【问题标题】:What is the difference between MLP implementation from scratch and in PyTorch?从头开始实现 MLP 和在 PyTorch 中实现有什么区别?
【发布时间】:2019-06-12 07:33:57
【问题描述】:

跟进How to update the learning rate in a two layered multi-layered perceptron?的问题

鉴于 XOR 问题:

X = xor_input = np.array([[0,0], [0,1], [1,0], [1,1]])
Y = xor_output = np.array([[0,1,1,0]]).T

还有一个简单的

  • 两层多层感知器 (MLP) 与
  • 它们和
  • 之间的 sigmoid 激活
  • 均方误差 (MSE) 作为损失函数/优化标准

如果我们这样从头开始训练模型:

from itertools import chain
import matplotlib.pyplot as plt
import numpy as np
np.random.seed(0)

def sigmoid(x): # Returns values that sums to one.
    return 1 / (1 + np.exp(-x))

def sigmoid_derivative(sx):
    # See https://math.stackexchange.com/a/1225116
    return sx * (1 - sx)

# Cost functions.
def mse(predicted, truth):
    return 0.5 * np.mean(np.square(predicted - truth))

def mse_derivative(predicted, truth):
    return predicted - truth

X = xor_input = np.array([[0,0], [0,1], [1,0], [1,1]])
Y = xor_output = np.array([[0,1,1,0]]).T

# Define the shape of the weight vector.
num_data, input_dim = X.shape
# Lets set the dimensions for the intermediate layer.
hidden_dim = 5
# Initialize weights between the input layers and the hidden layer.
W1 = np.random.random((input_dim, hidden_dim))

# Define the shape of the output vector. 
output_dim = len(Y.T)
# Initialize weights between the hidden layers and the output layer.
W2 = np.random.random((hidden_dim, output_dim))

# Initialize weigh
num_epochs = 5000
learning_rate = 0.3

losses = []

for epoch_n in range(num_epochs):
    layer0 = X
    # Forward propagation.

    # Inside the perceptron, Step 2. 
    layer1 = sigmoid(np.dot(layer0, W1))
    layer2 = sigmoid(np.dot(layer1, W2))

    # Back propagation (Y -> layer2)

    # How much did we miss in the predictions?
    cost_error = mse(layer2, Y)
    cost_delta = mse_derivative(layer2, Y)

    #print(layer2_error)
    # In what direction is the target value?
    # Were we really close? If so, don't change too much.
    layer2_error = np.dot(cost_delta, cost_error)
    layer2_delta = cost_delta *  sigmoid_derivative(layer2)

    # Back propagation (layer2 -> layer1)
    # How much did each layer1 value contribute to the layer2 error (according to the weights)?
    layer1_error = np.dot(layer2_delta, W2.T)
    layer1_delta = layer1_error * sigmoid_derivative(layer1)

    # update weights
    W2 += - learning_rate * np.dot(layer1.T, layer2_delta)
    W1 += - learning_rate * np.dot(layer0.T, layer1_delta)
    #print(np.dot(layer0.T, layer1_delta))
    #print(epoch_n, list((layer2)))

    # Log the loss value as we proceed through the epochs.
    losses.append(layer2_error.mean())
    #print(cost_delta)


# Visualize the losses
plt.plot(losses)
plt.show()

从 epoch 0 开始,我们的损失急剧下降,然后迅速饱和:

但是如果我们用pytorch训练一个类似的模型,训练曲线在饱和之前损失会逐渐下降:

从头开始的 MLP 和 PyTorch 代码有什么区别?

为什么会在不同点收敛?

