How to combine vectors based on value?

Good Afternoon,

Another quick question. Say I have a matrix,

   B = [732145      732151      732168           0           0           0           0
         284         281         297           0           0           0           0]

and

   C = 
    [ 0   281   284     0     0   297     0
     0     0     0     0     0     0     0]

I am trying to get it so that I have:

   D = [0    281      284    0   0   297     0
        0  732151    732145  0   0  732168   0].

Should I be using the [r c] kind of terminology. I keep almost getting there, I think, then I get confused.

Any thoughts would be greatly appreciated! or reading!

Thank you very much for any time!

Answer 1

[Guillaume]

Regarding your latest comment, a much simpler way of obtaining your D result, assumming that elements of B and C are uniques is:

assert(numel(unique(C)) == numel(C) && numel(unique(B)) == numel(B)); %does not work if elements are not unique
[values, ic, ib] = intersect(C, B);
D = zeros(numel(C), numel(B));
D(sub2ind(size(D), ic, ib)) = values;

If the values of B and C are not unique, you need an additional level of indirection through unique.

[Chameleon17]

Thank you for that, i've not used the intersect before, that's good to know!

The values of B are all unique - but I may have repeats of the C values at certain times.

Answer 2

[Andrei Bobrov]

    B = [4 5 6 7; 0 0 0 0; 0 0 0 0; 0 0 0 0];
    C = [0; 5; 4];

    D = B;
    [ll,idx] = ismember(B(1,:),C);
    D(sub2ind(size(D),1+idx(ll),find(ll))) = C(idx(ll));

Answer 3

[Geoff Hayes]

Do you wish to simply iterate over each element in the first row of C and, for those that are non-zero, check to see if it exists in the second row of matrix B and copy the the value from its first row into D? If so, try the following

 D = C;
 for k=1:size(C,2)
     if C(1,k) ~= 0
         [idx] = find(B(2,:)==C(1,k));
         if ~isempty(idx)
             D(2,k) = B(1,idx);
         end
     end
 end

In the above, we just iterate over each column in the first row of C. If the element is non-zero then we use find to find that element of the second row of B that matches the value in C. If the returned index, idx, is non-empty then we set this value to the corresponding element of D.

[Chameleon17]

Thank you very much! That's exactly what I need to do, and what I need to learn more about. Can I trouble you for another related problem?

Say I have:

B = [4 5 6 7; 0 0 0 0; 0 0 0 0; 0 0 0 0] C = [0; 5; 4]

and I'm trying to get to D which would be:

D = [4 5 6 7; 0 0 0 0; 0 5 0 0; 4 0 0 0]

I've been playing around with it for ages, but I'm still stumped.. I know it will be quite an easy solution...

Thank you again!

[Geoff Hayes]

First you need to define the rules or steps on how you start with B and C and get to D. What are the steps? Once you know what they are, you should be able code it up. (I'm asking only because it isn't clear to me how D is created. Why is 5 in the second element of the third row of D? Why is 4 in the first element of the last row of D?)

[Chameleon17]

Thanks for the advice! It's a really simplified version of something I want to apply for a lot more data, I did it though!

Sometimes someone else commenting on it helps a lot, I really appreciate the advice on this forum and the time people contribute! You go round in circles in your own head sometimes with the simplest of things.

k = 1: length(B)

g = 1:length(C)

for h = 1:length(g)

for m = 1:length(k)

if C(h,1) == B(1,m)

D(h,m) = C(h,1)

end
end
end

[Guillaume]

I assume you've made a typo and that the for loops are actually:

for h = 1:g
   for m = 1:k

[Chameleon17]

Ah yes, I had a bunch of them, it is also D(h+1,m) too!