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

This is another example of using str.find() but a lot of the answers make it more complicated than necessary:

def occurrences(text, sub):
    c, n = 0, text.find(sub)
    while n != -1:
        c += 1
        n = text.find(sub, n+1)
    return c

In []:
occurrences('1011101111', '11')

Out[]:
5

Answer 2

Given

sequence = '1011101111'
sub = "11"

Code

In this particular case:

sum(x == tuple(sub) for x in zip(sequence, sequence[1:]))
# 5

More generally, this

windows = zip(*([sequence[i:] for i, _ in enumerate(sequence)][:len(sub)]))
sum(x == tuple(sub) for x in windows)
# 5

or extend to generators:

import itertools as it


iter_ = (sequence[i:] for i, _ in enumerate(sequence))
windows = zip(*(it.islice(iter_, None, len(sub))))
sum(x == tuple(sub) for x in windows)

---

Alternative

You can use more_itertools.locate:

import more_itertools as mit


len(list(mit.locate(sequence, pred=lambda *args: args == tuple(sub), window_size=len(sub))))
# 5

Answer 3

s = "bobobob"
sub = "bob"
ln = len(sub)
print(sum(sub == s[i:i+ln] for i in xrange(len(s)-(ln-1))))

Answer 4

A fairly pythonic way would be to use list comprehension here, although it probably wouldn't be the most efficient.

sequence = 'abaaadcaaaa'
substr = 'aa'

counts = sum([
    sequence.startswith(substr, i) for i in range(len(sequence))
])
print(counts)  # 5

The list would be [False, False, True, False, False, False, True, True, False, False] as it checks all indexes through the string, and because int(True) == 1, sum gives us the total number of matches.

Answer 5

If strings are large, you want to use Rabin-Karp, in summary:

• a rolling window of substring size, moving over a string
• a hash with O(1) overhead for adding and removing (i.e. move by 1 char)
• implemented in C or relying on pypy

Answer 6

Well, this might be faster since it does the comparing in C:

def occurrences(string, sub):
    count = start = 0
    while True:
        start = string.find(sub, start) + 1
        if start > 0:
            count+=1
        else:
            return count