除了权重初始化、代码中的np.random.rand() 和默认的手电筒初始化之外,我似乎看不出模型有什么不同。

PyTorch 代码:

from tqdm import tqdm
import numpy as np

import torch
from torch import nn
from torch import tensor
from torch import optim

import matplotlib.pyplot as plt

torch.manual_seed(0)
device = 'gpu' if torch.cuda.is_available() else 'cpu'

# XOR gate inputs and outputs.
X = xor_input = tensor([[0,0], [0,1], [1,0], [1,1]]).float().to(device)
Y = xor_output = tensor([[0],[1],[1],[0]]).float().to(device)


# Use tensor.shape to get the shape of the matrix/tensor.
num_data, input_dim = X.shape
print('Inputs Dim:', input_dim) # i.e. n=2 

num_data, output_dim = Y.shape
print('Output Dim:', output_dim) 
print('No. of Data:', num_data) # i.e. n=4

# Step 1: Initialization. 

# Initialize the model.
# Set the hidden dimension size.
hidden_dim = 5
# Use Sequential to define a simple feed-forward network.
model = nn.Sequential(
            # Use nn.Linear to get our simple perceptron.
            nn.Linear(input_dim, hidden_dim),
            # Use nn.Sigmoid to get our sigmoid non-linearity.
            nn.Sigmoid(),
            # Second layer neurons.
            nn.Linear(hidden_dim, output_dim),
            nn.Sigmoid()
        )
model

# Initialize the optimizer
learning_rate = 0.3
optimizer = optim.SGD(model.parameters(), lr=learning_rate)

# Initialize the loss function.
criterion = nn.MSELoss()

# Initialize the stopping criteria
# For simplicity, just stop training after certain no. of epochs.
num_epochs = 5000 

losses = [] # Keeps track of the loses.

# Step 2-4 of training routine.

for _e in tqdm(range(num_epochs)):
    # Reset the gradient after every epoch. 
    optimizer.zero_grad() 
    # Step 2: Foward Propagation
    predictions = model(X)

    # Step 3: Back Propagation 
    # Calculate the cost between the predictions and the truth.
    loss = criterion(predictions, Y)
    # Remember to back propagate the loss you've computed above.
    loss.backward()

    # Step 4: Optimizer take a step and update the weights.
    optimizer.step()

    # Log the loss value as we proceed through the epochs.
    losses.append(loss.data.item())


plt.plot(losses)

【问题讨论】:

  • 如果您能告诉我们在手动编码示例中潜水有多锐利,这可能会有所帮助。 2个时代? 20?对我来说,图表的明显解释是学习率在某种程度上非常不同。 (另外,作为单独的说明:MSE 损失在这里可能不是适当的错误函数,并且您将希望在实践中对 $[0, 1]$ 中的输出使用负对数损失/交叉熵损失,但是对于这么简单的问题,它并不重要,当然它与问题并不特别相关。)
  • 你的从头开始的代码抛出---> 60 layer1_error = np.dot(layer2_delta, W2.T) ..... ValueError: shapes (4,50) and (1,5) not aligned: 50 (dim 1) != 1 (dim 0)
  • @alvas 在训练结束时,理想情况下你的损失应该在0.0 左右,对吧?这是否意味着 PyTorch 代码有问题? @coldspeed 我能够从从头开始的代码中重现 OP 的结果。当你运行它时,layer2_delta 似乎以某种方式以(4, 50) 结束(对我来说layer2_delta.shape(4,1))。
  • 好的,花了一点时间,但我想出了如何让您的手动代码产生与 Pytorch 代码相同的结果。有 4 个显着差异需要说明。这都是一些小的调整,所以看起来你的手工代码的核心是好的(除了必须加倍学习率的东西:这可能是某个地方的数学错误)。

标签: python numpy neural-network deep-learning pytorch


【解决方案1】:

手卷代码和 PyTorch 代码的区别列表

事实证明,您的手动代码和 PyTorch 代码所做的事情之间存在很多差异。以下是我发现的内容,大致按对输出的影响从大到小的顺序列出:

