Neural Network Regression Score

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John
John el 1 de Sept. de 2014
Respondida: Greg Heath el 22 de Sept. de 2014
Hey!
So, I'm a little confused about the reported performance of my algorithm that uses the matlab neural network toolbox.
After training/testing with my dataset, I get a great Mean Square Error performance value, and a reasonably high R value on the regression plot (R = ~0.88).
However, when I look at the actual mapping of target and predicted values, it's not quite right. See this plot:
The diagonal dotted line is obviously the ideal output, and the black circles (and black line showing line of best fit) is my actual output. As you can see, all my outputs are negative and not on the diagonal line. However, there does seem to be a decent correlation between the target and actual output scores, hence the decent R value.
Am I just not mapping/scaling the output values correctly? Any tips or insight into this?

Respuesta aceptada

Greg Heath
Greg Heath el 22 de Sept. de 2014
In general,
Linear regression model
y = m*t + b + e % e random error, uncorrelated with t
y0 = m*t0+b +e0 % average
y-y0 = m*(t-t0)+(e-e0) % avg((t-t0)*(e-e0))=0
var(y) = m^2*var(t) + var(e)
Rsquare = var(y)/var(t)
= m^2 + var(e)/var(t)
~ m^2 % var(e) << var(t)
Remember, this is just one way to view the fit
y = net(x);
obtained from
[ net tr y err ] = train(net,x,t) % err=t-y
Hope this helps.
Thank you for formally accepting my answer
Greg

Más respuestas (1)

Greg Heath
Greg Heath el 2 de Sept. de 2014
You say you get a great MSE value however
R^2 = 1-MSE/MSE00 = 0.77
where
MSE00 = mean(var(target',1))
Therefore, your model only accounts for 77% of the mean target variance.
The regression plot looks like the R=0.88 refers to the dark black line and not the line y=t.
Hard to say more without seeing your code.

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