目录
1. 使用Numpy实现SRN
2. 在1的基础上,增加激活函数tanh
3. 分别使用nn.RNNCell、nn.RNN实现SRN
4. 分析“二进制加法” 源代码
5. 实现“Character-Level Language Models”源代码
6. 分析“序列到序列”源代码
7. “编码器-解码器”的简单实现
心得体会
import numpy as np
inputs = np.array([[1., 1.],
[1., 1.],
[2., 2.]]) # 初始化输入序列
print('inputs is ', inputs)
state_t = np.zeros(2, ) # 初始化存储器
print('state_t is ', state_t)
w1, w2, w3, w4, w5, w6, w7, w8 = 1., 1., 1., 1., 1., 1., 1., 1.
U1, U2, U3, U4 = 1., 1., 1., 1.
print('--------------------------------------')
for input_t in inputs:
print('inputs is ', input_t)
print('state_t is ', state_t)
in_h1 = np.dot([w1, w3], input_t) + np.dot([U2, U4], state_t)
in_h2 = np.dot([w2, w4], input_t) + np.dot([U1, U3], state_t)
state_t = in_h1, in_h2
output_y1 = np.dot([w5, w7], [in_h1, in_h2])
output_y2 = np.dot([w6, w8], [in_h1, in_h2])
print('output_y is ', output_y1, output_y2)
print('---------------')
inputs is [[1. 1.]
[1. 1.]
[2. 2.]]
state_t is [0. 0.]
--------------------------------------
inputs is [1. 1.]
state_t is [0. 0.]
output_y is 4.0 4.0
---------------
inputs is [1. 1.]
state_t is (2.0, 2.0)
output_y is 12.0 12.0
---------------
inputs is [2. 2.]
state_t is (6.0, 6.0)
output_y is 32.0 32.0
---------------
改变输入序列顺序:
inputs = np.array([[1., 1.],
[2., 2.],
[1., 1.]]) # 初始化输入序列
inputs is [[1. 1.]
[2. 2.]
[1. 1.]]
state_t is [0. 0.]
--------------------------------------
inputs is [1. 1.]
state_t is [0. 0.]
output_y is 4.0 4.0
---------------
inputs is [2. 2.]
state_t is (2.0, 2.0)
output_y is 16.0 16.0
---------------
inputs is [1. 1.]
state_t is (8.0, 8.0)
output_y is 36.0 36.0
---------------
可以看出输出序列也会发生改变,这就是RNN与FNN的区别
import numpy as np
inputs = np.array([[1., 1.],
[1., 1.],
[2., 2.]]) # 初始化输入序列
print('inputs is ', inputs)
state_t = np.zeros(2, ) # 初始化存储器
print('state_t is ', state_t)
w1, w2, w3, w4, w5, w6, w7, w8 = 1., 1., 1., 1., 1., 1., 1., 1.
U1, U2, U3, U4 = 1., 1., 1., 1.
print('--------------------------------------')
for input_t in inputs:
print('inputs is ', input_t)
print('state_t is ', state_t)
in_h1 = np.tanh(np.dot([w1, w3], input_t) + np.dot([U2, U4], state_t))
in_h2 = np.tanh(np.dot([w2, w4], input_t) + np.dot([U1, U3], state_t))
state_t = in_h1, in_h2
output_y1 = np.dot([w5, w7], [in_h1, in_h2])
output_y2 = np.dot([w6, w8], [in_h1, in_h2])
print('output_y is ', output_y1, output_y2)
print('---------------')
inputs is [[1. 1.]
[1. 1.]
[2. 2.]]
state_t is [0. 0.]
--------------------------------------
inputs is [1. 1.]
state_t is [0. 0.]
output_y is 1.9280551601516338 1.9280551601516338
---------------
inputs is [1. 1.]
state_t is (0.9640275800758169, 0.9640275800758169)
output_y is 1.9984510891336251 1.9984510891336251
---------------
inputs is [2. 2.]