  • 您的代码和 PyTorch 代码使用两个不同的函数来报告丢失。
  • 您的代码和 PyTorch 代码设置的初始权重非常不同。您在问题中提到了这一点,但事实证明它对结果产生了相当大的影响。
  • 默认情况下,torch.nn.Linear 层会为模型添加额外的一堆“偏差”权重。因此,Pytorch 模型的第一层有效地具有3x5 权重,而第二层具有6x1 权重。手卷代码中的层分别具有2x55x1 权重。
    • 偏差似乎有助于模型更快地学习和适应。如果关闭偏差,Pytorch 模型需要大约两倍的训练 epoch 才能达到0 损失附近。
  • 奇怪的是,看起来 Pytorch 模型使用的学习率实际上是您指定的一半。或者,也可能是 2 的杂散因子在您的手卷数学/代码中某处被发现。

如何从手卷和 Pytorch 代码中得到相同的结果

通过仔细考虑以上 4 个因素,可以实现手卷代码和 Pytorch 代码的完全对等。通过正确的调整和设置,两个 sn-ps 将产生相同的结果:

最重要的调整 - 使损失报告功能匹配

关键的区别在于你最终使用两个完全不同的函数来测量两个代码 sn-ps 中的损失:

  • 在手卷代码中,您将损失测量为layer2_error.mean()。如果你解压变量,你可以看到 layer2_error.mean() 是一个有点古怪和毫无意义的值:

    layer2_error.mean()
    == np.dot(cost_delta, cost_error).mean()
    == np.dot(mse_derivative(layer2, Y), mse(layer2, Y)).mean()
    == np.sum(.5 * (layer2 - Y) * ((layer2 - Y)**2).mean()).mean()
    
  • 另一方面,在 PyTorch 代码中,损失是根据 mse 的传统定义来衡量的,即等同于 np.mean((layer2 - Y)**2)。您可以通过如下方式修改 PyTorch 循环来向自己证明这一点:

    def mse(x, y):
        return np.mean((x - y)**2)
    
    torch_losses = [] # Keeps track of the loses.
    torch_losses_manual = [] # for comparison
    
    # Step 2-4 of training routine.
    
    for _e in tqdm(range(num_epochs)):
        # Reset the gradient after every epoch. 
        optimizer.zero_grad() 
        # Step 2: Foward Propagation
        predictions = model(X)
    
        # Step 3: Back Propagation 
        # Calculate the cost between the predictions and the truth.
        loss = criterion(predictions, Y)
        # Remember to back propagate the loss you've computed above.
        loss.backward()
    
        # Step 4: Optimizer take a step and update the weights.
        optimizer.step()
    
        # Log the loss value as we proceed through the epochs.
        torch_losses.append(loss.data.item())
        torch_losses_manual.append(mse(predictions.detach().numpy(), Y.detach().numpy()))
    
    plt.plot(torch_losses, lw=5, label='torch_losses')
    plt.plot(torch_losses_manual, lw=2, label='torch_losses_manual')
    plt.legend()
    

输出:

同样重要 - 使用相同的初始权重

PyTorch 使用它自己的特殊例程来设置初始权重,这会产生与np.random.rand 截然不同的结果。我还不能完全复制它,但为了下一个最好的事情,我们可以劫持 Pytorch。这是一个函数,它将获得与 Pytorch 模型使用的相同的初始权重:

import torch
from torch import nn
torch.manual_seed(0)

def torch_weights(nodes_in, nodes_hidden, nodes_out, bias=None):
    model = nn.Sequential(
        nn.Linear(nodes_in, nodes_hidden, bias=bias),
        nn.Sigmoid(),
        nn.Linear(nodes_hidden, nodes_out, bias=bias),
        nn.Sigmoid()
    )

    return [t.detach().numpy() for t in model.parameters()]

最后 - 在 Pytorch 中,关闭所有偏置权重并将学习率加倍

最终,您可能希望在自己的代码中实现偏差权重。现在,我们将关闭 Pytorch 模型中的偏差,并将手卷模型的结果与无偏差 Pytorch 模型的结果进行比较。

此外,为了使结果匹配,您需要将 Pytorch 模型的学习率提高一倍。这有效地沿 x 轴缩放结果(即,将速率加倍意味着达到损失曲线上的某些特定特征需要一半的 epoch)。

