extractBefore with matches string

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eko supriyadi 2022-6-4

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https://ww2.mathworks.cn/matlabcentral/answers/1733820-extractbefore-with-matches-string

评论： Voss 2022-6-6

采纳的回答： Voss

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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 2022-6-4

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1733820-extractbefore-with-matches-string#answer_978955

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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 2022-6-6

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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 2022-6-6

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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     }