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

My answer, to the bob question on the course:

s = 'azcbobobegghaklbob'
total = 0
for i in range(len(s)-2):
    if s[i:i+3] == 'bob':
        total += 1
print 'number of times bob occurs is: ', total

Answer 2

That can be solved using regex.

import re
def function(string, sub_string):
    match = re.findall('(?='+sub_string+')',string)
    return len(match)

Answer 3

For a duplicated question i've decided to count it 3 by 3 and comparing the string e.g.

counted = 0

for i in range(len(string)):

    if string[i*3:(i+1)*3] == 'xox':
       counted = counted +1

print counted

Answer 4

If you want to count permutation counts of length 5 (adjust if wanted for different lengths):

def MerCount(s):
  for i in xrange(len(s)-4):
    d[s[i:i+5]] += 1
return d

Answer 5

>>> import re
>>> text = '1011101111'
>>> len(re.findall('(?=11)', text))
5

If you didn't want to load the whole list of matches into memory, which would never be a problem! you could do this if you really wanted:

>>> sum(1 for _ in re.finditer('(?=11)', text))
5

As a function (re.escape makes sure the substring doesn't interfere with the regex):

>>> def occurrences(text, sub):
        return len(re.findall('(?={0})'.format(re.escape(sub)), text))

>>> occurrences(text, '11')
5

Answer 6

How to find a pattern in another string with overlapping

This function (another solution!) receive a pattern and a text. Returns a list with all the substring located in the and their positions.

def occurrences(pattern, text):
    """
    input: search a pattern (regular expression) in a text
    returns: a list of substrings and their positions 
    """
    p = re.compile('(?=({0}))'.format(pattern))
    matches = re.finditer(p, text)
    return [(match.group(1), match.start()) for match in matches]

print (occurrences('ana', 'banana'))
print (occurrences('.ana', 'Banana-fana fo-fana'))

[('ana', 1), ('ana', 3)]
[('Bana', 0), ('nana', 2), ('fana', 7), ('fana', 15)]