How to compare multiple strings and assign values to the comparison?

I have data from 6 different sources in a form of strings. I want to compare which strings are the same.

So if I have:

d1='111'
d2='223'
d3='111'
d4='125'
d5='111'
d6='125'

d1,d3,d5 are the same, d4 and d6 are the same too and d2 is different to all, so I will like to receive a vector where similar strings have the same number, like this:

[1 2 1 4 1 4]

if all the strings are different I will like to receive something like this:

[1 2 3 4 5 6]

or all the same

[1 1 1 1 1 1]

so far I've use a very ugly program with a lot of IF and ISEQUAL, any suggestions I will really appreciate it!

Answer 1

[José-Luis]

Copying beginning from Siva because I'm lazy:

 K = [{'111'} ; {'223'} ; {'111'} ; {'125'} ;{'111'} ; {'125'}] ;

Solving your problem more efficiently:

 [val,ia,ib] = unique(K,'stable') ; % get unique values
 your_answer = ia(ib);

[German Preciat Gonzalez]

Thanks!

[José-Luis]

My pleasure.

[Guillaume]

I find that method of creating a cell array very odd!

K = {'111'; '223'; '111'; '125'; '111'; '125'};

is less to type and a lot simpler.

Answer 2

[Guillaume]

Dr. Siva Srinivas Kolukula has the right idea: use unique. His answer is unnecessarily complicated though.

First though, do not use numbered variables. If they're numbered it means that they've got some sort of relationship. That relationship is much better expressed by putting the content of these variables together into the same container: matrix, cell array, map, table, etc. In your case, a cell array seems best.

Anyway, the third return value of unique is what you're after:

d = {'111', '223', '111', '125', '111', '125'}
[~, ~, uid] = unique(d, 'stable')  %'stable' is optional. If not present unique ids are assigned alphabetically

[José-Luis]

Except that will give the index to the unique vector, not to the first occurrence.

[Guillaume]

Well, the question was "I will like to receive a vector where similar strings have the same number", which the above does.

[José-Luis]

Indeed, but that's not what the example result showed. Anyway, interpreting questions is an art sometimes...

Answer 3

[KSSV]

clc; clear all ;
K = [{'111'} ; {'223'} ; {'111'} ; {'125'} ;{'111'} ; {'125'}] ;
[val,idx] = unique(K,'stable') ; % get unique values
iwant = zeros(size(K)) ;  % initialize what you want 
count = 0 ;    
% loop for each equal string 
for i = 1:length(idx)
    count = count+1 ;
    Index = strfind(K,val{i}) ;
    Index = find(not(cellfun('isempty', Index))) ;
    iwant(Index) = count ;
end

Answer 4

[KSSV]

Jose-Luis idea is good....but only one step is needed.

[val,ia,your_answer] = unique(K,'stable') ; % get unique values

[José-Luis]

No, what the op wanted, if I understood correctly, is the index to the first occurrence. Therefore that wouldn't work.