Mercurial > pylearn

annotate weights.py @ 524:317a052f9b14

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better main, allow to debug in a debugger.
author Frederic Bastien <bastienf@iro.umontreal.ca>
date Fri, 14 Nov 2008 16:46:03 -0500
parents 4f3c66146f17
children
rev   line source
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1 """
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2 Routine to initialize weights.
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3
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4 @note: We assume that numpy.random.seed() has already been performed.
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5 """
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6
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7 from math import pow, sqrt
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8 import numpy.random
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9
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10 sqrt3 = sqrt(3.0)
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11 def random_weights(nin, nout, scale_by=1./sqrt3, power=0.5):
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12 """
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13 Generate an initial weight matrix with nin inputs (rows) and nout
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14 outputs (cols).
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15 Each weight is chosen uniformly at random to be in range:
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16 [-scale_by*sqrt(3)/pow(nin,power), +scale_by*sqrt(3)/pow(nin,power)]
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17 @note: Play with scale_by, but reasonable values are <=1, maybe 1./sqrt3
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18 power=0.5 is strongly recommanded (see below).
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19
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20 Suppose these weights w are used in dot products as follows:
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21 output = w' input
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22 If w ~ Uniform(-r,r) and Var[input_i]=1 and x_i's are independent, then
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23 Var[w]=r2/3
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24 Var[output] = Var[ sum_{i=1}^d w_i input_i] = d r2 / 3
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25 To make sure that variance is not changed after the dot product,
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26 we therefore want Var[output]=1 and r = sqrt(3)/sqrt(d). This choice
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27 corresponds to the default values scale_by=sqrt(3) and power=0.5.
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28 More generally we see that Var[output] = Var[input] * scale_by.
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29
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30 Now, if these are weights in a deep multi-layer neural network,
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31 we would like the top layers to be initially more linear, so as to let
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32 gradients flow back more easily (this is an explanation by Ronan Collobert).
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33 To achieve this we want scale_by smaller than 1.
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34 Ronan used scale_by=1/sqrt(3) (by mistake!) and got better results than scale_by=1
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35 in the experiment of his ICML'2008 paper.
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36 Note that if we have a multi-layer network, ignoring the effect of the tanh non-linearity,
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37 the variance of the layer outputs would go down roughly by a factor 'scale_by' at each
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38 layer (making the layers more linear as we go up towards the output).
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39 """
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40 return (numpy.random.rand(nin, nout) * 2.0 - 1) * scale_by * sqrt3 / pow(nin,power)