Overlapping Count Of Substring In A String In Python

I want to find all the counts (overlapping and non-overlapping) of a sub-string in a string. I found two answers one of which is using regex which is not my intention and the othe

Solution 1:

defsliding(a, n):
    return (a[i:i+n] for i in xrange(len(a) - n + 1))

defsubstring_count(a, b):
    returnsum(s == b for s in sliding(a, len(b)))

assertlist(sliding('abcde', 3)) == ['abc', 'bcd', 'cde']    
assert substring_count('ababaa', 'aba') == 2

Solution 2:

Does this do the trick?

def count(string, substring):
    n = len(substring)
    cnt = 0for i in range(len(string) - n):
        ifstring[i:i+n] == substring:
            cnt += 1return cnt

print count('ababaa', 'aba') # 2

I don't know if there's a more efficient solution, but this should work.

Solution 3:

count = len(set([string.find('aba',x) for x in range(len(string)) ifstring.find('aba',x) >= 0]))

Solution 4:

Here, using re.finditer() is the best way to achieve what you want.

import re 

defget_substring_count(s, sub_s):
    returnsum(1for m in re.finditer('(?=%s)' % sub_s, s))

get_substring_count('ababaa', 'aba')
# 2 as response

Solution 5:

Here's a function you could use:

defcount(haystack, needle):
    returnlen([x for x in [haystack[i:j+1] for i in xrange(len(haystack)) for j in xrange(i,len(haystack))] if x == needle])

Then:

>>>count("ababaa", "aba")
2