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

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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!

Accepted Answer

José-Luis
José-Luis on 28 Jun 2016
Edited: José-Luis on 28 Jun 2016
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);
  3 Comments
Guillaume
Guillaume on 28 Jun 2016
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.

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More Answers (3)

Guillaume
Guillaume on 28 Jun 2016
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
  3 Comments
Guillaume
Guillaume on 28 Jun 2016
Well, the question was "I will like to receive a vector where similar strings have the same number", which the above does.
José-Luis
José-Luis on 28 Jun 2016
Indeed, but that's not what the example result showed. Anyway, interpreting questions is an art sometimes...

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KSSV
KSSV on 28 Jun 2016
Edited: KSSV on 28 Jun 2016
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

KSSV
KSSV on 28 Jun 2016
Jose-Luis idea is good....but only one step is needed.
[val,ia,your_answer] = unique(K,'stable') ; % get unique values
  1 Comment
José-Luis
José-Luis on 28 Jun 2016
No, what the op wanted, if I understood correctly, is the index to the first occurrence. Therefore that wouldn't work.

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