how create coordinate array?

48 vues (au cours des 30 derniers jours)
huda nawaf
huda nawaf le 8 Sep 2012
hi,
I found this function to plot the graph of adjacency matrix.
The input is adj matrix and coordinates array, xy. I have adj matrix , but I do not know how create coordinate array.
please can anybody help me in this regarding?
thanks
  1 commentaire
Oleg Komarov
Oleg Komarov le 8 Sep 2012
Which is already part of MATLAB: gplot().

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Réponses (2)

nanren888
nanren888 le 8 Sep 2012
huda nawaf,
Interesting that the Mathworks copyright is still in the code you are using. It is not clear what is added to gplot by the addition.
gplot has an example in doc: try doc gplot
It seems as a graph has no inherent shape, that you must provide the xy points for each node for it. The example in doc shows a Buckyball node layout.
I've never used gplot before. I read the help, read doc gplot, played a few minutes & have this.
x = 1:2;
y = 1:3;
[XX,YY] = meshgrid(x,y);
a = (rand([2*3,2*3])>0.8);
P=[XX(:),YY(:)];
gplot(a,P,'-*');
axis([0,7,0,7]);
a
P
a =
0 0 0 0 0 1
0 0 1 1 0 1
0 0 1 1 0 0
0 0 0 0 0 0
0 0 1 0 0 1
0 0 1 0 1 1
P =
1 1
1 2
1 3
2 1
2 2
2 3
Maybe it will help a little?
  4 commentaires
huda nawaf
huda nawaf le 9 Sep 2012
no walter, my erlier question is see the cluster visually. Now, I need see the nodes of adjacancy matrix how connect to each other.
that figure is not relate with the above matrix.
please, inform me how I can send image , then others see it. I think I sent image in previous time but nobody saw it.
thanks

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Walter Roberson
Walter Roberson le 9 Sep 2012
If all you have is the adjacency matrix, then even if the weights of the vertices can be interpreted as distances, it is difficult to reconstruct a planar graph. Oh, algebraically it can be done, but it involves solving (N^2)/2 simultaneous non-linear equations, and that gets very time consuming extremely quickly.
So, you might as well just use rand() to generate random x and y coordinates and graph using those.
  4 commentaires
huda nawaf
huda nawaf le 13 Sep 2012
thanks walter, please show me how do that with random coordinates
Walter Roberson
Walter Roberson le 13 Sep 2012
xy = rand(NumberOfNodes, 2);
I suspect that likely NumberOfNodes = size(AdjacencyMatrix, 1);

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