state_t is (0.9992255445668126, 0.9992255445668126)
output_y is 1.9999753470497836 1.9999753470497836
---------------
nn.RNN(input_size, hidden_size, num_layers=1, nonlinearity=tanh, bias=True, batch_first=False, dropout=0, bidirectional=False)
参数说明
input_size输入特征的维度, 一般rnn中输入的是词向量,那么 input_size 就等于一个词向量的维度
hidden_size隐藏层神经元个数,或者也叫输出的维度(因为rnn输出为各个时间步上的隐藏状态)
num_layers网络的层数
nonlinearity激活函数
bias是否使用偏置
batch_first输入数据的形式,默认是 False,就是这样形式,(seq(num_step), batch, input_dim),也就是将序列长度放在第一位,batch 放在第二位
dropout是否应用dropout, 默认不使用,如若使用将其设置成一个0-1的数字即可
birdirectional是否使用双向的 rnn,默认是 False
注意某些参数的默认值在标题中已注明
输入输出shape
1、input_shape = [时间步数, 批量大小, 特征维度] = [num_steps(seq_length), batch_size, input_dim]
2、在前向计算后会分别返回输出和隐藏状态h,其中输出指的是隐藏层在各个时间步上计算并输出的隐藏状态,它们通常作为后续输出层的输⼊。需要强调的是,该“输出”本身并不涉及输出层计算,形状为(时间步数, 批量大小, 隐藏单元个数);隐藏状态指的是隐藏层在最后时间步的隐藏状态:当隐藏层有多层时,每⼀层的隐藏状态都会记录在该变量中;对于像⻓短期记忆(LSTM),隐藏状态是⼀个元组(h, c),即hidden state和cell state(此处普通rnn只有一个值)隐藏状态h的形状为(层数, 批量大小,隐藏单元个数)
import torch
batch_size = 1
seq_len = 3
input_size = 2
hidden_size = 2
num_layers = 1
output_size = 2
cell = torch.nn.RNN(input_size=input_size, hidden_size=hidden_size, num_layers=num_layers)
for name, param in cell.named_parameters(): # 初始化参数
if name.startswith("weight"):
torch.nn.init.ones_(param)
else:
torch.nn.init.zeros_(param)
# 线性层
liner = torch.nn.Linear(hidden_size, output_size)
liner.weight.data = torch.Tensor([[1, 1], [1, 1]])
liner.bias.data = torch.Tensor([0.0])
inputs = torch.Tensor([[[1, 1]],
[[1, 1]],
[[2, 2]]])
hidden = torch.zeros(num_layers, batch_size, hidden_size)
out, hidden = cell(inputs, hidden)
print('Input :', inputs[0])
print('hidden:', 0, 0)
print('Output:', liner(out[0]))
print('--------------------------------------')
print('Input :', inputs[1])
print('hidden:', out[0])
print('Output:', liner(out[1]))
print('--------------------------------------')
print('Input :', inputs[2])
print('hidden:', out[1])
print('Output:', liner(out[2]))
Input : tensor([[1., 1.]])
hidden: 0 0
Output: tensor([[1.9281, 1.9281]], grad_fn=)
--------------------------------------
Input : tensor([[1., 1.]])
hidden: tensor([[0.9640, 0.9640]], grad_fn=)
Output: tensor([[1.9985, 1.9985]], grad_fn=)
--------------------------------------
Input : tensor([[2., 2.]])
hidden: tensor([[0.9992, 0.9992]], grad_fn=)
Output: tensor([[2.0000, 2.0000]], grad_fn=)
具体计算过程
[batch_size, input_dim] * [input_dim, num_hiddens] + [batch_size, num_hiddens] *[num_hiddens, num_hiddens] +bias
可以发现每个隐藏状态形状都是[batch_size, num_hiddens], 起始输出也是一样的
注意:上面为了方便假设num_step=1
nn.RNNCell
相比一步到位的nn.RNN,也可以使用nn.RNNCell,它将序列上的每个时刻分开来处理。
也就是说,如果要处理的是3个句子,每个句子10个单词,每个单词用长100的向量,那么送入nn.RNN的Tensor的shape就是[10,3,100]。
但如果使用nn.RNNCell,则将每个时刻分开处理,送入的Tensor的shape是[3,100],但要将此计算单元运行10次。显然这种方式比较麻烦,但使用起来也更灵活。
构造方法
构造方法和nn.RNN类似,依次传入feature_len和hidden_len,因为这只是一个计算单元,所以不涉及层数。
import torch
batch_size = 1
seq_len = 3 # 序列长度
input_size = 2 # 输入序列维度
hidden_size = 2 # 隐藏层维度
output_size = 2 # 输出层维度
# RNNCell
cell = torch.nn.RNNCell(input_size=input_size, hidden_size=hidden_size)
# 初始化参数 https://zhuanlan.zhihu.com/p/342012463
for name, param in cell.named_parameters():
if name.startswith("weight"):
torch.nn.init.ones_(param)
else:
torch.nn.init.zeros_(param)
# 线性层
liner = torch.nn.Linear(hidden_size, output_size)
liner.weight.data = torch.Tensor([[1, 1], [1, 1]])
liner.bias.data = torch.Tensor([0.0])
seq = torch.Tensor([[[1, 1]],
[[1, 1]],
[[2, 2]]])
hidden = torch.zeros(batch_size, hidden_size)
output = torch.zeros(batch_size, output_size)
for idx, input in enumerate(seq):
print('=' * 20, idx, '=' * 20)
print('Input :', input)
print('hidden :', hidden)
hidden = cell(input, hidden)
output = liner(hidden)
print('output :', output)
==================== 0 ====================
Input : tensor([[1., 1.]])
hidden : tensor([[0., 0.]])
output : tensor([[1.9281, 1.9281]], grad_fn=)
==================== 1 ====================
Input : tensor([[1., 1.]])
