Radial Basis Function with "newrb" dimensional problem
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I'm trying to use the function "newrb" for generating an output considering a set of initial inputs and the RBF.
My problem is that as input x I have 5 set of 8 elements (8 coordinates in a 8D space), while as output y I would like to obtain just 1 value (the corresponding 9th coordinates). However, the function said that y must have the same dimension of the value used for x.
To better explain myself, if we consider just 1 set, which has 8 coordinates, I want to use the neural network for obtaining just 1 final value, but I am not allowed to do this since "newrb" required an y with the same dimension of x.
How can I solve this issue??
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nick
2024 年 10 月 8 日
The "newrb" function requires the target matrix T to have the same number of columns as the input matrix P. This means that if your input matrix P has dimensions [R x Q], your target matrix T should have dimensions [S x Q], where Q is the number of samples.
The following MATLAB code shows RBF network to predict a single value (the 9th coordinate) for 5 sets using 8 elements (8 coordinates in an 8D space) of each set:
% Sample input data
P = rand(8, 5); % 5 sets of 8 elements (8 coordinates in an 8D space)
% Sample output data
T = rand(1, 5); % 9th coordinate for each set
net = newrb(P, T);
view(net);
You may refer to the following documentation to learn more about "newrb" function :
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