Python: Count overlapping substring in a string

Question

Say I have string = 'hannahannahskdjhannahannah' and I want to count the number of times the string hannah occurs, I can't simply use count, because that only counts the substring once in each case.

Ie.

I am expecting to return 4 but only returns 2 when I run this in pyCharm with string.count('hannah')

score 9 · Accepted Answer · answered Sep 08 '14 at 10:16

9

You could use a running index to fetch the next occurance:

bla = 'hannahannahskdjhannahannah'
cnt = 0
idx = 0
while True:
    idx = bla.find('hannah', idx)
    if idx >= 0:
        cnt += 1
        idx += 1
    else:
        break
print(cnt)

Gives:

>> 4

answered Sep 08 '14 at 10:16

RickyA

14,690
5
65
93

This is perfect Ricky! Thank you very much for your help. Is there any more specific documentation around running index approach? – Ryan Drake Sep 08 '14 at 10:25
2

Not that I know of. Just common sense. – RickyA Sep 08 '14 at 10:27

score 2 · Answer 2 · answered Sep 08 '14 at 10:19

How about something like this?

>>> d = {}
>>> string = 'hannahannahskdjhannahannah'
>>> for i in xrange(0,len(string)-len('hannah')+1):
...     if string[i:i+len('hannah')] == 'hannah':
...             d['hannah'] = d.get('hannah',0)+1
... 
>>> d
{'hannah': 4}
>>>

This searches the string for hannah by splicing the string iteratively from index 0 all the way up to the length of the string minus the length of hannah

gopi m · Answer 3 · 2015-06-21T06:03:20.200

1

'''
s: main string
sub: sub-string
count: number of sub-strings found
p: use the found sub-string's index in p for finding the next occurrence of next sub-string
'''
count=0
p=0
for letter in s:
    p=s.find(sub,p)   
    if(p!=-1):
        count+=1
        p+=1
print count

edited Jun 21 '15 at 06:03

answered Jun 21 '15 at 05:52

gopi m

11
2

score 0 · Answer 4 · answered Nov 16 '16 at 10:14

If you want to count also nonconsecutive substrings, this is the way to do it

def subword(lookup,whole):
    if len(whole)<len(lookup):
          return 0
    if lookup==whole:
          return 1
    if lookup=='':
          return 1
    if lookup[0]==whole[0]:
         return subword(lookup[1:],whole[1:])+subword(lookup,whole[1:])
    return subword(lookup,whole[1:])

score 0 · Answer 5 · answered May 27 '21 at 20:33

0

def Count_overlap(string, substring):   
    count = 0
    start = 0
 
    while start < len(string):
        pos = string.find(substring, start)
  
        if pos != -1:
            start = pos + 1
            count += 1
        else:
            break
    return count
string = "hannahannahskdjhannahannah"
print(Count_overlap(string, "hannah"))

answered May 27 '21 at 20:33

Vedant Bahel

3
2

Could you provide a more detailed explanation for your code and how this resolves the question being asked? – Skully May 27 '21 at 21:34

score -2 · Answer 6 · answered Sep 08 '14 at 10:14

-2

Don't want to answer this for you as it's simple enough to work out yourself.

But if I were you I'd use the string.find() method which takes the string you're looking for and the position to start looking from, combined with a while loop which uses the result of the find method as it's condition in some way.

That should in theory give you the answer.

answered Sep 08 '14 at 10:14

NDevox

3,916
4
20
33

3

Either give a complete answer, or make a comment. – RickyA Sep 08 '14 at 10:17
Agreed Ricky, not helpful at all. – Ryan Drake Sep 08 '14 at 10:25
2

I'll make a comment next time but still don't think it's a good idea to write someones code for them, otherwise no one would learn anything. That said the accepted answer was literally an implementation of what I said. – NDevox Sep 08 '14 at 10:49
Yeah because Stackoverflow isn't a teaching resource, it's a Q&A. Sometimes a man just needs a fish, and has no desire to get up to their waist in water trying to figure out how a fishing pole works. – J.J May 01 '17 at 13:15

Python: Count overlapping substring in a string

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