如果想了解一下，手动计算的结果能否和pytorch计算的结果对应起来，那么下面的代码可能参考一下。
这个例子来自官网，因为是示教的，网络结构极为简单
https://pytorch.org/tutorials/beginner/examples_nn/two_layer_net_nn.html
我对其中的部分源码进行了修改，加了一些手动计算的源码，并且每一个输出都写出了torch.nn网络计算结果和手动计算结果的比较，一步一步调试，很容易明白其back-propagation 是如何实现的，如果串级有更多的网络层，原理也相同，只不过把上一步计算的结果，放到前一层中避免重复计算而已。
我刻意找了一个以前看到过的贴子，供大家参考，这里back-proparation描述给人印象深刻，
https://ml-cheatsheet.readthedocs.io/en/latest/backpropagation.html
源码如下，运行时最好单步调试，不然是学不到什么的；共计500步是源码中的，运行时最好改成了5步或3步，对了解原理来说跑一两次就够了，详细的计算步骤在源码里我作了注释，
# -*- coding: utf-8 -*-
"""
PyTorch: nn
-----------
A fully-connected ReLU network with one hidden layer, trained to predict y from x
by minimizing squared Euclidean distance.
This implementation uses the nn package from PyTorch to build the network.
PyTorch autograd makes it easy to define computational graphs and take gradients,
but raw autograd can be a bit too low-level for defining complex neural networks;
this is where the nn package can help. The nn package defines a set of Modules,
which you can think of as a neural network layer that has produces output from
input and may have some trainable weights.
"""
import torch
# N is batch size; D_in is input dimension;
# H is hidden dimension; D_out is output dimension.
N, D_in, H, D_out = 2, 3, 4, 5
# Create random Tensors to hold inputs and outputs
x = torch.randn(N, D_in)
y = torch.randn(N, D_out)
#x = torch.tensor( [[-0.2497, 2.0979, 1.7150], [ 0.6786, 0.4429, 0.7582]])
#y = torch.tensor([[-0.0217, 0.8911], [-1.0743, -1.1462]])
print(x)
print(y)
# Use the nn package to define our model as a sequence of layers. nn.Sequential
# is a Module which contains other Modules, and applies them in sequence to
# produce its output. Each Linear Module computes output from input using a
# linear function, and holds internal Tensors for its weight and bias.
model = torch.nn.Sequential(
torch.nn.Linear(D_in, H),
torch.nn.ReLU(),
torch.nn.Linear(H, D_out),
)
print("list model parameters...")
print(list(model.parameters())) # print(list(model.parameters()))
# The nn package also contains definitions of popular loss functions; in this
# case we will use Mean Squared Error (MSE) as our loss function.
loss_fn = torch.nn.MSELoss(reduction='sum')
learning_rate = 1e-4
for t in range(500):
# Forward pass: compute predicted y by passing x to the model. Module objects
# override the __call__ operator so you can call them like functions. When
# doing so you pass a Tensor of input data to the Module and it produces
# a Tensor of output data.
y_pred = model(x)
print(y_pred)
# Compute and print loss. We pass Tensors containing the predicted and true
# values of y, and the loss function returns a Tensor containing the
# loss.
loss = loss_fn(y_pred, y)
print(t,"loss_fn", loss.item())
# Zero the gradients before running the backward pass.
model.zero_grad()
# Backward pass: compute gradient of the loss with respect to all the learnable
# parameters of the model. Internally, the parameters of each Module are stored
# in Tensors with requires_grad=True, so this call will compute gradients for
# all learnable parameters in the model.
loss.backward()
re = (y_pred-y)
res = sum(sum(re*re))
print(loss)
print(re)
print("------------------------------------------------------------------")
