I have seen one can define a custom loss layer for example EuclideanLoss in caffe like this:
import caffe
import numpy as np
class EuclideanLossLayer(caffe.Layer):
"""
Compute the Euclidean Loss in the same manner as the C++
EuclideanLossLayer
to demonstrate the class interface for developing layers in Python.
"""
def setup(self, bottom, top):
# check input pair
if len(bottom) != 2:
raise Exception("Need two inputs to compute distance.")
def reshape(self, bottom, top):
# check input dimensions match
if bottom[0].count != bottom[1].count:
raise Exception("Inputs must have the same dimension.")
# difference is shape of inputs
self.diff = np.zeros_like(bottom[0].data, dtype=np.float32)
# loss output is scalar
top[0].reshape(1)
def forward(self, bottom, top):
self.diff[...] = bottom[0].data - bottom[1].data
top[0].data[...] = np.sum(self.diff**2) / bottom[0].num / 2.
def backward(self, top, propagate_down, bottom):
for i in range(2):
if not propagate_down[i]:
continue
if i == 0:
sign = 1
else:
sign = -1
bottom[i].diff[...] = sign * self.diff / bottom[i].num
However, I have a few question regarding that code:
If I want to customise this layer and change the computation of the loss in this line:
top[0].data[...] = np.sum(self.diff**2) / bottom[0].num / 2.
Lets say to:
channelsAxis = bottom[0].data.shape[1]
self.diff[...] = np.sum(bottom[0].data, axis=channelAxis) - np.sum(bottom[1].data, axis=channelAxis)
top[0].data[...] = np.sum(self.diff**2) / bottom[0].num / 2.
How do I have to change the backward function? For EuclideanLoss it is:
bottom[i].diff[...] = sign * self.diff / bottom[i].num
How does it have to look for my described loss?
What is the sign for?
Although it can be a very educating exercise to implement the loss you are after as a "Python"
layer, you can get the same loss using existing layers. All you need is to add a "Reduction"
layer for each of your blobs before calling the regular "EuclideanLoss"
layer:
layer {
type: "Reduction"
name: "rx1"
bottom: "x1"
top: "rx1"
reduction_param { axis: 1 operation: SUM }
}
layer {
type: "Reduction"
name: "rx2"
bottom: "x2"
top: "rx2"
reduction_param { axis: 1 operation: SUM }
}
layer {
type: "EuclideanLoss"
name: "loss"
bottom: "rx1"
bottom: "rx2"
top: "loss"
}
Update:
Based on your comment , if you only want to sum over the channel dimension and leave all other dimensions unchanged, you can use fixed 1x1 conv (as you suggested):
layer {
type: "Convolution"
name: "rx1"
bottom: "x1"
top: "rx1"
param { lr_mult: 0 decay_mult: 0 } # make this layer *fixed*
convolution_param {
num_output: 1
kernel_size: 1
bias_term: 0 # no need for bias
weight_filler: { type: "constant" value: 1 } # sum
}
}
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