本文参考自https://github.com/fastai/course-nlp。
使用 RNN 预测数字的英文单词版本
在上一课中，我们将 RNN 用作语言模型的一部分。今天，我们将深入了解 RNN 是什么以及它们如何工作。我们将使用尝试预测数字的英文单词版本的问题来实现这一点。
让我们预测这个序列中接下来应该是什么：
eight thousand one , eight thousand two , eight thousand three , eight thousand four , eight thousand five , eight thousand six , eight thousand seven , eight thousand eight , eight thousand nine , eight thousand ten , eight thousand eleven , eight thousand twelve…
Jeremy 创建了这个合成数据集，以便有更好的方法来检查事情是否正常、调试和了解发生了什么。在尝试新想法时，最好有一个较小的数据集来这样做，以便快速了解你的想法是否有前途（有关其他示例，请参阅 Imagenette 和 Imagewoof）这个英文单词数字将作为学习 RNN 的良好数据集。我们今天的任务是预测计数时接下来会出现哪个单词。
在深度学习中，有两种类型的数字
参数是学习到的数字。激活是计算出的数字（通过仿射函数和元素非线性）。
当你学习深度学习中的任何新概念时，问问自己：这是一个参数还是一个激活？
提醒自己：指出隐藏状态，从没有 for 循环的版本转到 for 循环。这是人们感到困惑的步骤。
Data
from fastai.text import *
bs=64
path = untar_data(URLs.HUMAN_NUMBERS)
path.ls()
[PosixPath('/home/racheltho/.fastai/data/human_numbers/models'),PosixPath('/home/racheltho/.fastai/data/human_numbers/valid.txt'),PosixPath('/home/racheltho/.fastai/data/human_numbers/train.txt')]
def readnums(d): return [', '.join(o.strip() for o in open(path/d).readlines())]
train.txt 为我们提供了以英文单词写出的数字序列：
train_txt = readnums('train.txt'); train_txt[0][:80]
'one, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirt'
valid_txt = readnums('valid.txt'); valid_txt[0][-80:]
' nine thousand nine hundred ninety eight, nine thousand nine hundred ninety nine'
train = TextList(train_txt, path=path)
valid = TextList(valid_txt, path=path)src = ItemLists(path=path, train=train, valid=valid).label_for_lm()
data = src.databunch(bs=bs)
train[0].text[:80]
'xxbos one , two , three , four , five , six , seven , eight , nine , ten , eleve'
len(data.valid_ds[0][0].data)
13017
bptt 代表时间反向传播。这告诉我们正在考虑多少历史步骤。
data.bptt, len(data.valid_dl)
(70, 3)
我们的验证集中有 3 个批次：
13017 个标记，每行文本中约有 ~70 个标记，每批次有 64 行文本。
13017/70/bs
2.905580357142857
我们将每个批次存储在单独的变量中，这样我们就可以通过这个过程更好地理解 RNN 在每个步骤中的作用：
it = iter(data.valid_dl)
x1,y1 = next(it)
x2,y2 = next(it)
x3,y3 = next(it)
it.close()
x1
tensor([[ 2, 19, 11, ..., 36, 9, 19],[ 9, 19, 11, ..., 24, 20, 9],[11, 27, 18, ..., 9, 19, 11],...,[20, 11, 20, ..., 11, 20, 10],[20, 11, 20, ..., 24, 9, 20],[20, 10, 26, ..., 20, 11, 20]], device='cuda:0')
numel() is a PyTorch method to return the number of elements in a tensor:
x1.numel()+x2.numel()+x3.numel()
13440
x1.shape, y1.shape
(torch.Size([64, 70]), torch.Size([64, 70]))
x2.shape, y2.shape
(torch.Size([64, 70]), torch.Size([64, 70]))
x3.shape, y3.shape
(torch.Size([64, 70]), torch.Size([64, 70]))
v = data.valid_ds.vocab
v.itos
['xxunk','xxpad','xxbos','xxeos','xxfld','xxmaj','xxup','xxrep','xxwrep',',','hundred','thousand','one','two','three','four','five','six','seven','eight','nine','twenty','thirty','forty','fifty','sixty','seventy','eighty','ninety','ten','eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen']
x1[:,0]
tensor([ 2, 9, 11, 12, 13, 11, 10, 9, 10, 14, 19, 25, 19, 15, 16, 11, 19, 9,10, 9, 19, 25, 19, 11, 19, 11, 10, 9, 19, 20, 11, 26, 20, 23, 20, 20,24, 20, 11, 14, 11, 11, 9, 14, 9, 20, 10, 20, 35, 17, 11, 10, 9, 17,9, 20, 10, 20, 11, 20, 11, 20, 20, 20], device='cuda:0')
y1[:,0]
tensor([19, 19, 27, 10, 9, 12, 32, 19, 26, 10, 11, 15, 11, 10, 9, 15, 11, 19,26, 19, 11, 18, 11, 18, 9, 18, 21, 19, 10, 10, 20, 9, 11, 16, 11, 11,13, 11, 13, 9, 13, 14, 20, 10, 20, 11, 24, 11, 9, 9, 16, 17, 20, 10,20, 11, 24, 11, 19, 9, 19, 11, 11, 10], device='cuda:0')
v.itos[9], v.itos[11], v.itos[12], v.itos[13], v.itos[10]
(',', 'thousand', 'one', 'two', 'hundred')
v.textify(x1[0])
