# column number extract using find function..

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인수 송 on 6 May 2022
Commented: Jon on 6 May 2022
A = [ 0 500 1000 1500 2000 2500 3000 3500 x1 x2 x3 5500 6000 6500 7000 7500 8000 8500 9000 9500 10000 ] ;
i want to know the column number using find function about 500,1000,2000,3000, x2, 4000, 5000, 6000, 7000, 8000, 9000, 10000 ].
: B = [ 2 3 5 7 9 11 13 15 17 19 21 23 25 27 29 ] ; (example : it's not answer)
The x1, x2, x3 is variable value.

dpb on 6 May 2022
Use the optional second output form of find
[r,c]=find(....);
Jon on 6 May 2022
I think this seems like it would only help you if you had a 2d array of logicals or 0's and 1's. The OP just has a 1d array. Unless maybe I am completely missing something. The issue is how to create the corresponding vector of logicals that find can be applied to. One way is using ismember as shown in my answer below.

Jon on 6 May 2022
A = [ 0 500 1000 1500 2000 2500 3000 3500 x1 x2 x3 5500 6000 6500 7000 7500 8000 8500 9000 9500 10000 ] ;
matchVals = [500,1000,2000,3000, x2, 4000, 5000, 6000, 7000, 8000, 9000, 10000 ];
B = find(ismember(A,matchVals))
Jon on 6 May 2022
In your case the values seem to follow a very regular progression. I'm not sure if this is just the case for the example you give or if it always holds. If you have such a regular progression you could also write a formulae, e.g
B = matchVals/500 + 1

Voss on 6 May 2022
[~,idx] = ismember([500 1000 2000],[0 500 1000 1500 2000 2500])
idx = 1×3
2 3 5
Jon on 6 May 2022
Good point, you don't need to use find as I did after calling ismember, just use the second output argument to get the indices