Asked by Alan
on 3 Dec 2011

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?

Answer by Sven
on 3 Dec 2011

Edited by MathWorks Support Team
on 9 Nov 2018

Accepted Answer

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)

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
on 3 Dec 2011

I am wondering though if "intersect" and "sort" would be expensive on large arrays.

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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Answer by 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);

Sven
on 4 Dec 2011

Iftikhar Ali
on 18 Oct 2015

Method 3 has solved my problem, thanks.

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Answer by 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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Answer by Stephen Politzer-Ahles
on 8 Jul 2014

Edited by 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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Answer by Junhong YE
on 21 Jul 2014

I think find(ismember(A,B)) would do it.

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Answer by 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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