hidden : tensor([[0.9640, 0.9640]], grad_fn=)
output : tensor([[1.9985, 1.9985]], grad_fn=)
==================== 2 ====================
Input : tensor([[2., 2.]])
hidden : tensor([[0.9992, 0.9992]], grad_fn=)
output : tensor([[2.0000, 2.0000]], grad_fn=)
这个没啥好说的,就是逢二进一,计算机组成原理课上讲的很明白。
RNN主要学两件事,一个是前一位的进位,一个是当前位的加法操作。只告诉当前阶段和前一阶段的计算结果,让网络自己学习加法和进位操作。
'利用RNN网络特性(能够记忆之前的事物),使RNN网络学会二进制加法,即能正确完成加法和进位两种操作'
"激活函数为sigmoid"
import copy, numpy as np
np.random.seed(0)
#定义sigmoid函数
def sigmoid(x):
output = 1 / (1 + np.exp(-x))
return output
#计算sigmoid函数的导数
def sigmoid_output_to_derivative(output):
return output * (1 - output)
# 生成要计算的二进制数据
int2binary = {} # 用于将输入的整数转为计算机可运行的二进制数用
binary_dim = 8 # 定义了二进制数的长度=8
largest_number = pow(2, binary_dim) # 二进制数最大能取的数就=256喽
binary = np.unpackbits(
np.array([range(largest_number)], dtype=np.uint8).T, axis=1)
for i in range(largest_number): # 将二进制数与十进制数做个一一对应的字典
int2binary[i] = binary[i]
# 初始参数
alpha = 0.1 # 反向传播时参数w更新的速度
input_dim = 2 # 输入数据的维度,程序是实现两个数相加的
hidden_dim = 16 # 隐藏层神经元个数=16
output_dim = 1 # 输出结果值是1维的
# 初始化神经网络的权重参数
synapse_0 = 2 * np.random.random((input_dim, hidden_dim)) - 1 # 输入层权值,维度为2X16,取值约束在[-1,1]间
synapse_1 = 2 * np.random.random((hidden_dim, output_dim)) - 1 # 隐层权值,维度为16X1,取值约束在[-1,1]间
synapse_h = 2 * np.random.random((hidden_dim, hidden_dim)) - 1 # 循环层,维度为16X16,取值约束在[-1,1]间
synapse_0_update = np.zeros_like(synapse_0) # 初始化增量矩阵
synapse_1_update = np.zeros_like(synapse_1)
synapse_h_update = np.zeros_like(synapse_h)
# training logic
for j in range(10000): # 模型迭代次数,可自行更改
# 随机生成相加的数,并将其转换为二进制数
# a_int 为十进制 且小于128, a为二进制
a_int = np.random.randint(largest_number / 2)
a = int2binary[a_int]
b_int = np.random.randint(largest_number / 2)
b = int2binary[b_int]
# c 为实际值
c_int = a_int + b_int # 真实和
c = int2binary[c_int]
# d 为预测值
d = np.zeros_like(c)
overallError = 0 # 打印显示误差
layer_2_deltas = list() # 反向求导用
layer_1_values = list()
# 先对隐藏层前一时刻状态初始化为 [0,0,0,,,,*16]
layer_1_values.append(np.zeros(hidden_dim))
# 前向传播;二进制求和,低位在右,高位在左 以此方向为正向
for position in range(binary_dim):
# 从最右边的数开始求和,所以索引要倒着写(从第七个开始求和)
X = np.array([[a[binary_dim - position - 1], b[binary_dim - position - 1]]])
# 输入的a与b(二进制形式) 1*2
y = np.array([[c[binary_dim - position - 1]]]).T # 真实label值 二进制
# 隐层输出 1*2 * 2*16 + 1*16 * 16*16 = 1*16
layer_1 = sigmoid(np.dot(X, synapse_0) + np.dot(layer_1_values[-1], synapse_h)) # X*w0+RNN前一时刻状态值*wh
# 输出层 1*16 * 16*1 = 1*1
layer_2 = sigmoid(np.dot(layer_1, synapse_1))
# 求误差
layer_2_error = y - layer_2
# 将layer_2_deltas 算出来 并存入列表( y - y_p )*f'(z) 其结果是一个数
layer_2_deltas.append((layer_2_error) * sigmoid_output_to_derivative(layer_2))
overallError += np.abs(layer_2_error[0]) # 误差,打印显示用
# a[7]+b[7]=d[7] 预测的和 循环结束后就会得到完整的二进制加法结果
d[binary_dim - position - 1] = np.round(layer_2[0][0])
# 深拷贝,将前向传播隐层输出保存起来
layer_1_values.append(copy.deepcopy(layer_1))
# 给记忆细胞赋初值 1*16 个0
future_layer_1_delta = np.zeros(hidden_dim)
# 反向传播,计算从左到右,即二进制高位到低位
for position in range(binary_dim):
X = np.array([[a[position], b[position]]]) # a[0],b[0]
# 因为从右往左是正向,所以此时拿前向传播中的隐层中第七位的值
layer_1 = layer_1_values[-position - 1]
# 拿到前向传播中的前一个值 layer_1_+1 便于后面对循环层的矩阵进行跟新
prev_layer_1 = layer_1_values[-position - 2]
# 拿出第七位的 layer_2_delta ,用于计算 layer_1_delta
layer_2_delta = layer_2_deltas[-position - 1]
