Python, specific count of chars in string

Question

I am trying to count the number of occurrences in a string in python. I would like to take a binary input, say '001101'. Then count the number of 1s, 0s, 11s, 00s etc.

I have tried to implement this by using count, but this will output that there are 3 1s, when i only want it to output 1 1, and 1 11s and for it to not count them individually, unless they are on their own.

I have also tried to implement this with find, but i am having the same problem.

Any help would be appreciated, thanks.

So you only want to count occurrences of individual sequences? Such as the number of single 1's, the number of 11's, the number of 111's, etc..? — pstatix
– pstatix, Commented Nov 14, 2017 at 13:18
"Then count the number of 1s, 0s, 11s, 00s etc" etc as in all eternity? — Ma0
– Ma0, Commented Nov 14, 2017 at 13:20
Hint: when looking for 1s for example, before counting them, pad them with 0s as in 010 plus the edge cases 10 and 01 for the beginnig and end. — Ma0
– Ma0, Commented Nov 14, 2017 at 13:22
yes individual sequences, but the sequence of 111 does not result into 3 1s, rather 1 111. Kounis thanks, didnt think of that. — Tom S
– Tom S, Commented Nov 14, 2017 at 13:25

user2390182 · Accepted Answer · 2017-11-14 13:29:30Z

1

You can do the following, using itertools.groupby and collections.Counter:

from itertools import groupby
from collections import Counter

s = '001101011'
c = Counter(''.join(g) for _, g in groupby(s))

c.get('11')
# 2
c.get('1')
# 1
c.get('111', 0)  # use default value to capture count 0 properly
# 0

This groups the string into substrings consisting only of equal chars and performs the counting on those substrings.

edited Nov 14, 2017 at 13:29

answered Nov 14, 2017 at 13:23

user2390182

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1 Comment

Ma0 Over a year ago

that is simply poetry

timgeb · Accepted Answer · 2017-11-14 13:24:10Z

0

You could tackle this with regular expressions:

>>> import re
>>> s='001101'

Single ones:

>>> sum(1 for _ in re.finditer('(?<!1)1(?!1)', s))
1

Pairs of ones:

>>> sum(1 for _ in re.finditer('(?<!1)11(?!1)', s))
1

and the same approach applies for groups of zeros.

answered Nov 14, 2017 at 13:24

timgeb

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Comments

petny · Accepted Answer · 2017-11-14 15:03:47Z

Generic solution if you dont want to specify which sequences of characters to look for.

def count_unique_chars(string):
    char_count = {}
    char = ''

    for pair in zip(list(string), list(string[1:]) + [None]):

        char += pair[0]

        if pair[0] == pair[1]:
            continue

        else:

            if char in char_count.keys():
                char_count[char] += 1
            else:
                char_count[char] = 1

            char = ''

    return char_count

Outputs a dict with the count of unique chars.

count_unique_chars('001101')

{'0': 1, '00': 1, '1': 1, '11': 1}

or

count_unique_chars('001101011000100111101000')

{'0': 3, '00': 2, '000': 2, '1': 3, '11': 2, '1111': 1}

count_unique_chars('hello world')

{' ': 1, 'd': 1, 'e': 1, 'h': 1, 'l': 1, 'll': 1, 'o': 2, 'r': 1, 'w': 1}

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