Finding the indices of the elements of one array in another
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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 个评论
采纳的回答
Sven
2011-12-3
编辑:MathWorks Support Team
2018-11-9
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 个评论
tc88
2016-8-22
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.
更多回答(6 个)
Alan
2011-12-6
2 个评论
John Sogade
2020-1-2
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)).
Iftikhar Ali
2015-10-18
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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Teja Muppirala
2011-12-3
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
2014-7-8
编辑:Stephen Politzer-Ahles
2014-7-8
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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