Indexing a value from other cell arrays into a new column

Hi everyone. I am trying to do some multiple arrays indexing in MATLAB. I've attached a workspace that contains these matrices to this question ("indexing.mat").

What I would like to do is to read each of the rows of the first column of the 'distances' matrix. Each of the values refers to that specific row of the 'indeces' matrix that contains an (x,y) pair in two columns. I would then like to abstract the value in the 'arrays' matrix based on that (x,y) pair and place it in the third column of the 'distances' matrix in that same row.

For example: row 12, first column of 'distances' is 411. Row 411 in 'indeces' is (30,92). That pair corresponds to 105.91090 in 'arrays'. I'd then like to place that value in row 12, third column of 'distances'.

Is there a way to automate this? I can try to do it manually but it's becoming labor-intensive. Thank you for your insight!

Answer 1

[David Hill]

for k=1:size(distances,1)
  a=flip(indeces(distances(k,1),:));
  distances(k,3)=arrays(a(1),a(2));
end

[Aaron Devanathan]

@David Hill Thank you so much for this! I'm so sorry to bother you, but is there a way to incorporate such that if it returns a 0, it should then return the value from 'arrays' associated with the 'indeces' (x,y) coordinate + the 'distances' value? For instance, rows 1 and 2 would return 54.3804 [(7,31) and (8,31)] from 'arrays', respectively?

[David Hill]

You are going to have to explain more. rows 1 and 2 give the same distance (2) which gives [6,33] in arrays. I am not following how rows 1 and 2 would be [7,31] and [8,31] by your explaination.

[Aaron Devanathan]

Sorry for the confusion. What I mean is if distances(k,3) = 0 then instead insert the closest non-zero value away from that coordinate in 'arrays'.

[David Hill]

[a,b]=find(arrays);
for k=1:size(distances,1)
  c=flip(indeces(distances(k,1),:));
  if isequal(arrays(c(1),c(2)),0)
     r=[a,b]-[c(1),c(2)];
     R=hypot(r(:,1),r(:,2));
     [~,idx]=min(R);
     distances(k,3)=arrays(a(idx),b(idx));
  else
     distances(k,3)=arrays(c(1),c(2));
  end
end

[Aaron Devanathan]

Thank you! I really appreciate your help.