print values in one matrix to another
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Walter Roberson
am 16 Apr. 2020
Is it mandatory to use a for loop? This is something that is easily vectorized.
Akzeptierte Antwort
Tommy
am 16 Apr. 2020
Bearbeitet: Tommy
am 16 Apr. 2020
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 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!