair is (value, sequence of the same values). The sequences are generators, not lists, but for simplicity I show them here as lists:
itertools.groupby('110011100011') ->
[('1', ['1', '1']), ('0', ['0', '0']), ('1', ['1', '1', '1']), ('0', ['0', '0', '0']), ('1', ['1', '1'])]
The count function is needed to count the length of a sequence. For generators, len does not work. We could just go with len(list(...)), but it creates an intermediate list that we throw away after counting its items. Insted I opted to generate a sequence of 1s, one for each item in the source sequence: (1 for _ in seq) Here _ is a valid identifier; it usually stands for "a value I don't care about".
Then, using a list comprehension, we've turned our list of pairs above into a list of lengths, only for sequences that contain '1's (or actually what we'd like).
Now Counter takes a sequence of values and counts the numbers of occurrences of each. This is what we finally need. Its interface is like dict's, where keys are the things we count, and values are the counts of the things.
See, no loops, no state machines, standard library does most things. Batteries included, as they say.