# 计算 layer_1_delta , future_layer_1_delta初始值为0 与 Whh 相乘
layer_1_delta = (future_layer_1_delta.dot(synapse_h.T) + layer_2_delta.dot(
synapse_1.T)) * sigmoid_output_to_derivative(layer_1)
# 跟新权值增量 (atleast_2d 避免列向量导致无法计算的问题)
synapse_1_update += np.atleast_2d(layer_1).T.dot(layer_2_delta) # 对w1进行更新
synapse_h_update += np.atleast_2d(prev_layer_1).T.dot(layer_1_delta) # 对wh进行更新
synapse_0_update += X.T.dot(layer_1_delta) # 对w0进行更新
# 跟新记忆细胞中的值
future_layer_1_delta = layer_1_delta
# 跟新权值
synapse_0 += synapse_0_update * alpha
synapse_1 += synapse_1_update * alpha
synapse_h += synapse_h_update * alpha
synapse_0_update *= 0
synapse_1_update *= 0
synapse_h_update *= 0
# print out progress
if (j % 1000 == 0): # 每1000次打印结果
print("Error:" + str(overallError))
print("Pred:" + str(d))
print("True:" + str(c))
out = 0
for index, x in enumerate(reversed(d)):
out += x * pow(2, index)
print(str(a_int) + " + " + str(b_int) + " = " + str(out))
print("------------")
Error:[3.45638663]
Pred:[0 0 0 0 0 0 0 1]
True:[0 1 0 0 0 1 0 1]
9 + 60 = 1
------------
Error:[3.63389116]
Pred:[1 1 1 1 1 1 1 1]
True:[0 0 1 1 1 1 1 1]
28 + 35 = 255
------------
Error:[3.91366595]
Pred:[0 1 0 0 1 0 0 0]
True:[1 0 1 0 0 0 0 0]
116 + 44 = 72
------------
Error:[3.72191702]
Pred:[1 1 0 1 1 1 1 1]
True:[0 1 0 0 1 1 0 1]
4 + 73 = 223
------------
Error:[3.5852713]
Pred:[0 0 0 0 1 0 0 0]
True:[0 1 0 1 0 0 1 0]
71 + 11 = 8
------------
Error:[2.53352328]
Pred:[1 0 1 0 0 0 1 0]
True:[1 1 0 0 0 0 1 0]
81 + 113 = 162
------------
Error:[0.57691441]
Pred:[0 1 0 1 0 0 0 1]
True:[0 1 0 1 0 0 0 1]
81 + 0 = 81
------------
Error:[1.42589952]
Pred:[1 0 0 0 0 0 0 1]
True:[1 0 0 0 0 0 0 1]
4 + 125 = 129
------------
Error:[0.47477457]
Pred:[0 0 1 1 1 0 0 0]
True:[0 0 1 1 1 0 0 0]
39 + 17 = 56
------------
Error:[0.21595037]
Pred:[0 0 0 0 1 1 1 0]
True:[0 0 0 0 1 1 1 0]
11 + 3 = 14
------------
翻译Character-Level Language Models 相关内容
The Unreasonable Effectiveness of Recurrent Neural Networks
字符级别的语言模型
好了,我们对什么是RNN有了一个概念,为什么它们超级令人兴奋,以及它们是如何工作的。现在,我们将在一个有趣的应用中把它作为基础。我们将训练RNN的字符级语言模型。也就是说,我们将给RNN一大块文本,并要求它在给定先前字符序列的情况下,对序列中下一个字符的概率分布进行建模。这将使我们能够一次生成一个字符的新文本。
作为一个工作例子,假设我们只有四个可能的字母 "helo "的词汇,并想在训练序列 "hello "上训练一个RNN。这个训练序列实际上是4个独立的训练例子的来源。1. 鉴于 "h "的上下文,"e "的概率应该是很高的,2. "l "在 "he "的上下文中应该是很高的,3.鉴于 "hel "的上下文,"l "也应该是很高的,最后4.鉴于 "hell "的上下文,"o "应该是很高的。
具体来说,我们将使用1-of-k编码将每个字符编码成一个向量(即除了词汇表中字符索引处的一个1之外,其他都是0),并通过阶梯函数将它们一个个送入RNN。然后,我们将观察到一连串的四维输出向量(每个字符一个维度),我们将其解释为RNN目前对序列中接下来的每个字符所赋予的信心。这里有一张图。
一个具有4维输入和输出层的RNN例子,以及一个由3个单元(神经元)组成的隐藏层。该图显示了当RNN被输入字符 "hell "时,在前向通道中的激活情况。输出层包含RNN为下一个字符(词汇为 "h,e,l,o")分配的置信度;我们希望绿色数字为高,红色数字为低。
例如,我们看到,在第一个时间步骤中,当RNN看到字符 "h "时,它给下一个字母 "h "分配了1.0的信心,给字母 "e "分配了2.2的信心,给 "l "分配了3.0的信心,给 "o "分配了4.1的信心。由于在我们的训练数据(字符串 "hello")中,下一个正确的字符是 "e",我们希望增加它的信心(绿色),减少所有其他字母的信心(红色)。同样,我们在4个时间步骤中的每一个步骤都有一个理想的目标字符,我们希望网络能赋予它更大的信心。由于RNN完全由可微调的操作组成,我们可以运行反向传播算法(这只是微积分中链式规则的递归应用),以找出我们应该向哪个方向调整它的每一个权重,以增加正确目标的分数(绿色粗体数字)。然后我们可以进行参数更新,在这个梯度方向上对每个权重进行微小的调整。如果我们在参数更新后给RNN提供相同的输入,我们会发现正确的字符(例如第一个时间步骤中的 "e")的得分会略高(例如2.3而不是2.2),而错误的字符的得分会略低。然后,我们多次重复这个过程,直到网络收敛,其预测结果最终与训练数据一致,即正确的字符总是被预测在下一个。
一个更技术性的解释是,我们在每个输出向量上同时使用标准的Softmax分类器(也通常被称为交叉熵损失)。RNN是用小型批次的随机梯度下降法训练的,我喜欢用RMSProp或Adam(每参数自适应学习率方法)来稳定更新。
还请注意,第一次输入字符 "l "时,目标是 "l",但第二次输入时,目标是 "o"。因此,RNN不能仅仅依靠输入,必须使用其递归连接来跟踪上下文以实现这一任务。