# Update the weights using gradient descent. Each parameter is a Tensor, so
# we can access its gradients like we did before.
pv = [] # parameter values
pgrad = [] # parameter values gradient
pvt = [] # parameter values transposed
with torch.no_grad():
for param in model.parameters():
pv.append(param)
pgrad.append(param.grad)
param -= learning_rate * param.grad
#print(param.grad)
#print(param)
# ====================================================================
# ---- now calculate the back propagation of the network manually ----
pvt.append(torch.transpose(pv[0],0,1))
pvt.append(pv[1].view(1,H))
pvt.append(torch.transpose(pv[2],0,1))
pvt.append(pv[3].view(1,D_out))
"""
assume the previous network is represented by the following parameters
x*P0 + P1 = x2
Relu(x2) = x4
x4*p2 + p3 = x6
y_pred == x6
"""
x1 = torch.matmul(x,pvt[0]) # the same as torch.mm matrix mulplication
x2 = x1 + pvt[1]
x3 = (x2>0)
x3 = x3.float()
x4 = x3*x2 # dot prduct (inner product) #don't use torch.mm(x3,x2)
x5 = torch.matmul(x4,pvt[2])
x6 = x5 + pvt[3]
print("-----x6 ", x6)
print("--y_pred: ", y_pred)
"""
d(x6)/d(p3) = [1,1,...] ===>>> pgrad[3] = res
d(x6)/d(p2) = x4 ===>>> pgrad[2] = rex4 = re * x4
"""
re = 2*(x6-y) # Loss = (y_pred-y)^2, there is no factor 1/2, so the derivative is Loss'=(y_pred-y)
# So, there is a 2 factor before the calculation
res = sum(re)
print("-----res ", res)
print("--pgrad[3]: ", pgrad[3])
rex4 = torch.mm(re.transpose(0,1), x4) # = 2*torch.mm(x4.transpose(0,1), re).transpose(0,1)
#rer = res.repeat(2).view(2,2).transpose(0,1) # you can test this line if needed
print("-----rex4 ", rex4)
print("--pgrad[2]: ", pgrad[2])
"""
in back-proparation: x2 = x4
d(x2)/d(p1) = [1,1, ....]
d(x6)/d(x4) * d(x2)/d(p1) = pvt[2] ===>>> pgrad[1]
d(x2)/d(p0) = x
d(x6)/d(x4) * d(x2)/d(p0) = pvt[2]*x ===>>> pgrad[0] = re*pvt[2]*x
"""
rex2 = torch.mm(re, pvt[2].transpose(0,1))*x3 # dims =(2x5.5x4)*2x4=2x4, * means dot product and . means matrix product
regx2 = torch.mm(rex2.transpose(0,1), x) # dims =4x2.2x3=4x3
print("-----regx2 ", regx2)
print("--pgrad[0]: ", pgrad[0])
print("-----sum(rex2) ", sum(rex2))
print("--pgrad[1]: ", pgrad[1])
# set a break point here to see the results
print("-----------------------")
"""
let's view this problem from the dimensios points of view
NOTE: * means dot product and . means matrix product
____________________________________________
func |dimension
------------------------|-------------------
x.p0 + p1 = x2 |2x3.3x4 + 2 = 2x4
x4 = relu(x2) |2x4 ==> 2x4
x4.p2 + p3 = x6 |2x4.4x5 + 2 = 2x5
re = x6- y |2x5
------------------------|-------------------
___________________________________________________________________________
grad(p2) = re^T.d(x6)/d(p2) = re^T.x4 |5x2.2x4=5x4
grad(p3) = sum_rows(re) |2x1
grad(p0) = re.d(x6)/d(x4).d(x2)/d(p0) |(2x5.4x5^T)^T.2x3 =(4x2).2x3 = 4x3
= (re.p2^T*Relu)^T.x | = (4x2).relu.2x3
grad(p1) = re.d(x6)/d(x4).d(x2)/d(p1) |(2x5.4x5^T)^T.2x3 =(4x2)
= (re.p2^T*ReLU)^T |
----------------------------------------|----------------------------------
"""
