用matlab matconvnet訓練網絡

Question

我想使用matlab和matconvnet-1.0-beta25訓練我的網絡。 我的問題是回歸，我使用pdist作為損失函數來獲取mse。 輸入數據為56*56*64*6000 ，目標數據為56*56*64*6000 ，網絡架構如下：

opts.networkType = 'simplenn' ;
opts = vl_argparse(opts, varargin) ;

lr = [.01 2] ;

% Define network CIFAR10-quick
net.layers = {} ;

% Block 1
net.layers{end+1} = struct('type', 'conv', ...
                           'weights', {{0.01*randn(5,5,64,32, 'single'), zeros(1, 32, 'single')}}, ...
                           'learningRate', lr, ...
                           'stride', 1, ...
                           'pad', 2) ;
net.layers{end+1} = struct('type', 'relu') ;
net.layers{end+1} = struct('type', 'conv', ...
                           'weights', {{0.05*randn(5,5,32,16, 'single'), zeros(1,16,'single')}}, ...
                           'learningRate', .1*lr, ...
                           'stride', 1, ...
                           'pad', 2) ;
net.layers{end+1} = struct('type', 'relu') ;
net.layers{end+1} = struct('type', 'conv', ...
                           'weights', {{0.01*randn(5,5,16,8, 'single'), zeros(1, 8, 'single')}}, ...
                           'learningRate', lr, ...
                           'stride', 1, ...
                           'pad', 2) ;
net.layers{end+1} = struct('type', 'relu') ;
net.layers{end+1} = struct('type', 'conv', ...
                           'weights', {{0.05*randn(5,5,8,16, 'single'), zeros(1,16,'single')}}, ...
                           'learningRate', .1*lr, ...
                           'stride', 1, ...
                           'pad', 2) ;
net.layers{end+1} = struct('type', 'relu') ;
net.layers{end+1} = struct('type', 'conv', ...
                           'weights', {{0.01*randn(5,5,16,32, 'single'), zeros(1, 32, 'single')}}, ...
                           'learningRate', lr, ...
                           'stride', 1, ...
                           'pad', 2) ;
net.layers{end+1} = struct('type', 'relu') ;
net.layers{end+1} = struct('type', 'conv', ...
                           'weights', {{0.05*randn(5,5,32,64, 'single'), zeros(1,64,'single')}}, ...
                           'learningRate', .1*lr, ...
                           'stride', 1, ...
                           'pad', 2) ;
net.layers{end+1} = struct('type', 'relu') ;
% Loss layer
net.layers{end+1} = struct('type', 'pdist') ;

% Meta parameters
net.meta.inputSize = [56 56 64] ;
net.meta.trainOpts.learningRate = [0.0005*ones(1,30) 0.0005*ones(1,10) 0.0005*ones(1,5)] ;
net.meta.trainOpts.weightDecay = 0.0001 ;
net.meta.trainOpts.batchSize = 100 ;
net.meta.trainOpts.numEpochs = numel(net.meta.trainOpts.learningRate) ;

% Fill in default values

net = vl_simplenn_tidy(net) ;

我將ncnn_train （我的名字getSimpleNNBatch(imdb, batch)中的ncnn_train getSimpleNNBatch(imdb, batch)函數更改為：

function [images, labels] = getSimpleNNBatch(imdb, batch)
    images = imdb.images.data(:,:,:,batch) ;
    labels = imdb.images.labels(:,:,:,batch) ;
    if rand > 0.5, images=fliplr(images) ; 
end

因為我的標簽是多維的。 我也將errorFunction中的cnn_train從multiclasses更改為none ：

opts.errorFunction = 'none' ;

並從以下方式更改error變量：

% accumulate errors
error = sum([error, [...
  sum(double(gather(res(end).x))) ;
  reshape(params.errorFunction(params, labels, res),[],1) ; ]],2) ;

至：

% accumulate errors
error = sum([error, [...
  mean(mean(mean(double(gather(res(end).x))))) ;
  reshape(params.errorFunction(params, labels, res),[],1) ; ]],2) ;

我的第一個問題是，為什么上述命令中的res(end).x第三維是一個而不是64？ 這是56*56*1*100 （批量為100）。

我做錯了嗎？

結果如下：

train: epoch 01:   2/ 40: 10.1 (27.0) Hz objective: 21360.722
train: epoch 01:   3/ 40: 13.0 (30.0) Hz objective: 67328685.873
...
train: epoch 01:  39/ 40: 29.7 (29.6) Hz objective: 5179175.587
train: epoch 01:  40/ 40: 29.8 (30.6) Hz objective: 5049697.440
val: epoch 01:   1/ 10: 87.3 (87.3) Hz objective: 49.512
val: epoch 01:   2/ 10: 88.9 (90.5) Hz objective: 50.012
...
val: epoch 01:   9/ 10: 88.2 (88.2) Hz objective: 49.936
val: epoch 01:  10/ 10: 88.1 (87.3) Hz objective: 49.962
train: epoch 02:   1/ 40: 30.2 (30.2) Hz objective: 49.650
train: epoch 02:   2/ 40: 30.3 (30.4) Hz objective: 49.704
...
train: epoch 02:  39/ 40: 30.2 (31.6) Hz objective: 49.739
train: epoch 02:  40/ 40: 30.3 (31.0) Hz objective: 49.722
val: epoch 02:   1/ 10: 91.8 (91.8) Hz objective: 49.687
val: epoch 02:   2/ 10: 92.0 (92.2) Hz objective: 49.831
...
val: epoch 02:   9/ 10: 92.0 (88.5) Hz objective: 49.931
val: epoch 02:  10/ 10: 91.9 (91.1) Hz objective: 49.962
train: epoch 03:   1/ 40: 31.7 (31.7) Hz objective: 49.014
train: epoch 03:   2/ 40: 31.2 (30.8) Hz objective: 49.237
...

這是我的網絡架構圖

Answer 1

的兩個輸入端pdist已經得到nxmx64x100如下尺寸和作為此所提到的，輸出pdist已經得到了相同的高度和寬度，但深度等於一。 關於錯誤定義的正確性，您應該調試並准確檢查大小和定義。