neural network input .
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i have these code with these input to my neural network , my question is how does the NN take the input ? to be more clear is NN going to take p1[1,1], p2[1,1],p3[1,1].....p13[1,1]; or it will take the whole p1 and after that the whole p2 , p3 ...,p13?
my code is
clear;
clc;
load asp65 ;
load curv65;
load dfd65;
load dffl65;
load dfr65;
load gl65;
load lc65;
load prc65;
load plc65;
load pfc65;
load slop65;
load sot65;
load vec65;
load ele65;
load hlo65;
p1=transpose(asp65);
p2=transpose(curv65);
p3=transpose(dfd65);
p4=transpose(dffl65);
p5=transpose(dfr65);
p6=transpose(gl65);
p7=transpose(lc65);
p8=transpose(prc65);
p9=transpose(plc65);
p10=transpose(pfc65);
p11=transpose(slop65);
p12=transpose(vec65);
p13=transpose(ele65);
t1=transpose(hlo65);
p=[p1;p2;p3;p4;p5;p6;p7;p8;p9;p10;p11;p12;p13];
%p=[p1,p2,p3,p4,p5,p6,p7,p8,p9,p10,p11,p12,p13]
%p15=[p7;p8;p9;p10;p11;12;13];
t=[t1];
%p = [-1 -1 2 2 ;0 5 0 5 ]
%t = [-1 -1 1 1 ];
for i=1:1:10
net=newff(minmax(p),t,[15 , 4],{'tansig','purelin'},'trainrp');
net.performFcn='msereg';
net.performParam.ratio=0.5;
net.trainParam.show = 5;
net.trainParam.lr = 0.5; % learning rate
net.trainParam.epochs = 1000;
net.trainParam.goal = 1e-5;
RandStream.setDefaultStream(RandStream('mt19937ar','seed',1));
%rand('state',sum(100*clock)) % initialize therandom
net = init(net);
[net,tr]=train(net,p,t);
a = sim(net,p);
e = a-t;
rmse = sqrt(mse(e));
acc=1-rmse
%simpleclusterOutputs = sim(net,p);
%plotroc(t,simpleclusterOutputs);
end
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更多回答（7 个）
Greg Heath
2011-12-14
0 个投票
See comments, examples and sample code by searching tne newsgroup using
heath newff close clear
heath newff Neq Nw Ntrials
Hope this helps.
Greg
Greg Heath
2011-12-15
>i have these code with these input to my neural network , my question is how does the
>NN take the input ? to be more clear is NN going to take p1[1,1], p2[1,1],p3[1,1].....
>p13[1,1]; or it will take the whole p1 and after that the whole p2 , p3 ...,p13?
It is hard to answer your question without knowing the sizes of the pi (i=1:13).
For an I-H-O node topology and Ntrn I/O training pairs,
[I Ntrn] = size(p)
[O Ntrn] = size(t)
The I/O vectors are columns of p and t.
>my code is clear; clc; load asp65 ; load curv65; load dfd65; load dffl65; load dfr65; >load gl65; load lc65; load prc65; load plc65; load pfc65; load slop65; load sot65; >load vec65; load ele65; load hlo65;
>p1=transpose(asp65); p2=transpose(curv65); p3=transpose(dfd65); p4=transpose(dffl65); >p5=transpose(dfr65); p6=transpose(gl65); p7=transpose(lc65); p8=transpose(prc65); >p9=transpose(plc65); p10=transpose(pfc65); p11=transpose(slop65); p12=transpose(vec65); > p13=transpose(ele65); t1=transpose(hlo65);
>>p=[p1;p2;p3;p4;p5;p6;p7;p8;p9;p10;p11;p12;p13]; %p=[p1,p2,p3,p4,p5,p6,p7,p8,p9,p10,p11,p12,p13] >%p15=[p7;p8;p9;p10;p11;12;13]; t=[t1]; %p = [-1 -1 2 2 ;0 5 0 5 ] %t = [-1 -1 1 1 ];
Size(p)? size(t1)?
Normalization of p and t1?
H = 15, O = 4?
Why 15? Better to search for the smallest good value satisfying
H <= floor( (Ntrn*O-O)/(I+O+1) )
Add an outer loop over candidate H values.
