Comparing two cell arrays of strings of different sizes

Dear all,

I'm trying to compare two cell arrays of strings, a lot like the following:

A = [X_ABCDE;X_BCDEA;X_BCED]

B = [X_A;X_B]

and the resulting array should be:

index = [1;2;2]

as I have a matrix D = [B, C] with C containing the cells I want to end up with.

I wanted to do it using a for-loop and a nested for-loop, but, as A is 8000x1 and D is 2000x2, this will take forever. I tried using strcmp and ismember, but these only work when the cells (or strings) are identical. strfind doesn't work either, because when I use strfind I need to have a for-loop over array A, and strfind only works if the smallest of both is a string.

for iCell = 1:length(A)
    index = strfind(A(iCell), cell2mat(D(iCell,1)));
end

A string in B is ALWAYS shorter than in A and a string in B can occur more than once in A. I hope you can help me out :)

[Stephen23]

Your state that: "I have a matrix D = [B, C] with C containing the cells I want to end up with". but if the matrix D is an output (i.e. is defined by what you "end up with"), then how can D be accessed within the loop?: D(iCell,1). How does D get defined if C is an output?

[Jasper Admiraal]

Sorry, I got it wrong. I want to end up with a certain order of the strings from C, indicated by the array index. In my code, a cell in B contains a code which corresponds to a specification in C (in the same row).

Answer 1

[Stephen23]

Use strncmp (not strcmp) to make the code simpler. Loop over the smaller cell array B, and pass the larger cell array A whole to strncmp:

A = {'X_ABCDE';'X_BCDEA';'X_BCED'}
B = {'X_A';'X_B'}
%
X = zeros(size(A));
for k = 1:numel(B)
    X(strncmp(A,B{k},3)) = k;
end

produces:

>> X
X =
   1
   2
   2

Do not use arrayfun or cellfun if you want fast code: they will be slower than a for-loop.

[Jasper Admiraal]

Thank you Stephen, works like a charm!

Answer 2

[Jos (10584)]

assuming B is all unique

A = {'X_ABCDE' ; 'X_BCDEA' ; 'X_BCE'}
B = {'X_A' ; 'X_B'}
index = arrayfun(@(k) find(strncmp(A{k},B,3)), 1:number(A))

[Stephen23]

My solution is much faster than this one (100 iterations)

Elapsed time is 0.0370002 seconds. % my answer
Elapsed time is 1.623 seconds.     % this answer

Using arrayfun is slower than using a loop.

[Jos (10584)]

of course it is. I just wanted to show a different answer ... :-)