Specifying node numbers using digraph

I need to plot something that has nodes Numbers and Coordinates as in the attached file: Data_Coord.txt

the file has four columns and they correspond to: NodeNumber CoordX CoordY CoordZ, respectively

The conectivity between nodes is given by the file Data_Conec.txt

the file has three columns and they correspond to: ElementNumber Node1 Node2, respectively

Note that the number of the nodes is not sequential up to the total number of nodes, meaning that, for example, I could have node 1 conected to node 10001, even though I only have 200 nodes. but the coordinate of node 1 and 10001 is given on the Data_Coord.txt file and the first column specifies the number of the node.

How can I plot this 3D sketch?

I tried using digraph but because of the node numbers I wasnt sure how to proceed

thanks,

Answer 1

[Christine Tobler]

The easiest way to use graph/digraph when your node numbers are not sequential is to convert the node numbers to string, which will have them treated as node names instead. Then, you should be able to use digraph and hte plot command with 'XData', 'YData', and 'ZData' options.

[Conrado Neto]

Hi Christine, thanks for your reply

this is what I tried, but I thought it wouldnt work given I change the node number to strings after the graph and plot commands, but it was the only solution I found:

G = digraph(E1,E2);

h = plot(G,'XData',X_Coord,'YData',Y_Coord,'ZData',Z_Coord,'ArrowSize',10); % ,'NodeLabel',1:100

labelnode(h,1:1965,Node_Coord)

which gave me the error message:

Error using matlab.graphics.chart.primitive.GraphPlot

Expected XData to be an array with number of elements equal to 1.11013e+06.

what is the proper way to do what you suggested? thanks again

[Christine Tobler]

Hi Conrado,

If E1, E2 are the second and third column of Data_Conec.txt, and Node_Coord is the first column of Data_Coord.txt, use

G = digraph(string(E1), string(E2), [], string(Node_Coord));
plot(G,'XData',X_Coord,'YData',Y_Coord,'ZData',Z_Coord,...
    'ArrowSize',10,'NodeLabel',G.Nodes.Name);

[Conrado Neto]

this works perfectly, thanks Christine!