How can I find the number of overlapping sequences in a String with Python? [duplicate]

Posted on

Question :

How can I find the number of overlapping sequences in a String with Python? [duplicate]

I have a long sequence, and I would like to know how often some sub-sequences occur in this sequence.

I know string.count(s, sub), but it only counts non-overlapping sequences.

Does a similar function which also counts overlapping sequences exist?

Answer #1:

As an alternative to writing your own search function, you could use the re module:

In [22]: import re

In [23]: haystack = 'abababa baba alibababa'

In [24]: needle = 'baba'

In [25]: matches = re.finditer(r'(?=(%s))' % re.escape(needle), haystack)

In [26]: print [m.start(1) for m in matches]
[1, 3, 8, 16, 18]

The above prints out the starting positions of all (potentially overlapping) matches.

If all you need is the count, the following should do the trick:

In [27]: len(re.findall(r'(?=(%s))' % re.escape(needle), haystack))
Out[27]: 5
Answered By: NPE

Answer #2:

A simple to understand way to do it is:

def count(sub, string):
    count = 0
    for i in xrange(len(string)):
        if string[i:].startswith(sub):
            count += 1
    return count

count('baba', 'abababa baba alibababa')
#output: 5

If you like short snippets, you can make it less readable but smarter:

def count(subs, s):
    return sum((s[i:].startswith(subs) for i in xrange(len(s))))

This uses the fact that Python can treat boolean like integers.

Answered By: e-satis

Answer #3:

This should help you :

matches =[]
st = 'abababa baba alibababa'
needle = 'baba'
for i in xrange(len(st)-len(needle)+1): 
   i = st.find(needle,i,i+len(needle))
   if(i >= 0):

see it here :

Did not benchmark it for long strings, see if its efficient enough for your use.

Answered By: DhruvPathak

Answer #4:

I learnt today that you could use a running index to fetch the next occurrence of your substring:

string = 'bobobobobobobob' # long string or variable here
count = 0
start = 0
while True:
    index = string.find('bob', start)
    if index >= 0:
        count += 1
        start += 1

Returns 7

Answered By: iamryandrake

Leave a Reply

Your email address will not be published. Required fields are marked *