'xxbos eight thousand one , eight thousand two , eight thousand three , eight thousand four , eight thousand five , eight thousand six , eight thousand seven , eight thousand eight , eight thousand nine , eight thousand ten , eight thousand eleven , eight thousand twelve , eight thousand thirteen , eight thousand fourteen , eight thousand fifteen , eight thousand sixteen , eight thousand seventeen , eight'
v.textify(x1[1])
', eight thousand forty six , eight thousand forty seven , eight thousand forty eight , eight thousand forty nine , eight thousand fifty , eight thousand fifty one , eight thousand fifty two , eight thousand fifty three , eight thousand fifty four , eight thousand fifty five , eight thousand fifty six , eight thousand fifty seven , eight thousand fifty eight , eight thousand fifty nine ,'
v.textify(x2[1])
'eight thousand sixty , eight thousand sixty one , eight thousand sixty two , eight thousand sixty three , eight thousand sixty four , eight thousand sixty five , eight thousand sixty six , eight thousand sixty seven , eight thousand sixty eight , eight thousand sixty nine , eight thousand seventy , eight thousand seventy one , eight thousand seventy two , eight thousand seventy three , eight thousand'
v.textify(y1[0])
'eight thousand one , eight thousand two , eight thousand three , eight thousand four , eight thousand five , eight thousand six , eight thousand seven , eight thousand eight , eight thousand nine , eight thousand ten , eight thousand eleven , eight thousand twelve , eight thousand thirteen , eight thousand fourteen , eight thousand fifteen , eight thousand sixteen , eight thousand seventeen , eight thousand'
v.textify(x2[0])
'thousand eighteen , eight thousand nineteen , eight thousand twenty , eight thousand twenty one , eight thousand twenty two , eight thousand twenty three , eight thousand twenty four , eight thousand twenty five , eight thousand twenty six , eight thousand twenty seven , eight thousand twenty eight , eight thousand twenty nine , eight thousand thirty , eight thousand thirty one , eight thousand thirty two ,'
v.textify(x3[0])
'eight thousand thirty three , eight thousand thirty four , eight thousand thirty five , eight thousand thirty six , eight thousand thirty seven , eight thousand thirty eight , eight thousand thirty nine , eight thousand forty , eight thousand forty one , eight thousand forty two , eight thousand forty three , eight thousand forty four , eight thousand forty five , eight thousand forty six , eight'
v.textify(x1[1])
', eight thousand forty six , eight thousand forty seven , eight thousand forty eight , eight thousand forty nine , eight thousand fifty , eight thousand fifty one , eight thousand fifty two , eight thousand fifty three , eight thousand fifty four , eight thousand fifty five , eight thousand fifty six , eight thousand fifty seven , eight thousand fifty eight , eight thousand fifty nine ,'
v.textify(x2[1])
'eight thousand sixty , eight thousand sixty one , eight thousand sixty two , eight thousand sixty three , eight thousand sixty four , eight thousand sixty five , eight thousand sixty six , eight thousand sixty seven , eight thousand sixty eight , eight thousand sixty nine , eight thousand seventy , eight thousand seventy one , eight thousand seventy two , eight thousand seventy three , eight thousand'
v.textify(x3[1])
'seventy four , eight thousand seventy five , eight thousand seventy six , eight thousand seventy seven , eight thousand seventy eight , eight thousand seventy nine , eight thousand eighty , eight thousand eighty one , eight thousand eighty two , eight thousand eighty three , eight thousand eighty four , eight thousand eighty five , eight thousand eighty six , eight thousand eighty seven , eight thousand eighty'