把它放在一起

为了从我的帖子开头的图中重现hand_rolled_losses 数据,您需要做的就是使用您的手动代码并将mse 函数替换为:

def mse(predicted, truth):
    return np.mean(np.square(predicted - truth))

初始化权重的行:

W1,W2 = [w.T for w in torch_weights(input_dim, hidden_dim, output_dim)]

以及跟踪损失的行:

losses.append(cost_error)

你应该很高兴。

为了从图中重现 torch_losses 数据,我们还需要在 Pytorch 模型中关闭偏置权重。为此,您只需像这样更改定义 Pytorch 模型的行:

model = nn.Sequential(
    # Use nn.Linear to get our simple perceptron.
    nn.Linear(input_dim, hidden_dim, bias=None),
    # Use nn.Sigmoid to get our sigmoid non-linearity.
    nn.Sigmoid(),
    # Second layer neurons.
    nn.Linear(hidden_dim, output_dim, bias=None),
    nn.Sigmoid()
)

您还需要更改定义learning_rate 的行:

learning_rate = 0.3 * 2

完整的代码清单

手卷代码

这是我的手卷神经网络代码版本的完整列表,以帮助重现我的结果:

from itertools import chain
import matplotlib.pyplot as plt
import numpy as np
import scipy as sp
import scipy.stats
import torch
from torch import nn

np.random.seed(0)
torch.manual_seed(0)

def torch_weights(nodes_in, nodes_hidden, nodes_out, bias=None):
    model = nn.Sequential(
        nn.Linear(nodes_in, nodes_hidden, bias=bias),
        nn.Sigmoid(),
        nn.Linear(nodes_hidden, nodes_out, bias=bias),
        nn.Sigmoid()
    )

    return [t.detach().numpy() for t in model.parameters()]

def sigmoid(x): # Returns values that sums to one.
    return 1 / (1 + np.exp(-x))

def sigmoid_derivative(sx):
    # See https://math.stackexchange.com/a/1225116
    return sx * (1 - sx)

# Cost functions.
def mse(predicted, truth):
    return np.mean(np.square(predicted - truth))

def mse_derivative(predicted, truth):
    return predicted - truth

X = xor_input = np.array([[0,0], [0,1], [1,0], [1,1]])
Y = xor_output = np.array([[0,1,1,0]]).T

# Define the shape of the weight vector.
num_data, input_dim = X.shape
# Lets set the dimensions for the intermediate layer.
hidden_dim = 5
# Define the shape of the output vector. 
output_dim = len(Y.T)

W1,W2 = [w.T for w in torch_weights(input_dim, hidden_dim, output_dim)]

num_epochs = 5000
learning_rate = 0.3
losses = []

for epoch_n in range(num_epochs):
    layer0 = X
    # Forward propagation.

    # Inside the perceptron, Step 2. 
    layer1 = sigmoid(np.dot(layer0, W1))
    layer2 = sigmoid(np.dot(layer1, W2))

    # Back propagation (Y -> layer2)

    # In what direction is the target value?
    # Were we really close? If so, don't change too much.
    cost_delta = mse_derivative(layer2, Y)
    layer2_delta = cost_delta *  sigmoid_derivative(layer2)

    # Back propagation (layer2 -> layer1)
    # How much did each layer1 value contribute to the layer2 error (according to the weights)?
    layer1_error = np.dot(layer2_delta, W2.T)
    layer1_delta = layer1_error * sigmoid_derivative(layer1)

    # update weights
    W2 += - learning_rate * np.dot(layer1.T, layer2_delta)
    W1 += - learning_rate * np.dot(layer0.T, layer1_delta)

