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

def count_substring(string, sub_string):
    count = 0
    for pos in range(len(string)):
        if string[pos:].startswith(sub_string):
            count += 1
    return count

This could be the easiest way.

Answer 2

def count_substring(string, sub_string):
    counter = 0
    for i in range(len(string)):
        if string[i:].startswith(sub_string):
        counter = counter + 1
    return counter

Above code simply loops throughout the string once and keeps checking if any string is starting with the particular substring that is being counted.

Answer 3

Function that takes as input two strings and counts how many times sub occurs in string, including overlaps. To check whether sub is a substring, I used the in operator.

def count_Occurrences(string, sub):
    count=0
    for i in range(0, len(string)-len(sub)+1):
        if sub in string[i:i+len(sub)]:
            count=count+1
    print 'Number of times sub occurs in string (including overlaps): ', count

Answer 4

A simple way to count substring occurrence is to use count():

>>> s = 'bobob'
>>> s.count('bob')
1

You can use replace () to find overlapping strings if you know which part will be overlap:

>>> s = 'bobob'
>>> s.replace('b', 'bb').count('bob')
2

Note that besides being static, there are other limitations:

>>> s = 'aaa'
>>> count('aa') # there must be two occurrences
1 
>>> s.replace('a', 'aa').count('aa')
3

Answer 5

def occurance_of_pattern(text, pattern):
    text_len , pattern_len = len(text), len(pattern)
    return sum(1 for idx in range(text_len - pattern_len + 1) if text[idx: idx+pattern_len] == pattern) 

Answer 6

Python's str.count counts non-overlapping substrings:

In [3]: "ababa".count("aba")
Out[3]: 1

Here are a few ways to count overlapping sequences, I'm sure there are many more :)

Look-ahead regular expressions

How to find overlapping matches with a regexp?

In [10]: re.findall("a(?=ba)", "ababa")
Out[10]: ['a', 'a']

Generate all substrings

In [11]: data = "ababa"
In [17]: sum(1 for i in range(len(data)) if data.startswith("aba", i))
Out[17]: 2