Finding repeating values in an array

Hello all,

I have 2 arrays

A = [ 0.3 0.6 1 0.6 0.3] 
B = [ 3 2 3 6 11 ]

I need to find the position of same elements in B and then the max value of the elements on the corresponding position in A.

In this case the number 3 is repeated in B on positions 1 and 3 so the corresponding values in A are 0.3 and 1 => max( 0.3 , 1 ) = 1

The end resault should be:

A1 = [ 0.6 1 0.6 0.3 ]
B1 = [ 2 3 6 11 ]

Any help is appreciated

[andrea]

maybe :

pos_in_b = B == 3 ;

max ( A ( pos_in_b) )

[dpb]

The find part is easy enough, the logic of how to build the A1, B1 vectors from A,B and the lookups escapes me entirely, though...???

[Vladimir Kostic]

dpb

If you look at the elements of A as the numerator and elements of B as the denominator i need to find fractions with the same denominator and use the one with the higher numerator.

In my example i have

0.3 0.6 1 0.6 0.3

3 2 3 6 11

So there are 2 fractions with the denominator of 3 (it isn't predetermined that it is 3 just happened to be this way, it can be any number and I can be more that just 1 number that repeats) and the numerators of those 2 fractions are 0.3 and 1, where the higher number is 1 ( 1>0.3). So i need to erase the fraction with the lower numerator.

0.6 1 0.6 0.3

2 3 6 11

Hope this clears it up

Answer 1

[aleksa markovic]

You ca try something like this:

Xa = [3 2 3 6 11];
mua = [.3 .6 1 .6 .3];
tmpX = [];
tmpmu = [];
for i = 1:size(Xa,2)
    if(sum(tmpX == Xa(i)) > 0)
        tmpmu(tmpX == Xa(i)) = max(mua(i),tmpmu(tmpX == Xa(i)));
    else
        tmpX = [tmpX Xa(i)];
        tmpmu = [tmpmu mua(i)];
    end
end
Xa = tmpX;
mua = tmpmu;

Answer 2

[Rik]

You can use the outputs of the unique function to achieve this.

A = [ 0.3 0.6 1 0.6 0.3];
B = [ 3 2 3 6 11];

[B1,~,ind]=unique(B);
A1=accumarray([ones(numel(A),1) ind],A(:),[],@max);

A1
B1

[Vladimir Kostic]

Could you explain further into detail, I'm still pretty new at MATLAB

[Rik]

A bit late, but here you go, no loops required.

Answer 3

[andrea]

maybe i do not understand the problem but anyway

[val, ind] = min ( A ( pos_in_b) )

A(ind) = []