Wrong result in str.count() method

Question

I'm confusing with a very simple string count operation:

s = 'BANANA'
s.count('ANA')

This should result in 2, right? Since the substring, ANA appears 2 times in BANANA.

But I've got 1 as a result.

>>> s = 'BANANA'
>>> s.count('ANA')
1

No idea why the wrong result. It is such a simple operation!

Appreciate any help.

PS: How can I solve this problem?

[Python Docs](https://docs.python.org/3/library/stdtypes.html#str.count) say that `.count` only counts non-overlapping occurrences. — Anson Biggs, Apr 21 '20 at 23:27
So, as a result, it's bringing only the first `ANA` because the second `A` is overlapped? — Henrique Branco, Apr 21 '20 at 23:28

Ehsan · Accepted Answer · 2020-04-21T23:47:11.170

5

string.count() does not count the overlapping occurrences.

If you would like to count overlapped occurrences, a simple loop over the string will count it:

s = 'BANANA'
i = 0
cnt = 0
while True:
    i = s.find('ANA', i)
    if i >= 0:
        i += 1
        cnt += 1
    else:
        break

Alternatively you can use regex too as in @Henrique's answer below.

edited Apr 21 '20 at 23:47

answered Apr 21 '20 at 23:27

Ehsan

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1

But only the `new regex` library, as I answered below... – Henrique Branco Apr 21 '20 at 23:44
@HenriqueBranco Yes. I was going to add that. But saw you added it already. Upvote. – Ehsan Apr 21 '20 at 23:45
2

This look ahead regex also works (old and new libs) - `re.findall(r'(?=(ANA))', s)` and shouldn't the `new regex` solution return 2 instead of 3 @HenriqueBranco? – DaveStSomeWhere Apr 21 '20 at 23:49
It works and it returns 2. Probably typo in the answer if you would like to edit @HenriqueBranco. I cannot edit since it is lower than 6 chars. – Ehsan Apr 21 '20 at 23:51
1

I made a mistake and already edited. Check editing in my answer! Thanks all! – Henrique Branco Apr 22 '20 at 00:33

FoggyDay · Answer 2 · 2020-04-21T23:30:18.427

4

Good question. But Python string count() doesn't "backtrack". Once it finds the first "ANA", it looks forward through the remaining two letters" "NA".

This kind of "forward search" is the same as most programming languages, for example Java indexOf(), C strstr() and VB.Net InStr()

edited Apr 21 '20 at 23:30

answered Apr 21 '20 at 23:27

FoggyDay

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Should I use regex instead? – Henrique Branco Apr 21 '20 at 23:29
I think a regex might help: you'd a) define the search pattern (e.g. "ANA"), and then b) count the number of returned "groups". But you'd have to experiment - I'm not sure off the top of my head. Please post back what you find :) – FoggyDay Apr 21 '20 at 23:32

score 4 · Answer 3 · answered Apr 21 '20 at 23:29

4

In 'BANANA', there is only a single complete 'ANA'. Count() is returning 1 because after it finds 'ANA', all that is left is 'NA'.

answered Apr 21 '20 at 23:29

Joseph Julian

71
6

score 2 · Answer 4 · edited Jan 26 '23 at 18:45

2

Solved the problem using the new regex library. It has a new parameter overlapped -- extremely useful.

>>> import regex as re
>>> len(re.findall("ANA", "BANANA", overlapped=True))
2

I found the solution at this question here in SO.

edited Jan 26 '23 at 18:45

wjandrea

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answered Apr 21 '20 at 23:38

Henrique Branco

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You don't really need `regex` for this. `re` can do it with lookaround. See [this answer](/a/48889434/4518341). – wjandrea Jan 26 '23 at 18:46

Wrong result in str.count() method

4 Answers4