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Finding the indices of the elements of one array in another

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Alan
Alan on 3 Dec 2011
Commented: David on 18 Jan 2020
Given two vectors A and B, find the index, idx into A of the element of B so that
A(idx)=B.
Now I know there must be many ways it can be done, but is there a one-liner?
For example if
A=[3 4 5 6 7];
B=[6 4 7];
then
[tf,loc]=ismember(A,B);
idx=[1:length(A)];
idx=idx(tf);
idx=idx(loc(tf));
disp(A(idx))
will do it but that is four steps. Is there a more elegant way?

  3 Comments

Alan
Alan on 3 Dec 2011
Actually, my solution doesn't work in general. Sometimes but not always.
Philip
Philip on 26 Sep 2014
MATLAB supports logical indexing. No need to use "find":
A = A( ismember( A, B ) );

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Accepted Answer

Sven
Sven on 3 Dec 2011
Edited: MathWorks Support Team on 9 Nov 2018
There are a few options to get the indices you are looking for. The following output indices (idx) preserve the order in A of the shared values:
[sharedvals,idx] = intersect(A,B,stable)
You can also use the following command if the order in A is not necessary:
[tf,idx] = ismember(B,A)

  3 Comments

Alan
Alan on 3 Dec 2011
That's nice. Two lines is better than four. I hadn't thought of using the index of sort.
Alan
Alan on 3 Dec 2011
I am wondering though if "intersect" and "sort" would be expensive on large arrays.
tc88
tc88 on 22 Aug 2016
meanwhile, the functionality of intersect has changed and a one-line solution is also possible using intersect:
[sharedVals,idxsIntoA] = intersect(B,A,'stable')
Be aware that the order of A and B must be changed, since the order of the first argument is retained.

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More Answers (6)

Alan
Alan on 3 Dec 2011
Thanks for the three solutions. Here is a function to test all three. "method_1" wins as the most elegant and fastest, but the other two taught other useful ways of looking at the problem.
function test_solution
% set up the problem
N=1e6;
% A is a random vector of the integers to N
A=randperm(N);
% B is another vector of integers
B=[ 2 1 4 3 ];
% I'd like to find the indicies of the element of B in A
% such that A(idx)=B;
tic
idx=method_1(A,B);
toc
disp(A(idx));
tic
idx=method_2(A,B);
toc
disp(A(idx));
tic
idx=method_3(A,B);
toc
disp(A(idx));
function idx = method_1(A,B)
idx = arrayfun(@(x)find(A==x,1),B);
function idx = method_2(A,B)
[~,idx1] = intersect(A,B);
[~,idx2] = sort(B);
idx=idx1(idx2);
function idx = method_3(A,B)
[A,idx1] = sort(A);
[~,idx2] = histc(B,A);
idx = idx1(idx2);

  3 Comments

Sven
Sven on 4 Dec 2011
Alan, well done in asking a question clearly (with code), and in particular taking the time to give feedback on the results above
David
David on 18 Jan 2020
Method one also works if there are multiple occurences of B in A. Intersect fails in this case.

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Alan
Alan on 6 Dec 2011
Thinking again, I realize that ismember is all I need
[~,idx]=ismember(B,A)
will do it.

  2 Comments

John Sogade
John Sogade on 2 Jan 2020
obviously this will fail to get A(idx), if any elements of idx are 0 (i.e. B not in A) and robust usage should be clarified to A(idx(idx ~= 0)).

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Teja Muppirala
Teja Muppirala on 3 Dec 2011
If A is sorted, then I think this is probably the easiest (and also fastest?) way to do it.
[~,idx] = histc(B,A)
If A is not sorted, then:
[As,s_idx] = sort(A);
[~,tmp] = histc(B,As);
idx = s_idx(tmp)

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Stephen Politzer-Ahles
Stephen Politzer-Ahles on 8 Jul 2014
Edited: Stephen Politzer-Ahles on 8 Jul 2014
The following should also work for your situation, and just needs one line:
A=[3 4 5 6 7];
B=[6 4 7];
idx = arrayfun( @(x)( find(A==x) ), B );

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Junhong YE
Junhong YE on 21 Jul 2014
I think find(ismember(A,B)) would do it.

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Iftikhar Ali
Iftikhar Ali on 18 Oct 2015
I am facing an issue finding indices of element matching in two arrays.
xpts = [0 0.0004 0.0011 0.0018 0.0025 0.003]; x = 0:0.0001:0.003; index1 = find(ismember(x, xpts));
It returns index1 = [1 5 12 26 31]
but there is one more element '0.0018' in x which also belongs xpts, and not including in the answer.
Similarly when I increase the number of points in x, there are few elements that are missed or not recognized by the find command. What's going wrong here.

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