Compare two arrays based on indexes

Hello,

Please help me with the following:

consider two arrays

A=[19;20;19;20;21;22]; B=[1;1;4;4;4;4];

A user should provide two values, i.e. v1, v2.

A condition (maybe with logical if) should search if v1 belong to A and v2 belong to B.

if those values both belong to A and B, another operation should take place, for example C=100 or C=200;

For example,

If v1=19; && v2=2;

Those values do not belong both in A and B, so C=200

If v1=19; && v2=1;

Those values belong both in A and B, so C=100

The indexes of v1 and v2 in A and B should be the same in order for C=100;

That means that the operation C=100; would take place only in the following combinations:

v1=19; v2=1;

v1=20; v2=1;

v1=19; v2=4;

v1=20; v2=4;

v1=21; v2=4;

v1=22; v2=4;

The condition

If v1=21; && v2=1; theoritically gives C=100;

but the indexes of v1 and v2 in A and B are not equal (5th index in A for v1 and 1st,2nd indexes for v2 in B).

Than you very much,

Best,

Pavlos

Answer 1

[Andrei Bobrov]

M = [A,B];
V = [v1,v2];
C = [200,100];

out = C(ismember(V,M,'rows')+1);

[pavlos]

Thank you, but how v1 and v2 are taken into account?

[Andrei Bobrov]

I'm fixed.

[pavlos]

Thank you again, but your code does not consider the equalities of the indexes of v1 and v2 in A and B.

In order for C=100; v1 and v2 should only belong to A and B but also, refer to the same index in A and B.

For instance, v1=19 and v2=1 gives C=100 because the index of v1 in A is 1 and the index of v2 in B is also 1.

[Guillaume]

your code does not consider the equalities of the indexes

Yes, it does. That's exactly how ismember with the 'rows' option work. Try it with V = [22, 1] and see that it returns 200.

Andrei's answer is the most efficient way to do exactly what you asked.