Default number of epochs
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Is it true that the default number of epochs in
[net, tr] = train(net,trainV,trainT)
is 1000?
When looking at help information on other training functions like trainlm it shows that default values of its training parameters are:
net.trainParam.epochs 100 Maximum number of epochs to train
Who can give a clear answer on this question. Thanks in advance, Ton S.
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Thomas Kanzig
2016년 1월 9일
Its true my friend. You must insert a biggest number for obtain a better trainings.
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Greg Heath
2012년 9월 20일
편집: Walter Roberson
2016년 7월 12일
In the good ol' days it used to be 100. Now it seems to be 1000.
The info in the documentation of trainlm, trainscg and probably others, needs to be updated.
clear all, clc,
x = randn(1,100);
t = x.^2;
net = newff(x,t,10);
trainfunc = net.trainFcn % trainlm
numepochs = net.trainParam.epochs % 1000
net = newfit(x,t,10);
trainfunc = net.trainFcn % trainlm
numepochs = net.trainParam.epochs % 1000
net = newpr(x,t,10);
trainfunc = net.trainFcn % trainscg
numepochs = net.trainParam.epochs % 1000
net = fitnet(10);
trainfunc = net.trainFcn % trainlm
numepochs = net.trainParam.epochs % 1000
net = patternnet(10);
trainfunc = net.trainFcn % trainscg
numepochs = net.trainParam.epochs % 1000
net = feedforwardnet(10);
trainfunc = net.trainFcn % trainlm
numepochs = net.trainParam.epochs % 1000
Hope this helps.
Thank you for officially accepting my answer.
Greg
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Greg Heath
2012년 9월 20일
The documentation in help trainlm and help trainscg contains these two statements:
1. Training occurs according to training parameters, with default values
2. Any or all of these can be overridden with parameter name/value argument pairs appended to the input argument list, or by appending a structure argument with fields having one or more of these names.
Athough doc traincsg and doc trainlm do not contain the second statement, they have the CORRECT VERSION of the first statement:
1. Training occurs according to training parameters, shown here with default values
Hope this helps.
Greg
P.S. Don't forget to formally accept my answer
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