extractBefore with matches string

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eko supriyadi
eko supriyadi 2022-6-4
评论： Voss 2022-6-6
Ran in:
Dear all,
i'm stuck use extractBefore with this situation:
a={'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625'}
a = 1×1 cell array
{'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625'}
matches333= extractBefore(a,'333')
matches333 = 1×1 cell array
{'32563 '}
This is because 33308 contains 333.. My goal is to retrieve the strings before pure 333 only. The result what i want is:
matches333 =
1×1 cell array
{'32563 33308 10314 20227 30113 40117 52008 81531 '}
Any idea to solve it! Tks

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采纳的回答

Voss
Voss 2022-6-4
Ran in:
In this case, you can include spaces around '333' to match the pure one:
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625'}
a = 1×1 cell array
{'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625'}
matches333 = extractBefore(a,' 333 ')
matches333 = 1×1 cell array
{'32563 33308 10314 20227 30113 40117 52008 81531'}
But that won't work if '333' is at the end:
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333'}
a = 1×1 cell array
{'32563 33308 10314 20227 30113 40117 52008 81531 333'}
matches333 = extractBefore(a,' 333 ')
matches333 = 1×1 cell array
{0×0 char}
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eko supriyadi
eko supriyadi 2022-6-6
Ran in:
Hi Voos how about if i want take take after number 333?
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
Therefore i make this code, would you check this looping to be more clear and intuitive (especially for regexp part), actually i'm newbie with regexp :)
after333=repmat({''},length(a),1);
for i=1:length(a)
idx = regexp(a{i},'(?<=333\s+)(\d+)','once');
after333{i} = a{i}(idx:end);
end
disp(after333)
{'56000 81625'} {1×0 char }
tks
Voss
Voss 2022-6-6
Ran in:
If you want to take the remainder of each char array after '333' (where '333' is its own "word"), I would use the same regular expression as before, and adjust the indexing:
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
after333 = repmat({''},length(a),1);
idx = regexp(a,'\<333\>','once');
for ii = 1:numel(a)
if ~isempty(idx{ii})
after333{ii} = a{ii}(idx{ii}+4:end);
end
end
disp(after333);
{'56000 81625'} {1×0 char }

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