在测试时,我们将一个字符送入RNN,得到一个关于下一个可能出现的字符的分布。我们从这个分布中取样,并将其直接送回以获得下一个字母。重复这个过程,你就可以对文本进行取样了! 现在让我们在不同的数据集上训练一个RNN,看看会发生什么。
为了进一步说明问题,出于教育目的,我还用Python/numpy写了一个最小的字符级RNN语言模型。它只有大约100行,如果你更擅长阅读代码而不是文字的话,希望它能对上述内容做一个简明、具体和有用的总结。我们现在将深入研究用更有效的Lua/Torch代码库产生的例子结果。
## reference page [The Unreasonable Effectiveness of Recurrent Neural Networks](http://karpathy.github.io/2015/05/21/rnn-effectiveness/)
import numpy as np
# data I/O
data = open('input.txt', 'r').read() # should be simple plain text file
# use set() to count the vacab size
chars = list(set(data))
data_size, vocab_size = len(data), len(chars)
print ('data has %d characters, %d unique.' % (data_size, vocab_size))
# dictionary to convert char to idx, idx to char
char_to_ix = { ch:i for i,ch in enumerate(chars) }
ix_to_char = { i:ch for i,ch in enumerate(chars) }
# hyperparameters
hidden_size = 100 # size of hidden layer of neurons
seq_length = 25 # number of steps to unroll the RNN for
learning_rate = 1e-1
# model parameters
Wxh = np.random.randn(hidden_size, vocab_size)*0.01 # input to hidden
Whh = np.random.randn(hidden_size, hidden_size)*0.01 # hidden to hidden
Why = np.random.randn(vocab_size, hidden_size)*0.01 # hidden to output
bh = np.zeros((hidden_size, 1)) # hidden bias
by = np.zeros((vocab_size, 1)) # output bias
def lossFun(inputs, targets, hprev):
xs, hs, ys, ps = {}, {}, {}, {}
hs[-1] = np.copy(hprev)
loss = 0
for t in range(len(inputs)):
xs[t] = np.zeros((vocab_size, 1)) # encode in 1-of-k representation
xs[t][inputs[t]] = 1
hs[t] = np.tanh(np.dot(Wxh, xs[t]) + np.dot(Whh, hs[t-1]) + bh)
ys[t] = np.dot(Why, hs[t]) + by
ps[t] = np.exp(ys[t]) / np.sum(np.exp(ys[t]))
loss += -np.log(ps[t][targets[t], 0])
# backward pass
dWxh, dWhh, dWhy = np.zeros_like(Wxh), np.zeros_like(Whh), np.zeros_like(Why)
dbh, dby = np.zeros_like(bh), np.zeros_like(by)
dhnext = np.zeros_like(hs[0])
for t in reversed(range(len(inputs))):
dy = np.copy(ps[t])
dy[targets[t]] -= 1 # backprop into y
dWhy += np.dot(dy, hs[t].T)
dby += dy
## backprop into h
## derivative of error with regard to the output of hidden layer
## derivative of H, come from output layer y and also come from H(t+1), the next time H
dh = np.dot(Why.T, dy) + dhnext
## backprop through tanh nonlinearity
## derivative of error with regard to the input of hidden layer
## dtanh(x)/dx = 1 - tanh(x) * tanh(x)
dhraw = (1 - hs[t] * hs[t]) * dh
dbh += dhraw
## derivative of the error with regard to the weight between input layer and hidden layer
dWxh += np.dot(dhraw, xs[t].T)
dWhh += np.dot(dhraw, hs[t-1].T)
## derivative of the error with regard to H(t+1)
## or derivative of the error of H(t-1) with regard to H(t)
dhnext = np.dot(Whh.T, dhraw)
for dparam in [dWxh, dWhh, dWhy, dbh, dby]:
np.clip(dparam, -5, 5, out=dparam) # clip to mitigate exploding gradients
return loss, dWxh, dWhh, dWhy, dbh, dby, hs[len(inputs)-1]
## given a hidden RNN state, and a input char id, predict the coming n chars
def sample(h, seed_ix, n):
"""
sample a sequence of integers from the model
h is memory state, seed_ix is seed letter for first time step
"""
## a one-hot vector
x = np.zeros((vocab_size, 1))
x[seed_ix] = 1
ixes = []
for t in range(n):
## self.h = np.tanh(np.dot(self.W_hh, self.h) + np.dot(self.W_xh, x))
h = np.tanh(np.dot(Wxh, x) + np.dot(Whh, h) + bh)
## y = np.dot(self.W_hy, self.h)
y = np.dot(Why, h) + by
