How to ignore punctuation in a user string while scanning for words (textscan())?

조회 수: 3(최근 30일)
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'];
[y, fs] = wavread(thisfid);
sound(y, fs);
fprintf(1,'Failed to process file wave "%s" because: ', thisfid);
  댓글 수: 1
Oleg Komarov
Oleg Komarov 2011년 9월 1일
How do you get the user-input?

댓글을 달려면 로그인하십시오.

채택된 답변

Lucas García
Lucas García 2011년 9월 1일
This removes the specified punctuation in your word:
regexprep(word, '[-!,.?]', '')
  댓글 수: 9
Jan Donyada
Jan Donyada 2011년 9월 1일
ah the array dimensions were slightly different. i modified the for loop so it works. thanks guys! updated code is above if ur interested

댓글을 달려면 로그인하십시오.

추가 답변(2개)

Oleg Komarov
Oleg Komarov 2011년 9월 1일
Use regexp.
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')

Walter Roberson
Walter Roberson 2011년 9월 1일
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')) = ' ';

Community Treasure Hunt

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

Start Hunting!

Translated by