replace multiple "1" with only one "1"

first, I have a strig array with all kinds of symbols, for example:

FF1 = ('eE_?@   wwqy   W');

then, I want to first turn FF2 into ASCII: FF2 =abs(FF1) . replace all the symbols with 32 ( the corresponding number for empty space). then I wanto to delete every continuous 32s into one 32. After replacement,

FF3  = ' 101    69    95    32   32    32    32    32   119   119   113   121    32    32    32    87'

the result i got should be

FF3  = ' 101    69    95    32  119   119   113   121    32   87'

this question is similar to " replace multiple space with one space", which could be done by

txt_new = regexprep(text,' +',' '). Instead, I use FF2 = regexprep(FF2, '   32+','   32'). But I failed.so how should I do?I attached the codes

another problem, it is only to use regexprep for str, can any function directly deal with number array?

FF2 = abs('eE_?@   wwqy   W');
FF3 = num2str(FF2);
lol = regexprep(FF3, '\w*32+', '   32');

[Jos (10584)]

Your title is not matching your question ...

[John D'Errico]

Andy - why flag an answer? Just add a comment if all you wanted to do was to thank a responder. Flags serve a different purpose.

Answer 1

[Jos (10584)]

Why not operate on the string array and then convert it to ascii? Like this:

    FF1 = 'eE_?@ wwqy W'
    FF3 = double(regexprep(FF1,'[^a-zA-Z0-9_]+',' '))

[Andy]

Thank you for your time. No, And in order to speed up calculation speed, I will convert all the string to number. then I will replace symbols with space.

Besides, It is not possible to use ur code to solve the problem.

[dpb]

If speed is an issue, regexp is a bottleneck, for sure...I don't believe there's any basis at all for the assumption that character substitution isn't just as fast as conversion first--after all, in Matlab they're simply an array of char() internally, anyway.

What is the problem that requires solving that the means by which one gets from FF1 to FF3 matters?

[Jos (10584)]

Apparently, you have a different problem than the one you posted here. What is it exactly you want to accomplish?

Answer 2

[Azzi Abdelmalek]

FF1 = 'eE_?@   wwqy   W'
FF2=abs(FF1)
FF3=FF2
a=[0 FF2==32 0];
ii1=strfind(a,[0 1])
ii2=strfind(a,[1 0])-1
FF3(cell2mat(arrayfun(@(x,y) x+1:y,ii1,ii2,'un',0)))=[]

Answer 3

[dpb]

I'd go at it differently, for sure...no real need for regexp() altho it'll work

>> ff3=FF1;
>> ff3(~isletter(ff3))=' ';
>> ff3=double(unique(ff3,'stable'))
ff3 =
 101    69    32   119   113   121    87
>>

Looks to me like your solution missed replacing the underscore unless it's to be special-cased?

As for the question asked about dealing with it as numeric I don't see why you'd want to until the end, but unique will do the same thing there as did above.

[Azzi Abdelmalek]

The expected result is

FF3  = ' 101    69    95    32  119   119   113   121    32   87'

[dpb]

Actually, though unique probably isn't the answer as it'll do the reduction globally rather than just for the blanks unless you restrict to the subset of FF3(FF3==' ')

Answer 4

[Azzi Abdelmalek]

Try this

FF1 = 'eE_?@   wwqy   W'
FF3=strjoin(strsplit(FF1))

[Azzi Abdelmalek]

Andy, why have you flagged this answer?

[Jan]

I've removed the flag. There is obviously no reason for a flag.