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

If you want to count all the sub-string (including overlapped) then use this method.

import re
def count_substring(string, sub_string):
    regex = '(?='+sub_string+')'
    # print(regex)
    return len(re.findall(regex,string))

Answer 2

Here's the solution in Python 3 and case insensitive:

s = 'foo bar foo'.upper()
sb = 'foo'.upper()
results = 0
sub_len = len(sb)
for i in range(len(s)):
    if s[i:i+sub_len] == sb:
        results += 1
print(results)

Answer 3

The current best answer involving method count doesn't really count for overlapping occurrences and doesn't care about empty sub-strings as well. For example:

>>> a = 'caatatab'
>>> b = 'ata'
>>> print(a.count(b)) #overlapping
1
>>>print(a.count('')) #empty string
9

The first answer should be 2 not 1, if we consider the overlapping substrings. As for the second answer it's better if an empty sub-string returns 0 as the asnwer.

The following code takes care of these things.

def num_of_patterns(astr,pattern):
    astr, pattern = astr.strip(), pattern.strip()
    if pattern == '': return 0

    ind, count, start_flag = 0,0,0
    while True:
        try:
            if start_flag == 0:
                ind = astr.index(pattern)
                start_flag = 1
            else:
                ind += 1 + astr[ind+1:].index(pattern)
            count += 1
        except:
            break
    return count

Now when we run it:

>>>num_of_patterns('caatatab', 'ata') #overlapping
2
>>>num_of_patterns('caatatab', '') #empty string
0
>>>num_of_patterns('abcdabcva','ab') #normal
2

Answer 4

Overlapping occurences:

def olpcount(string,pattern,case_sensitive=True):
    if case_sensitive != True:
        string  = string.lower()
        pattern = pattern.lower()
    l = len(pattern)
    ct = 0
    for c in range(0,len(string)):
        if string[c:c+l] == pattern:
            ct += 1
    return ct

test = 'my maaather lies over the oceaaan'
print test
print olpcount(test,'a')
print olpcount(test,'aa')
print olpcount(test,'aaa')

Results:

my maaather lies over the oceaaan
6
4
2

Answer 5

Depending what you really mean, I propose the following solutions:

1. You mean a list of space separated sub-strings and want to know what is the sub-string position number among all sub-strings:

   s = 'sub1 sub2 sub3'
   s.split().index('sub2')
   >>> 1
   

2. You mean the char-position of the sub-string in the string:

   s.find('sub2')
   >>> 5
   

3. You mean the (non-overlapping) counts of appearance of a su-bstring:

   s.count('sub2')
   >>> 1
   s.count('sub')
   >>> 3
   

Answer 6

           def count_substring(string, sub_string):
               k=len(string)
               m=len(sub_string)
               i=0
               l=0
               count=0
               while l<k:
                   if string[l:l+m]==sub_string:
                        count=count+1
                   l=l+1     
               return count

           if __name__ == '__main__':
                string = input().strip()
                sub_string = input().strip()

                count = count_substring(string, sub_string)
                print(count)