Find count of repeated letters (sequence)

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Jothi
Jothi on 8 Oct 2013
Answered: Walter Roberson on 1 Dec 2016
Sir,
How to find the no. of repeated sequence (letters) in the given sentence.
for example, a="I want THAAAAAT APPPPPLE ):):): totally unprepared";
The No. of repeated sequences are: 3
ie.,
1. THAAAAAT
2. APPPPPLE
3. ):):):
thanks
  4 Comments
Walter Roberson
Walter Roberson on 9 Oct 2013
Your #3, ):):): does not have continuously repeated symbols.
If the repeated sequences are to be identified, then why would all of THAAAAAT be output, and not just AAAAA ?
Jothi
Jothi on 9 Oct 2013
:) this symbol indicates one type of emotion symbol (positive emotion).
I don't want the output as string just find the no. of repeated sequences are appeared in the given sentence. ie.,
input is,
a="I want THAAAAAT APPPPPLE ):):): totally unprepared";
output is,
No. of repeated sequences are: 3

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Answers (2)

Cedric
Cedric on 8 Oct 2013
Edited: Cedric on 8 Oct 2013
Try to understand the following and fine-tune it to your needs:
n = sum( diff([0, diff(a)==0]) == 1 )
In particular, evaluate
diff(a)==0
and see how your problem actually translates into counting clusters of the outcome of diff(a)==0.
  4 Comments
Cedric
Cedric on 10 Oct 2013
Edited: Cedric on 10 Oct 2013
You seem to indicate that one repeated sequence is '):'. As far as I am concerned, there is no simple generic solution if you want to detect repeated, arbitrary patterns. To illustrate,
'AABBCCDDEEFFAABBCCDDEEFF'
Here, repeated patterns are
'AA', 'BB', .., 'FF', 'AABB', 'BBCC', .., 'AABBCC', 'BBCCDD', ..,
'AABBCCDD', 'BBCCDDEE', .., 'AABBCCDDEEFF'
Using regular expressions, we can probably get some solution but it will be prohibitively time consuming.
Sean de Wolski
Sean de Wolski on 10 Oct 2013
Yeah, every emoticon would have to be predefined. For the chatroom we use here there are even word emoticons like (b) which inserts a frosty beer mug or (ply) which inserts an image of a playing card.

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Walter Roberson
Walter Roberson on 1 Dec 2016
a='I want THAAAAAT APPPPPLE ):):): totally unprepared';
regexp(a, '(.+)\1{2,}', 'match')
ans =
1×3 cell array
'AAAAA' 'PPPPP' '):):):'

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