replace multiple "1" with only one "1"
显示 更早的评论
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');
2 个评论
Jos (10584)
2014-4-20
Your title is not matching your question ...
John D'Errico
2014-4-21
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.
回答（4 个）
Jos (10584)
2014-4-20
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_]+',' '))
3 个评论
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)
2014-4-21
Apparently, you have a different problem than the one you posted here. What is it exactly you want to accomplish?
Azzi Abdelmalek
2014-4-20
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)))=[]
dpb
2014-4-20
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.
2 个评论
Azzi Abdelmalek
2014-4-20
The expected result is
FF3 = ' 101 69 95 32 119 119 113 121 32 87'
dpb
2014-4-20
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==' ')
Azzi Abdelmalek
2014-4-20
Try this
FF1 = 'eE_?@ wwqy W'
FF3=strjoin(strsplit(FF1))
2 个评论
Azzi Abdelmalek
2014-4-21
Andy, why have you flagged this answer?
Jan
2014-4-21
I've removed the flag. There is obviously no reason for a flag.
类别
在 帮助中心 和 File Exchange 中查找有关 Language Support 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
