Filter löschen
Filter löschen

How can i shorten the duration this code takes? it takes 0.4 seconds to complete if you insert a longer string. I would prefer not using the for loop but it is still welcome

1 Ansicht (letzte 30 Tage)
this code scans a string for letters and ignores repeating letters, for example scans 'xyxzyx' and returns 'xyz' as there are only 3 letters present. the goal is to re-write in a way that takes less time to execute. thanks in advance... here is the code
str='shdydjsgdnsladhdksjkhdjksa'
fixed2=str; %ignore the redundancy
A1=fixed2;
for i=1:numel(fixed2)
k=fixed2(i);
A1=regexprep(A1,'(?<=(??@k).*)(??@k)','')
end

Akzeptierte Antwort

Fangjun Jiang
Fangjun Jiang am 1 Sep. 2016
what about A1=unique(str)?

Weitere Antworten (0)

Kategorien

Mehr zu Characters and Strings finden Sie in Help Center und File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by