Find elements of an array in another array

49 visualizzazioni (ultimi 30 giorni)
Hello,
Let's say I have 2 arrays of double, call then A and B. If both have unique entries and I want to find the position of each element of A in array B I can do:
[~, pos] = ismember(A,B);
What if the elements of A show up multiple times in B and I want to get the first time they show up or the last time they show up? I know I can do
pos = zeros(length(A),1);
for k = 1:length(A)
pos(k) = find(B == A(k),1,'first');
end;
But is there a better, more efficient way of doing it? For loops are not exactly in the spirit of Matlab as far as I know.
Thanks, Alex

Risposta accettata

Azzi Abdelmalek
Azzi Abdelmalek il 20 Set 2013
Modificato: Azzi Abdelmalek il 20 Set 2013
Maybe in your case, all element in A are present in B
A=[1 2 3 4 5 6 7]
B=[12 13 2 4 3 2 4 2 25 1 6 7 5]
pos=arrayfun(@(x) find(B==x,1),A)
  2 Commenti
Alexandru
Alexandru il 20 Set 2013
Thanks Azzi! If I do find(B==x,1,'first') or find(B==x,1,'last') in your code I get exactly what I want.
One more question. Suppose I want to get the second occurrence when it exists. So for the second element of A which is 2 the positions it shows in B are 3, 6 and 8. Suppose I want the code to return 6. How would I modify it?
Azzi Abdelmalek
Azzi Abdelmalek il 20 Set 2013
Modificato: Azzi Abdelmalek il 20 Set 2013
Use in the loop
id=find(B==A(k),2)
id=id(2)

Accedi per commentare.

Più risposte (1)

Azzi Abdelmalek
Azzi Abdelmalek il 20 Set 2013
Modificato: Azzi Abdelmalek il 20 Set 2013
I prefer this one. It should be much faster
A=[1 2 3 4 5 6 7]
B=[12 13 2 4 3 2 4 2 25 1 6 7 5];
[ii,jj]=unique(B,'stable');
n=numel(A);
pos = zeros(n,1);
for k = 1:n
pos(k)=jj(find(ii == A(k)));
end;
  2 Commenti
Alexandru
Alexandru il 20 Set 2013
Ok, so in this case it would come down to a for loop.
Thanks again!
Azzi Abdelmalek
Azzi Abdelmalek il 20 Set 2013
Modificato: Azzi Abdelmalek il 20 Set 2013
This answer is more efficient then the first one

Accedi per commentare.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by