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print values in one matrix to another

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k
k 2020-4-16
评论： Tommy 2020-4-16
Hi,
I have very large matricies but will abbreviate them as best as possible here.
I have a matrix called "connectivity" which contains 5884 rows (total_elements = 5884) and 5 columns. The first column is the element number, the second column is node 1 and the third column is node 2:
connectivity(total_elements, 5);
connectivity(:,1) = 1:total_elements;
connectivity(:,2) = N1;
connectivity(:,3) = N2;
making this matrix look like
1 900 175
2 175 200
3 200 145
4 145 300...
I have a second matrix "C" which has the x and y coordinates for each of the nodes.
NodeNumber Xcomponent Ycomponent
...
I need to write a loop that will search matrix "C" for the correct x and y coordinates for each node and create a new matrix "length," which looks like the following:
length = zeros(total_elements, 7);
length(:,1) = 1:total_elements
length(:,2) = N1;
length(:,3) = N2;
length(:,4) = x1; %x coordinate of node 1
length(:,5)=x2; %x coordinate of node 2
length(:,6)=y1; %y coordinate of node 1
legnth(:,7)=y2; %y coordinate of node 2
being:
element# N1 N2 x1 x2 y1 y2
  2 个评论
Walter Roberson
Walter Roberson 2020-4-16
Is it mandatory to use a for loop? This is something that is easily vectorized.
k
k 2020-4-16
No, a for loop isn't necessary. I essentially just need to match values from one matrix to another and print them into a final matrix.
Ex:
matrix A
node # xcoordinate ycoordinate
1 0.23 0.5
2 2.3 6.2
3 2.1 3.2
matrix B
element# N1 N2
1 2 3
2 3 1
3 2 1
matrix C, which I'm trying to create
element# N1 xcoordinate ycoordinate N2 xcoordinate yycoordinate
1 2 2.3 6.2 3 2.1 3.2
2 3 2.1 3.2 1 0.23 0.5
3 2 2.3 6.2 1 0.23 0.5
Hopefully this makes more sense! Again, I need to scale this to where the total number of elements is 5884 and the number of nodes is 2000.

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Tommy
Tommy 2020-4-16
编辑：Tommy 2020-4-16
Let me know if this works:
length = [connectivity, zeros(total_elements, 4)];
for i = 1:total_elements
length(i, [4 6]) = C(C(:,1)==connectivity(i,2),[2 3]);
length(i, [5 7]) = C(C(:,1)==connectivity(i,3),[2 3]);
end
(edit) If you want to avoid the loop, and to match the format you've given in the comments:
[~,i] = ismember(connectivity(:,[2 3]),C(:,1));
length = [connectivity(:,[1 2]), C(i(:,1),[2 3]), connectivity(:,3), C(i(:,2),[2 3])];
Either way should work for any number of elements and nodes, provided your connectivity and C matrices are formatted as you've described.
  8 个评论
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k
k 2020-4-16
It worked!! Thank you so so much!!

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