## softmax
p = np.exp(y) / np.sum(np.exp(y))
## sample according to probability distribution
ix = np.random.choice(range(vocab_size), p=p.ravel())
## update input x
## use the new sampled result as last input, then predict next char again.
x = np.zeros((vocab_size, 1))
x[ix] = 1
ixes.append(ix)
return ixes
## iterator counter
n = 0
## data pointer
p = 0
mWxh, mWhh, mWhy = np.zeros_like(Wxh), np.zeros_like(Whh), np.zeros_like(Why)
mbh, mby = np.zeros_like(bh), np.zeros_like(by) # memory variables for Adagrad
smooth_loss = -np.log(1.0/vocab_size)*seq_length # loss at iteration 0
## main loop
while True:
if p + seq_length + 1 >= len(data) or n == 0:
# reset RNN memory
hprev = np.zeros((hidden_size, 1))
# go from start of data
p = 0
inputs = [char_to_ix[ch] for ch in data[p : p + seq_length]]
targets = [char_to_ix[ch] for ch in data[p + 1 : p + seq_length + 1]]
# sample from the model now and then
if n % 100 == 0:
sample_ix = sample(hprev, inputs[0], 200)
txt = ''.join(ix_to_char[ix] for ix in sample_ix)
print ('---- sample -----')
print ('----\n %s \n----' % (txt, ))
loss, dWxh, dWhh, dWhy, dbh, dby, hprev = lossFun(inputs, targets, hprev)
smooth_loss = smooth_loss * 0.999 + loss * 0.001
if n % 100 == 0:
print ('iter %d, loss: %f' % (n, smooth_loss)) # print progress
for param, dparam, mem in zip([Wxh, Whh, Why, bh, by],
[dWxh, dWhh, dWhy, dbh, dby],
[mWxh, mWhh, mWhy, mbh, mby]):
mem += dparam * dparam
## learning_rate is adjusted by mem, if mem is getting bigger, then learning_rate will be small
## gradient descent of Adagrad
param += -learning_rate * dparam / np.sqrt(mem + 1e-8) # adagrad update
p += seq_length # move data pointer
n += 1 # iteration counter
# Model
class Seq2Seq(nn.Module):
def __init__(self):
super(Seq2Seq, self).__init__()
self.encoder = nn.RNN(input_size=n_class, hidden_size=n_hidden, dropout=0.5) # encoder
self.decoder = nn.RNN(input_size=n_class, hidden_size=n_hidden, dropout=0.5) # decoder
self.fc = nn.Linear(n_hidden, n_class)
def forward(self, enc_input, enc_hidden, dec_input):
# enc_input(=input_batch): [batch_size, n_step+1, n_class]
# dec_inpu(=output_batch): [batch_size, n_step+1, n_class]
enc_input = enc_input.transpose(0, 1) # enc_input: [n_step+1, batch_size, n_class]
dec_input = dec_input.transpose(0, 1) # dec_input: [n_step+1, batch_size, n_class]
# h_t : [num_layers(=1) * num_directions(=1), batch_size, n_hidden]
_, h_t = self.encoder(enc_input, enc_hidden)
# outputs : [n_step+1, batch_size, num_directions(=1) * n_hidden(=128)]
outputs, _ = self.decoder(dec_input, h_t)
model = self.fc(outputs) # model : [n_step+1, batch_size, n_class]
return model
model = Seq2Seq().to(device)
criterion = nn.CrossEntropyLoss().to(device)
optimizer = torch.optim.Adam(model.parameters(), lr=0.001)
#下面是训练,由于输出的 pred 是个三维的数据,所以计算 loss 需要每个样本单独计算,因此就有了下面 for 循环的代码
for epoch in range(5000):
for enc_input_batch, dec_input_batch, dec_output_batch in loader:
# make hidden shape [num_layers * num_directions, batch_size, n_hidden]
h_0 = torch.zeros(1, batch_size, n_hidden).to(device)
(enc_input_batch, dec_intput_batch, dec_output_batch) = (enc_input_batch.to(device), dec_input_batch.to(device), dec_output_batch.to(device))
