extractBefore with matches string

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

Answer 1

[Voss]

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}

[Voss]

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]

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]

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]

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]

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]

Isn't that the same a that's already there?

[eko supriyadi]

no, here:

a = {'32563 33308 10314 20227 30113 40117 52008 81531 333 56000 81625','32563 33308 10314 20227 30113 40117 52008 81531 333'};

[Voss]

% 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]

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]

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     }