    # Log the loss value as we proceed through the epochs.
    losses.append(mse(layer2, Y))

# Visualize the losses
plt.plot(losses)
plt.show()

Pytorch 代码

import matplotlib.pyplot as plt
from tqdm import tqdm
import numpy as np

import torch
from torch import nn
from torch import tensor
from torch import optim

torch.manual_seed(0)
device = 'gpu' if torch.cuda.is_available() else 'cpu'

num_epochs = 5000
learning_rate = 0.3 * 2

# XOR gate inputs and outputs.
X = tensor([[0,0], [0,1], [1,0], [1,1]]).float().to(device)
Y = tensor([[0],[1],[1],[0]]).float().to(device)

# Use tensor.shape to get the shape of the matrix/tensor.
num_data, input_dim = X.shape
num_data, output_dim = Y.shape

# Step 1: Initialization. 

# Initialize the model.
# Set the hidden dimension size.
hidden_dim = 5
# Use Sequential to define a simple feed-forward network.
model = nn.Sequential(
    # Use nn.Linear to get our simple perceptron.
    nn.Linear(input_dim, hidden_dim, bias=None),
    # Use nn.Sigmoid to get our sigmoid non-linearity.
    nn.Sigmoid(),
    # Second layer neurons.
    nn.Linear(hidden_dim, output_dim, bias=None),
    nn.Sigmoid()
)

# Initialize the optimizer
optimizer = optim.SGD(model.parameters(), lr=learning_rate)

# Initialize the loss function.
criterion = nn.MSELoss()

def mse(x, y):
    return np.mean((x - y)**2)

torch_losses = [] # Keeps track of the loses.
torch_losses_manual = [] # for comparison

# Step 2-4 of training routine.

for _e in tqdm(range(num_epochs)):
    # Reset the gradient after every epoch. 
    optimizer.zero_grad() 
    # Step 2: Foward Propagation
    predictions = model(X)

    # Step 3: Back Propagation 
    # Calculate the cost between the predictions and the truth.
    loss = criterion(predictions, Y)
    # Remember to back propagate the loss you've computed above.
    loss.backward()

    # Step 4: Optimizer take a step and update the weights.
    optimizer.step()

    # Log the loss value as we proceed through the epochs.
    torch_losses.append(loss.data.item())
    torch_losses_manual.append(mse(predictions.detach().numpy(), Y.detach().numpy()))

plt.plot(torch_losses, lw=5, c='C1', label='torch_losses')
plt.plot(torch_losses_manual, lw=2, c='C2', label='torch_losses_manual')
plt.legend()

注意事项

偏置权重

您可以在this tutorial 中找到一些非常有启发性的示例来说明偏差权重是什么以及如何实现它们。他们列出了一堆神经网络的纯 Python 实现,与您的手卷非常相似,因此您很可能可以调整他们的一些代码来实现您自己的偏差。

产生权重初始猜测的函数

这是我从相同的tutorial 改编的一个函数,它可以为权重生成合理的初始值。我认为 Pytorch 内部使用的算法有些不同,但这会产生类似的结果:

import scipy as sp
import scipy.stats

def tnorm_weights(nodes_in, nodes_out, bias_node=0):
    # see https://www.python-course.eu/neural_network_mnist.php
    wshape = (nodes_out, nodes_in + bias_node)
    bound = 1 / np.sqrt(nodes_in)
    X = sp.stats.truncnorm(-bound, bound)
    return X.rvs(np.prod(wshape)).reshape(wshape) 

【讨论】:

  • 感谢非常详细的回答!不过,简短的问题是,您已经使用 PyTorch 复制了“手动”NN。反过来会是什么样子,即使用“手动”numpy NN 复制 PyTorch?
  • 如果你愿意,请用另一个答案回答你是否可以因复制两种方式的答案而获得奖励=)
  • 啊等等,我设法通过将“手卷”版本中的 LR 减半并使用 MSE 及其派生词的正确定义来做相反的事情。mse = return np.square(predicted - truth).mean()mse_derivative = return 2 * (predicted - truth)
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