> RandStream.setDefaultStream(RandStream('mt19937ar','seed',1));
Only initialize once BEFORE the loops!
%rand('state',sum(100*clock)) % initialize the random net = init(net);
If you ever do this, replace 100 by 1e9. Don't need INIT because NEWFF is self-initializing.
>for i=1:1:10
>>net=newff(minmax(p),t,[H , O],{'tansig','purelin'},'trainrp');
>net.performFcn='msereg'; net.performParam.ratio=0.5;
Why MSEREG?? Why not use as many defaults as possible?
>net.trainParam.show = 5;
>net.trainParam.lr = 0.5; % learning rate
>net.trainParam.epochs = 1000;
>net.trainParam.goal = 1e-5;
Why 1e-5??
>[net,tr]=train(net,p,t); > a = sim(net,p); e = a-t;
Why not [net, tr, a, e] = ... ?
>rmse = sqrt(mse(e)); >acc=1-rmse
Useless performance index. Use R^2.
Hope this helps.
Greg
%simpleclusterOutputs = sim(net,p);
%plotroc(t,simpleclusterOutputs);
end
Greg Heath
2011-12-19
0 个投票
>1-i have these code with these input to my neural network , my question is how does the NN take the input ?
Once the matrix p is created, the net takes the first column of p, then the second column of p, etc. If
[ I N ] = size(p),
then p is interpreted as containing N I-dimensional column vectors. Since
size(pi) = [ 255 255] % i = 1:13
you need to columnize your pi matrices and form
p = [ p1(:), p2(:), ..., p13(:)];
THEN
size(p) = [ 255^2 13] = [ 65025 133 ]
AND
you need to investigate 2-D dimensionality reduction via averaging pixels.
====================================================
>2. p=[p1;p2;p3;p4;p5;p6;p7;p8;p9;p10;p11;p12;p13];
Incorrect. See above
Size(p)? size(t1)?
> the size of p is (13 x255)=3315 elements.
Incorrect. See above.
>the size of (t1) is 255
Incorrect. Size has two dimensions. See the documentation
help size
doc size
The correct answer is
size(t1) = [ O N ]
and is interpreted as N O-dimensional output column vectors.
H = 15, O = 4?
>one hidden lair (sp: layer) has 15 neuron based on previous researches . out put in the target has two values either 1 or 0 so once i put the output 4 i achieved much better regression and accuracy values .
I don't understand. What do the 4 outputs represent?
Oh! Oh! Is this classification and you are using a binary coded output?
For classification use
size(t1) = [ 13 N]
where each column is a column of the 13 dimensional unit matrix eye(13). The input image vector is then assigned to the class corresponding to the maximum output.
==================================================================
Greg Heath
2011-12-19
...continued
>3-net.performFcn='msereg'; net.performParam.ratio=0.5;
Why MSEREG?? Why not use as many defaults as possible?
>if i keep the default my accuracy and regression will R=0.333 and the accuracy 90 but if i use net.performFcn='msereg' the R =.6999 and the accuracy 96%
Something is wrong somewhere.
How is R defined? Square root of the coefficient of determination (See Wikipedia)?
Try the default again with the columnized image inputs and unit matrix column outputs
===================================
4-net.trainParam.goal = 1e-5;
Why 1e-5??
> its the best value i can get better performance result through .
No need to search for a suitable value. Just choose a goal for R. Then work backwords to find the corresponding goal for MSE.
======================================================
5->[net,tr]=train(net,p,t); > a = sim(net,p); e = a-t;
Why not [net, tr, a, e] = ... ?
what is the wrong with my code why yours is better ?
Nothing is wrong with your code. Just letting you know that the additional code and computations are unnecessary.
===================================================
6->rmse = sqrt(mse(e)); acc=1-rmse
Useless scale-dependent performance index. Use R^2.
> what is R^2?
A common statistical measure of fit. Look up coefficient of determination in Wikipedia.
============================
Hope this helps
Greg
Mo al
2011-12-20
0 个投票
2 个评论
Walter Roberson
2011-12-20
No. If your variable is the wrong way around, transpose it as you pass it in, using p.' or transpose(p)
Mo al
2012-8-31
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