Different length array comparison and replacements

If I have 2 different length array, and I want to compare the values between these two arrays. First I sort both arrays in 'ascend' manner. So I start comparing the 1st term of Array A with the 1st term of Array B. If A(i)<B(j) then replace in a C predefined matrix. Once a element is replace in C, I need to move to the second term of A. I don't need to compare every element of A with the rest of B, because I already make a replacement. Also it can be the case that any member of A is less than B.

I tried with the double "for" loop, but it is not working. Because It compares every element of A with every element of B, it does not matter if it make a replacement.

For example:

 if true
    A=[445;
       874]

    B= [265;
        446;
        744;
        872;
        875]

       for i=1:length(A)
            for j=1:length(B)
                if A(i)<B(j)
                   C(j)=A(i);
                end
            end
        end

end

In this example A(1) should replace C(3) and A(2) should replace C(5). But as mention, it can be the case where any member of A replace a member of C.

Thanks!

Answer 1

[Jos (10584)]

Add a break in the if so the function will break out of the inner-loop, and moves on to the next element of A:

help break

In addition, you can pre-allocate C before the loops

C = zeros(size(B))

Last, it can be the case that an element of C is overwritten (e.g., when A = [2 3], B = [1 4 10], which will result in C = [0 3 0]). Is this your intention?

[Patty]

Thanks! This is what I was looking for, to stop the inner "for". But I was not sure about the break function. :)

I just tried and it's working.

Answer 2

[George Papazafeiropoulos]

% data
A=[445;
    874];
B= [265;
    446;
    744;
    872;
    875];
% engine
AA=A(:,ones(1,size(B,1)))';
BB=B(:,ones(1,size(A,1)));
[r,c,~]=find(AA-BB<0);
ranges=histc(c,unique(c));
a=max(ranges);
ind1=reshape((1:a*length(ranges))',a,[]);
lb=0:a:a*(length(ranges)-1);
lb=lb(ones(a,1),:);
ub=ranges'+(0:length(ranges)-1)*a;
ub=ub(ones(a,1),:)+1;
ind2=ind1>lb & ind1<ub;
y(ind2)=r;
y(~ind2)=nan;
y=reshape(y,a,[]);
y=y(1,:);
u=diff(y)>0;
ind=[1 u.*(2:length(u)+1)];
A(~ind)=[];
ind(~ind)=[];
replacement_indices=y(1,ind);
% result
C=B;
C(replacement_indices)=A;
find(C-B)