# enc_input_batch : [batch_size, n_step+1, n_class]
# dec_intput_batch : [batch_size, n_step+1, n_class]
# dec_output_batch : [batch_size, n_step+1], not one-hot
pred = model(enc_input_batch, h_0, dec_intput_batch)
# pred : [n_step+1, batch_size, n_class]
pred = pred.transpose(0, 1) # [batch_size, n_step+1(=6), n_class]
loss = 0
for i in range(len(dec_output_batch)):
# pred[i] : [n_step+1, n_class]
# dec_output_batch[i] : [n_step+1]
loss += criterion(pred[i], dec_output_batch[i])
if (epoch + 1) % 1000 == 0:
print('Epoch:', '%04d' % (epoch + 1), 'cost =', '{:.6f}'.format(loss))
optimizer.zero_grad()
loss.backward()
optimizer.step()
seq2seq(sequence to sequence)模型是NLP中的一个经典模型,基于RNN网络模型构建,用途非常广泛:语言翻译,人机对话,问答系统等。
Seq2Seq,就如字面意思,输入一个序列,输出另一个序列,比如在机器翻译中,输入英文,输出中文。这种结构最重要的地方在于输入序列和输出序列的长度是可变的。而Seq2Seq模型也经常在输出的长度不确定时采用。
# code by Tae Hwan Jung(Jeff Jung) @graykode, modify by wmathor
import torch
import numpy as np
import torch.nn as nn
import torch.utils.data as Data
device = torch.device('cuda' if torch.cuda.is_available() else 'cpu')
# S: Symbol that shows starting of decoding input
# E: Symbol that shows starting of decoding output
# ?: Symbol that will fill in blank sequence if current batch data size is short than n_step
letter = [c for c in 'SE?abcdefghijklmnopqrstuvwxyz']
letter2idx = {n: i for i, n in enumerate(letter)}
seq_data = [['man', 'women'], ['black', 'white'], ['king', 'queen'], ['girl', 'boy'], ['up', 'down'], ['high', 'low']]
# Seq2Seq Parameter
n_step = max([max(len(i), len(j)) for i, j in seq_data]) # max_len(=5)
n_hidden = 128
n_class = len(letter2idx) # classfication problem
batch_size = 3
def make_data(seq_data):
enc_input_all, dec_input_all, dec_output_all = [], [], []
for seq in seq_data:
for i in range(2):
seq[i] = seq[i] + '?' * (n_step - len(seq[i])) # 'man??', 'women'
enc_input = [letter2idx[n] for n in (seq[0] + 'E')] # ['m', 'a', 'n', '?', '?', 'E']
dec_input = [letter2idx[n] for n in ('S' + seq[1])] # ['S', 'w', 'o', 'm', 'e', 'n']
dec_output = [letter2idx[n] for n in (seq[1] + 'E')] # ['w', 'o', 'm', 'e', 'n', 'E']
enc_input_all.append(np.eye(n_class)[enc_input])
dec_input_all.append(np.eye(n_class)[dec_input])
dec_output_all.append(dec_output) # not one-hot
# make tensor
return torch.Tensor(enc_input_all), torch.Tensor(dec_input_all), torch.LongTensor(dec_output_all)
'''
enc_input_all: [6, n_step+1 (because of 'E'), n_class]
dec_input_all: [6, n_step+1 (because of 'S'), n_class]
dec_output_all: [6, n_step+1 (because of 'E')]
'''
enc_input_all, dec_input_all, dec_output_all = make_data(seq_data)
class TranslateDataSet(Data.Dataset):
def __init__(self, enc_input_all, dec_input_all, dec_output_all):
self.enc_input_all = enc_input_all
self.dec_input_all = dec_input_all
self.dec_output_all = dec_output_all
def __len__(self): # return dataset size
return len(self.enc_input_all)
def __getitem__(self, idx):
return self.enc_input_all[idx], self.dec_input_all[idx], self.dec_output_all[idx]
loader = Data.DataLoader(TranslateDataSet(enc_input_all, dec_input_all, dec_output_all), batch_size, True)
# Model
class Seq2Seq(nn.Module):
def __init__(self):
super(Seq2Seq, self).__init__()
self.encoder = nn.RNN(input_size=n_class, hidden_size=n_hidden, dropout=0.5) # encoder
self.decoder = nn.RNN(input_size=n_class, hidden_size=n_hidden, dropout=0.5) # decoder
self.fc = nn.Linear(n_hidden, n_class)
