String count with overlapping occurrences

Question

What\'s the best way to count the number of occurrences of a given string, including overlap in Python? This is one way:

def function(string, str_to_search_f         


        

Answer 1

An alternative very close to the accepted answer but using while as the if test instead of including if inside the loop:

def countSubstr(string, sub):
    count = 0
    while sub in string:
        count += 1
        string = string[string.find(sub) + 1:]
    return count;

This avoids while True: and is a little cleaner in my opinion

Answer 2

You can also try using the new Python regex module, which supports overlapping matches.

import regex as re

def count_overlapping(text, search_for):
    return len(re.findall(search_for, text, overlapped=True))

count_overlapping('1011101111','11')  # 5

Answer 3

Here is my edX MIT "find bob"* solution (*find number of "bob" occurences in a string named s), which basicaly counts overlapping occurrences of a given substing:

s = 'azcbobobegghakl'
count = 0

while 'bob' in s:
    count += 1 
    s = s[(s.find('bob') + 2):]

print "Number of times bob occurs is: {}".format(count)

Answer 4

def count_overlaps (string, look_for):
    start   = 0
    matches = 0

    while True:
        start = string.find (look_for, start)
        if start < 0:
            break

        start   += 1
        matches += 1

    return matches

print count_overlaps ('abrabra', 'abra')