【动手学——多层感知机】day03_multilayer perceptron从零实现

multilayer perceptron多层感知机

一、多层感知机的基本知识

隐藏层
下图展示了一个多层感知机的神经网络图，它含有一个隐藏层，该层中有5个隐藏单元。

含单层隐藏层的多层感知机公式：

从联立后的式子可以看出，虽然神经网络引入隐藏层，但依然等价于一个单层神经网络。不难发现，即便添加再多的隐藏层，以上设计只能等价于仅含输出层的单层神经网络。所以，如果需要对隐藏层的输出作一些非线性变化，就需要引入激活函数。

*多层感知机中引入激活函数的原因：

将多个无激活函数的线性层叠加起来，其表达能力与单个线性层相同，所以，如果需要对隐藏层的输出作一些非线性变化，就需要引入激活函数

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二、使用多层感知机图像分类的从零开始的实现

import torch
import numpy as np
import sys
sys.path.append("/home/kesci/input")
import d2lzh1981 as d2l
print(torch.__version__)

获取数据集

batch_size = 256
train_iter, test_iter = d2l.load_data_fashion_mnist(batch_size,root='/home/kesci/input/FashionMNIST2065')

定义模型参数

num_inputs, num_outputs, num_hiddens = 784, 10, 256

W1 = torch.tensor(np.random.normal(0, 0.01, (num_inputs, num_hiddens)), dtype=torch.float)
b1 = torch.zeros(num_hiddens, dtype=torch.float)
W2 = torch.tensor(np.random.normal(0, 0.01, (num_hiddens, num_outputs)), dtype=torch.float)
b2 = torch.zeros(num_outputs, dtype=torch.float)

params = [W1, b1, W2, b2]
for param in params:
    param.requires_grad_(requires_grad=True)

定义激活函数

#使用ReLU激活函数，比0大则保留，比0小则对应0
def relu(X):
    return torch.max(input=X, other=torch.tensor(0.0))

定义网络

def net(X):
    X = X.view((-1, num_inputs))		#将X的形状改变
    H = relu(torch.matmul(X, W1) + b1)  #用X和权重W作矩阵乘法，加上偏移量b，输入激活函数ReLU中得到H
    return torch.matmul(H, W2) + b2		#再做一次线性的矩阵乘法，加上偏差作为输出

定义损失函数

loss = torch.nn.CrossEntropyLoss()

训练

num_epochs, lr = 5, 100.0
# def train_ch3(net, train_iter, test_iter, loss, num_epochs, batch_size,
#               params=None, lr=None, optimizer=None):
#     for epoch in range(num_epochs):
#         train_l_sum, train_acc_sum, n = 0.0, 0.0, 0
#         for X, y in train_iter:
#             y_hat = net(X)
#             l = loss(y_hat, y).sum()
#             
#             # 梯度清零
#             if optimizer is not None:
#                 optimizer.zero_grad()
#             elif params is not None and params[0].grad is not None:
#                 for param in params:
#                     param.grad.data.zero_()
#            
#             l.backward()
#             if optimizer is None:
#                 d2l.sgd(params, lr, batch_size)
#             else:
#                 optimizer.step()  # “softmax回归的简洁实现”一节将用到
#             
#             
#             train_l_sum += l.item()
#             train_acc_sum += (y_hat.argmax(dim=1) == y).sum().item()
#             n += y.shape[0]
#         test_acc = evaluate_accuracy(test_iter, net)
#         print('epoch %d, loss %.4f, train acc %.3f, test acc %.3f'
#               % (epoch + 1, train_l_sum / n, train_acc_sum / n, test_acc))

d2l.train_ch3(net, train_iter, test_iter, loss, num_epochs, batch_size, params, lr)

三、 使用pytorch的简洁实现

import torch
from torch import nn
from torch.nn import init
import numpy as np
import sys
sys.path.append("/home/kesci/input")
import d2lzh1981 as d2l

print(torch.__version__)

初始化模型和各个参数

num_inputs, num_outputs, num_hiddens = 784, 10, 256
    
net = nn.Sequential(
        d2l.FlattenLayer(),
        nn.Linear(num_inputs, num_hiddens),
        nn.ReLU(),
        nn.Linear(num_hiddens, num_outputs), 
        )
    
for params in net.parameters():
    init.normal_(params, mean=0, std=0.01)

训练

batch_size = 256
train_iter, test_iter = d2l.load_data_fashion_mnist(batch_size,root='/home/kesci/input/FashionMNIST2065')
loss = torch.nn.CrossEntropyLoss()

optimizer = torch.optim.SGD(net.parameters(), lr=0.5)

num_epochs = 5
d2l.train_ch3(net, train_iter, test_iter, loss, num_epochs, batch_size, None, None, optimizer)

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错题：

多层感知机

答：256256的图片总共有256256=65536个元素，与隐层单元元素两两相乘，得到655361000，然后隐层单元元素分别于输出类别个数两两相乘，即100010，最后两者相加得：

655361000 + 100010 = 65546000