def forward(self, enc_input, enc_hidden, dec_input):
# enc_input(=input_batch): [batch_size, n_step+1, n_class]
# dec_inpu(=output_batch): [batch_size, n_step+1, n_class]
enc_input = enc_input.transpose(0, 1) # enc_input: [n_step+1, batch_size, n_class]
dec_input = dec_input.transpose(0, 1) # dec_input: [n_step+1, batch_size, n_class]
# h_t : [num_layers(=1) * num_directions(=1), batch_size, n_hidden]
_, h_t = self.encoder(enc_input, enc_hidden)
# outputs : [n_step+1, batch_size, num_directions(=1) * n_hidden(=128)]
outputs, _ = self.decoder(dec_input, h_t)
model = self.fc(outputs) # model : [n_step+1, batch_size, n_class]
return model
model = Seq2Seq().to(device)
criterion = nn.CrossEntropyLoss().to(device)
optimizer = torch.optim.Adam(model.parameters(), lr=0.001)
for epoch in range(5000):
for enc_input_batch, dec_input_batch, dec_output_batch in loader:
# make hidden shape [num_layers * num_directions, batch_size, n_hidden]
h_0 = torch.zeros(1, batch_size, n_hidden).to(device)
(enc_input_batch, dec_intput_batch, dec_output_batch) = (enc_input_batch.to(device), dec_input_batch.to(device), dec_output_batch.to(device))
# enc_input_batch : [batch_size, n_step+1, n_class]
# dec_intput_batch : [batch_size, n_step+1, n_class]
# dec_output_batch : [batch_size, n_step+1], not one-hot
pred = model(enc_input_batch, h_0, dec_intput_batch)
# pred : [n_step+1, batch_size, n_class]
pred = pred.transpose(0, 1) # [batch_size, n_step+1(=6), n_class]
loss = 0
for i in range(len(dec_output_batch)):
# pred[i] : [n_step+1, n_class]
# dec_output_batch[i] : [n_step+1]
loss += criterion(pred[i], dec_output_batch[i])
if (epoch + 1) % 1000 == 0:
print('Epoch:', '%04d' % (epoch + 1), 'cost =', '{:.6f}'.format(loss))
optimizer.zero_grad()
loss.backward()
optimizer.step()
# Test
def translate(word):
enc_input, dec_input, _ = make_data([[word, '?' * n_step]])
enc_input, dec_input = enc_input.to(device), dec_input.to(device)
# make hidden shape [num_layers * num_directions, batch_size, n_hidden]
hidden = torch.zeros(1, 1, n_hidden).to(device)
output = model(enc_input, hidden, dec_input)
# output : [n_step+1, batch_size, n_class]
predict = output.data.max(2, keepdim=True)[1] # select n_class dimension
decoded = [letter[i] for i in predict]
translated = ''.join(decoded[:decoded.index('E')])
return translated.replace('?', '')
print('test')
print('man ->', translate('man'))
print('mans ->', translate('mans'))
print('king ->', translate('king'))
print('black ->', translate('black'))
print('up ->', translate('up'))
Epoch: 1000 cost = 0.002254
Epoch: 1000 cost = 0.002543
Epoch: 2000 cost = 0.000528
Epoch: 2000 cost = 0.000472
Epoch: 3000 cost = 0.000153
Epoch: 3000 cost = 0.000150
Epoch: 4000 cost = 0.000053
Epoch: 4000 cost = 0.000050
Epoch: 5000 cost = 0.000017
Epoch: 5000 cost = 0.000019
test
man -> women
mans -> women
king -> queen
black -> white
up -> down
Encoder–Decoder是一种框架,许多算法中都有该种框架,在这个框架下可以使用不同的算法来解决不同的任务。
seq2seq模型属于encoder-decoder框架的范围,Seq2Seq 强调目的,不特指具体方法,满足输入序列,输出序列的目的,都可以统称为 Seq2Seq 模型。
这节课学习了简单循环神经网络,与之前学习的FNN与CNN对比,他们都只能单独的取处理一个个的输入,前一个输入和后一个输入是完全没有关系的。但是,某些任务需要能够更好的处理序列的信息,即前面的输入和后面的输入是有关系的。
比如,当我们在理解一句话意思时,孤立的理解这句话的每个词是不够的,我们需要处理这些词连接起来的整个序列; 当我们处理视频的时候,我们也不能只单独的去分析每一帧,而要分析这些帧连接起来的整个序列。
所以为了解决一些这样类似的问题,能够更好的处理序列的信息,RNN的作用就体现在此处。