How to ignore punctuation in a user string while scanning for words (textscan())?
2 vues (au cours des 30 derniers jours)
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It works perfectly now. thanks to oleg and lucas for ur help! if ur interested, here's how it looks like in the end:
function pushbutton1_Callback(hObject, eventdata, handles)
words = get(handles.editbox, 'string'); %scans user input string from editbox
wavdirectory = 'C:\Program Files\MATLAB\R2010b\Recordings\';
wordsstring = regexp(words, '\w+', 'match') ; %reads string only, ignores punctuation
[j, k] = size(wordsstring); %stores number of words in user input string
for m = 1:k
thisfid = [wavdirectory wordsstring{m} '.wav'];
try
[y, fs] = wavread(thisfid);
sound(y, fs);
catch
fprintf(1,'Failed to process file wave "%s" because: ', thisfid);
lasterror
end
end
1 commentaire
Oleg Komarov
le 1 Sep 2011
How do you get the user-input? http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer
Réponse acceptée
Lucas García
le 1 Sep 2011
This removes the specified punctuation in your word:
regexprep(word, '[-!,.?]', '')
9 commentaires
Oleg Komarov
le 1 Sep 2011
word = {'this.-is..!a,test', 'it!!.works???-kwan-kwan'};
C = regexp(word, '\w+', 'match')
C{2}
C =
{1x4 cell} {1x4 cell}
ans =
'it' 'works' 'kwan' 'kwan'
Plus de réponses (2)
Oleg Komarov
le 1 Sep 2011
If you want more details provide some example inputs and required output.
From Lucas' example:
word = {'this.-is..!a,test', 'it!!.works???-'};
C = regexp(word, '\w+', 'match')
2 commentaires
Walter Roberson
le 1 Sep 2011
Ah, the poster broke this out in to a separate question, which I did not see before I answered in the original thread. My answer there was:
Before the textscan:
words = lower(words);
words(~ismember(words, ['a':'z' ' '])) = ' ';
then go ahead with the textscan
On second look, that could be shortened to
words = lower(words);
words(~ismember(words, 'a':'z')) = ' ';
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