extractBefore with matches string

2022 6 4
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更新时间：2022 6 6
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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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Voss
Voss 2022-6-4
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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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Voss
Voss 2022-6-4
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Here's a way that will work in both cases ('333' at the end and '333' not at the end) for this scalar cell array a:
% a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625'};
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333'};
idx = regexp(a{1},'\<333\>','once');
if isempty(idx)
matches333 = {''};
else
matches333 = {a{1}(1:idx-1)};
end
matches333
matches333 = 1×1 cell array
{'32563 33308 10314 20227 30113 40117 52008 81531 '}
eko supriyadi
eko supriyadi 2022-6-4
how about if we have 2 cell or more, like this:
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
my solution:
matches333={};
for i=1:length(a)
idx = regexp(a{i},'\<333\>','once');
if isempty(idx)
matches333= {''};
else
matches333{i} = {a{i}(1:idx-1)}; %produce 1*2 cell
end
end
the result is column oriented, not use :
matches333=matchess33'
to solve problem, i'm interest with the looping, can you fix the loop above so the result have 2*1 cell not 1*2 cell..tks
Voss
Voss 2022-6-4
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a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
% pre-allocate a column cell array
% containing empty char vectors
matches333 = repmat({''},numel(a),1);
% you can do regexp on a cell array of chars
idx = regexp(a,'\<333\>','once')
idx = 1×2 cell array
{[49]} {[49]}
for ii = 1:numel(a)
% now you only have to set matches333{ii} if there was a match
% (it's already the empty char vector otherwise)
if ~isempty(idx{ii})
matches333{ii} = a{ii}(1:idx{ii}-1);
end
end
disp(matches333);
{'32563 33308 10314 20227 30113 40117 52008 81531 '} {'32563 33308 10314 20227 30113 40117 52008 81531 '}
eko supriyadi
eko supriyadi 2022-6-4
yes, thank you Voos, i just forgot change the matches333 matrix preallocate.. many thanks for your help :)
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
matches333 = repmat({''},numel(a),1);
for i=1:length(a)
idx = regexp(a{i},'\<333\>','once');
if isempty(idx)
matches333= {''};
else
matches333{i} = {a{i}(1:idx-1)}
end
end
eko supriyadi
eko supriyadi 2022-6-4
edited: in early replay i'm use example
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
can, you edited your example in 3rd posting
Voss
Voss 2022-6-4
Isn't that the same a that's already there?
eko supriyadi
eko supriyadi 2022-6-4
no, here:
a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
Voss
Voss 2022-6-4
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% copied from my comment:
old_a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
% copied from your comment:
new_a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};
isequal(old_a,new_a)
ans = logical
1
Regardless of what a you want to use, the code is the same:
a = new_a;
matches333 = repmat({''},numel(a),1);
idx = regexp(a,'\<333\>','once');
for ii = 1:numel(a)
if ~isempty(idx{ii})
matches333{ii} = a{ii}(1:idx{ii}-1);
end
end
disp(matches333);
{'32563 33308 10314 20227 30113 40117 52008 81531 '} {'32563 33308 10314 20227 30113 40117 52008 81531 '}
eko supriyadi
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
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 }

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