Azzera filtri
Azzera filtri

delete elements from a cell array

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elisa ewin
elisa ewin il 21 Giu 2017
Modificato: Andrei Bobrov il 22 Giu 2017
Hi!
I have a cell array b (attached); in each cell of b I have an expression like this: 'Weather booming Chilli Relax https://t.co/pwp00Ndw3d' or expressions with @,#,$. I want to delete from these expressions all the characters like @,#,$ and the links like https://t.co/pwp00Ndw3d.
Example: if I have 'Weather booming @Chilli Relax# https://t.co/pwp00Ndw3d', I will want it becames 'Weather booming Chill Relax'
Can you help me? thanks
  3 Commenti
elisa ewin
elisa ewin il 22 Giu 2017
sorry, now I re-write the question
Jan
Jan il 22 Giu 2017
Modificato: Jan il 22 Giu 2017
Weather booming Chilli Relax https://t.co/pwp00Ndw3d
This looks strange. It reminds me to Google: Britney Spears Instagram account used by hackers.
Perhaps I'm too distrustful, but I've modified the URL slightly to be sure. This does not change the core of the question or the answer. Sorry, these are hard times in the world wide web. Please do not take this personally.

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Andrei Bobrov
Andrei Bobrov il 22 Giu 2017
regexprep(b,'[$#@]|\<https:/+\S*\>','')
  8 Commenti
Andrei Bobrov
Andrei Bobrov il 22 Giu 2017
Modificato: Andrei Bobrov il 22 Giu 2017
Hi Jan! Yes! "Russian rocket". :)
regexprep(b,'\<[^A-Za-z \?\,]|https:/+\S*\>','')

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Più risposte (1)

Jan
Jan il 22 Giu 2017
Modificato: Jan il 22 Giu 2017
S = 'Weather booming Chilli Relax https://t.co/pwp00Ndw3d';
C = strsplit(S, ' ');
C(contains(C, '/')) = []; % Or how you identify a link
for iC = 1:numel(C)
aC = C{iC};
C{iC} = aC(isstrprop(aC, 'alphanum'));
end
Result = sprintf('%s ', C{:});
Result(end) = [];
The command contains was introduced in R2016b. If you have an older version, use:
function Tf = contains(C, Patterm)
Tf = ~cellfun('isempty', strfind(C, Pattern));
end

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