Count number of occurrences of a given substring in a string

Question

How can I count the number of times a given substring is present within a string in Python?

For example:

>>> \'foo bar foo\'.numberOfOccurre         


        

Answer 1

s = 'arunununghhjj'
sb = 'nun'
results = 0
sub_len = len(sb)
for i in range(len(s)):
    if s[i:i+sub_len] == sb:
        results += 1
print results

Answer 2

One way is to use re.subn. For example, to count the number of occurrences of 'hello' in any mix of cases you can do:

import re
_, count = re.subn(r'hello', '', astring, flags=re.I)
print('Found', count, 'occurrences of "hello"')

Answer 3

Below logic will work for all string & special characters

def cnt_substr(inp_str, sub_str):
    inp_join_str = ''.join(inp_str.split())
    sub_join_str = ''.join(sub_str.split())

    return inp_join_str.count(sub_join_str)

print(cnt_substr("the sky is   $blue and not greenthe sky is   $blue and not green", "the sky"))

Answer 4

I will keep my accepted answer as the "simple and obvious way to do it" - however, that does not cover overlapping occurrences. Finding out those can be done naively, with multiple checking of the slices - as in: sum("GCAAAAAGH"[i:].startswith("AAA") for i in range(len("GCAAAAAGH")))

(which yields 3) - it can be done by trick use of regular expressions, as can be seen at Python regex find all overlapping matches? - and it can also make for fine code golfing - This is my "hand made" count for overlappingocurrences of patterns in a string which tries not to be extremely naive (at least it does not create new string objects at each interaction):

def find_matches_overlapping(text, pattern):
    lpat = len(pattern) - 1
    matches = []
    text = array("u", text)
    pattern = array("u", pattern)
    indexes = {}
    for i in range(len(text) - lpat):
        if text[i] == pattern[0]:
            indexes[i] = -1
        for index, counter in list(indexes.items()):
            counter += 1
            if text[i] == pattern[counter]:
                if counter == lpat:
                    matches.append(index)
                    del indexes[index]
                else:
                    indexes[index] = counter
            else:
                del indexes[index]
    return matches

def count_matches(text, pattern):
    return len(find_matches_overlapping(text, pattern))

Answer 5

j = 0
    while i < len(string):
        sub_string_out = string[i:len(sub_string)+j]
        if sub_string == sub_string_out:
            count += 1
        i += 1
        j += 1
    return count

Answer 6

To find overlapping occurences of a substring in a string in Python 3, this algorithm will do:

def count_substring(string,sub_string):
    l=len(sub_string)
    count=0
    for i in range(len(string)-len(sub_string)+1):
        if(string[i:i+len(sub_string)] == sub_string ):      
            count+=1
    return count  

I myself checked this algorithm and it worked.