Neural Network Regression Score

6 次查看（过去 30 天）
John
John 2014-9-1
回答： Greg Heath 2014-9-22
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:
http://i.imgur.com/flJ9a0p.png
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?

请先登录，再回答此问题。

采纳的回答

Greg Heath
Greg Heath 2014-9-22
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

更多回答（1 个）

Greg Heath
Greg Heath 2014-9-2
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.

类别

在 Help Center 和 File Exchange 中查找有关 Sequence and Numeric Feature Data Workflows 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by