Find char consecutive in a string

Hi! I have this string

stringa={stringa1; stringa2; stringa3; stringa4};
stringa1='{a(abc)(ac)d(cf)}';
stringa2='{(ad)c(bc)(ae)}';
stringa3='{(ef)(ab)(df)cb}';
stringa4='{eg(af)cbc}';

and I want to find per every string (stringa1, stringa2 ..) the number of 'ab' 'ba'. a and b not must be consecutive. I try strfind but it's not give me the right answer. Can you help me? Thanks

[Guillaume]

I assume you mean a and b must not be consecutive?

Can you show what the answer should be for each of your example string? Can a 'a' be part of several 'ab' string or is it restricted to the closest 'b'.

If you ignore all the other characters is 'a***a***b***b' 1, 2 or 4 occurences?

[pamela sulis]

yes, a e b must not b consecutive.

    For example 'ab'
    stringa1='{a(abc)(ac)d(cf)}'; --> I find 1 'ab'
    stringa2='{(ad)c(bc)(ae)}'; --> I find 1 'ab'
    stringa3='{(ef)(ab)(df)cb}';  --> I find 1 'ab'
    stringa4='{eg(af)cbc}'; ---> I find 1 'ab'

    For example 'ba'
    stringa1='{a(abc)(ac)d(cf)}'; --> I find 1 'ba'
    stringa2='{(ad)c(bc)(ae)}'; --> I find 1 'ba'
    stringa3='{(ef)(ab)(df)cb}';  --> I find 0 'ba'
    stringa4='{eg(af)cbc}'; ---> I find 0 'ba'

I want that the code makes the same... i try strfind but don't give me the answer that i want.

About your question: 'If you ignore all the other characters is 'a***a***b***b' 1, 2 or 4 occurences?' it is one occurences: i want know only if there are 'ab' or 'ba' not the number of time they are in the string.

Answer 1

[Guillaume]

Thanks for clarifying. You still haven't my answered my question about whether or not a character can be part of several matches or if the matches can intersect (see my 'a***a***b***b' example).

Assuming the answer is no to both questions, the simplest way is to use a regular expression:

stringa1='{a(abc)(ac)d(cf)}';
stringa2='{(ad)c(bc)(ae)}';
stringa3='{(ef)(ab)(df)cb}';
stringa4='{eg(af)cbc}';
stringa={stringa1; stringa2; stringa3; stringa4};
matchstarts = regexp(stringa, 'a.+?b'); %for 'ab'
matchcounts = cellfun(@numel, matchstarts)

The regular expression above says match 'a', followed by as little as necessary (the ?) but at least one (the +) characters, followed by a 'b'.

For 'ba', the regular expression would then be 'b.+?a'

Note that if the answer is yes to either question, regular expressions won't work.

[pamela sulis]

Thanks, now it's right! An other question: I have read the section about 'regular expression' but i don't understand how to find '(ab)'. If I have brackets, how can I modify the expression 'a.+?b'?

[Guillaume]

Brackets are special characters in regular expression so if you want to actually find a bracket character you need to escape it with '\'

Possibly, the expression you want is: '\(a.+?b\)'. This finds an opening bracket, followed immediately by an 'a' followed by at least one character and as few as possible, followed by a b, immediately followed by a closing bracket.

[pamela sulis]

I try '\(a.+?b\)' but it doesn't give me the right answer... I try also '(\a.+?b)\' because i have read an example in the section about regular expression and it doesn't give me right answer. For the previous strings, I want that it find 2 (ab): one in the first string and another in the second but both give me 0.

[Guillaume]

'(\a.+?b)\' is a meaningless regular expression.

I'm not clear on what you're trying to match anymore. Can you describe it clearly, in words, the same way I've explained my regular expression.

As I've said '\(a.+?b\)' will match an opening bracket, followed immediately by an 'a' followed by at least one character and as few as possible, followed by a 'b', immediately followed by a closing bracket. It will not match anything else.

[pamela sulis]

Thanks! I have solved!