Reputation: 426
I have trained NN for Regression problem. my data type is HDF5_DATA
that made of .jpg
images (3X256X256) and float-label array (3 labels). Data-Set create script:
import h5py, os
import caffe
import numpy as np
SIZE = 256 # images size
with open( '/home/path/trainingTintText.txt', 'r' ) as T :
lines = T.readlines()
X = np.zeros( (len(lines), 3, SIZE, SIZE), dtype='f4' )
labels = np.zeros( (len(lines),3), dtype='f4' )
for i,l in enumerate(lines):
sp = l.split(' ')
img = caffe.io.load_image( sp[0] )
img = caffe.io.resize( img, (SIZE, SIZE, 3) )
transposed_img = img.transpose((2,0,1))[::-1,:,:] # RGB->BGR
X[i] = transposed_img*255
print X[i]
labels[i,0] = float(sp[1])
labels[i,1] = float(sp[2])
labels[i,2] = float(sp[3])
with h5py.File('/home/path/train.h5','w') as H:
H.create_dataset('data', data=X)
H.create_dataset('label', data=labels)
with open('/home/path/train_h5_list.txt','w') as L:
L.write( '/home/path/train.h5' )
this is (not fullish) architecture:
name: "NN"
layers {
name: "NNd"
top: "data"
top: "label"
type: HDF5_DATA
hdf5_data_param {
source: "/home/path/train_h5_list.txt"
batch_size: 64
}
include: { phase: TRAIN }
}
layers {
name: "data"
type: HDF5_DATA
top: "data"
top: "label"
hdf5_data_param {
source: "/home/path/train_h5_list.txt"
batch_size: 100
}
include: { phase: TEST }
}
layers {
name: "conv1"
type: CONVOLUTION
bottom: "data"
top: "conv1"
convolution_param {
num_output: 32
kernel_size: 11
stride: 2
bias_filler {
type: "constant"
value: 0.1
}
}
}
layers {
name: "ip2"
type: INNER_PRODUCT
bottom: "ip1"
top: "ip2"
inner_product_param {
num_output: 3
bias_filler {
type: "constant"
value: 0.1
}
}
}
layers {
name: "relu22"
type: RELU
bottom: "ip2"
top: "ip2"
}
layers {
name: "loss"
type: EUCLIDEAN_LOSS
bottom: "ip2"
bottom: "label"
top: "loss"
}
when I train the NN I got very high loss values:
I1117 08:15:57.707001 2767 solver.cpp:337] Iteration 0, Testing net (#0)
I1117 08:15:57.707033 2767 net.cpp:684] Ignoring source layer fkp
I1117 08:15:59.111842 2767 solver.cpp:404] Test net output #0: loss = 256.672 (* 1 = 256.672 loss)
I1117 08:15:59.275205 2767 solver.cpp:228] Iteration 0, loss = 278.909
I1117 08:15:59.275255 2767 solver.cpp:244] Train net output #0: loss = 278.909 (* 1 = 278.909 loss)
I1117 08:15:59.275276 2767 sgd_solver.cpp:106] Iteration 0, lr = 0.01
I1117 08:16:57.115145 2767 solver.cpp:337] Iteration 100, Testing net (#0)
I1117 08:16:57.115486 2767 net.cpp:684] Ignoring source layer fkp
I1117 08:16:58.884704 2767 solver.cpp:404] Test net output #0: loss = 238.257 (* 1 = 238.257 loss)
I1117 08:16:59.026926 2767 solver.cpp:228] Iteration 100, loss = 191.836
I1117 08:16:59.026971 2767 solver.cpp:244] Train net output #0: loss = 191.836 (* 1 = 191.836 loss)
I1117 08:16:59.026993 2767 sgd_solver.cpp:106] Iteration 100, lr = 0.01
I1117 08:17:56.890614 2767 solver.cpp:337] Iteration 200, Testing net (#0)
I1117 08:17:56.890880 2767 net.cpp:684] Ignoring source layer fkp
I1117 08:17:58.665057 2767 solver.cpp:404] Test net output #0: loss = 208.236 (* 1 = 208.236 loss)
I1117 08:17:58.809150 2767 solver.cpp:228] Iteration 200, loss = 136.422
I1117 08:17:58.809248 2767 solver.cpp:244] Train net output #0: loss = 136.422 (* 1 = 136.422 loss)
when I divide the images and the label arrays by 255
I got very low loss results (neat to 0
). what is the reason for those loss results? am I doing something wrong? thanks
Upvotes: 1
Views: 359
Reputation: 3604
With the Euclidean loss, this is only to be expected. The Euclidean loss should be smaller by a factor of 256 if you divide all of the labels by 256 and re-train. It doesn't mean that dividing the labels by 256 makes the network become any better at predicting the labels; you've just changed the "scale" (the "units").
In particular, the Euclidean loss is (roughly) L = sqrt((x1 -y1)2 + (x2 -y2)2), where x is the correct answer and y is the output from the neural network. Suppose you divide every x by 256, then re-train. The neural network will learn to divide its output y by 256. How will this affect the Euclidean loss L? Well, if you work through the math, you'll find that L shrinks by a factor of 256.
It'd be like the difference between trying to predict a distance in feet, vs a distance in yards. The latter would involve dividing by 3. Conceptually, the overall accuracy of the network would remain the same; but the Euclidean loss would be divided by a factor of three, because you've changed the units from yards to meters. An average error of 0.1 feet would correspond to an average error of 0.0333 yards; but conceptually yield the "same" accuracy, even though 0.0333 looks like a smaller number than 0.1.
Dividing the images by 256 should be irrelevant. It's dividing the labels by 256 that caused the reduction in the loss function.
Upvotes: 2