v.textify(x3[-1])
'ninety , nine thousand nine hundred ninety one , nine thousand nine hundred ninety two , nine thousand nine hundred ninety three , nine thousand nine hundred ninety four , nine thousand nine hundred ninety five , nine thousand nine hundred ninety six , nine thousand nine hundred ninety seven , nine thousand nine hundred ninety eight , nine thousand nine hundred ninety nine xxbos eight thousand one , eight'
data.show_batch(ds_type=DatasetType.Valid)
我们将迭代地考虑一些不同的模型，构建更传统的 RNN。
单一全连接模型
data = src.databunch(bs=bs, bptt=3)
x,y = data.one_batch()
x.shape,y.shape
(torch.Size([64, 3]), torch.Size([64, 3]))
nv = len(v.itos); nv
39
nh=64
def loss4(input,target): return F.cross_entropy(input, target[:,-1])
def acc4 (input,target): return accuracy(input, target[:,-1])
x[:,0]
tensor([13, 13, 10, 9, 18, 9, 11, 11, 13, 19, 16, 23, 24, 9, 12, 9, 13, 14,15, 11, 10, 22, 15, 9, 10, 14, 11, 16, 10, 28, 11, 9, 20, 9, 15, 15,11, 18, 10, 28, 23, 24, 9, 16, 10, 16, 19, 20, 12, 10, 22, 16, 17, 17,17, 11, 24, 10, 9, 15, 16, 9, 18, 11])
Layer names:
i_h: input to hiddenh_h: hidden to hiddenh_o: hidden to outputbn: batchnorm
class Model0(nn.Module):def __init__(self):super().__init__()self.i_h = nn.Embedding(nv,nh) # green arrowself.h_h = nn.Linear(nh,nh) # brown arrowself.h_o = nn.Linear(nh,nv) # blue arrowself.bn = nn.BatchNorm1d(nh)def forward(self, x):h = self.bn(F.relu(self.i_h(x[:,0])))if x.shape[1]>1:h = h + self.i_h(x[:,1])h = self.bn(F.relu(self.h_h(h)))if x.shape[1]>2:h = h + self.i_h(x[:,2])h = self.bn(F.relu(self.h_h(h)))return self.h_o(h)
learn = Learner(data, Model0(), loss_func=loss4, metrics=acc4)
learn.fit_one_cycle(6, 1e-4)
循环也一样
让我们重构一下，使用 for 循环。它的作用和之前一样：
class Model1(nn.Module):def __init__(self):super().__init__()self.i_h = nn.Embedding(nv,nh) # green arrowself.h_h = nn.Linear(nh,nh) # brown arrowself.h_o = nn.Linear(nh,nv) # blue arrowself.bn = nn.BatchNorm1d(nh)def forward(self, x):h = torch.zeros(x.shape[0], nh).to(device=x.device)for i in range(x.shape[1]):h = h + self.i_h(x[:,i])h = self.bn(F.relu(self.h_h(h)))return self.h_o(h)
这是展开的 RNN 图（我们之前的 RNN 图）和卷起的 RNN 图（我们现在的 RNN 图）之间的区别：
learn = Learner(data, Model1(), loss_func=loss4, metrics=acc4)
learn.fit_one_cycle(6, 1e-4)
我们的准确性大致相同，因为我们做的事情与以前相同。
多重全连接模型
之前，我们只是预测一行文本中的最后一个单词。给定 70 个标记，标记 71 是什么？这种方法会丢弃大量数据。为什么不根据标记 1 预测标记 2，然后预测标记 3，然后预测标记 4，依此类推？我们将修改模型来做到这一点。
data = src.databunch(bs=bs, bptt=20)
x,y = data.one_batch()
x.shape,y.shape
(torch.Size([64, 20]), torch.Size([64, 20]))
class Model2(nn.Module):def __init__(self):super().__init__()self.i_h = nn.Embedding(nv,nh)self.h_h = nn.Linear(nh,nh)self.h_o = nn.Linear(nh,nv)self.bn = nn.BatchNorm1d(nh)def forward(self, x):h = torch.zeros(x.shape[0], nh).to(device=x.device)res = []for i in range(x.shape[1]):h = h + self.i_h(x[:,i])h = F.relu(self.h_h(h))res.append(self.h_o(self.bn(h)))return torch.stack(res, dim=1)
learn = Learner(data, Model2(), metrics=accuracy)
请注意，我们的准确率现在变差了，因为我们在做一项更艰巨的任务。当我们预测单词 k（k<70）时，我们能获得的历史记录比我们仅预测单词 71 时要少。（Model2每次forward调用时都重新初始化h）
维持状态
为了解决这个问题，让我们保留上一行文本的隐藏状态，这样我们就不会在每一行新文本上重新开始。
class Model3(nn.Module):def __init__(self):super().__init__()self.i_h = nn.Embedding(nv,nh)self.h_h = nn.Linear(nh,nh)self.h_o = nn.Linear(nh,nv)self.bn = nn.BatchNorm1d(nh)self.h = torch.zeros(bs, nh).cuda()def forward(self, x):res = []h = self.hfor i in range(x.shape[1]):h = h + self.i_h(x[:,i])h = F.relu(self.h_h(h))res.append(self.bn(h))self.h = h.detach()res = torch.stack(res, dim=1)res = self.h_o(res)return res
learn = Learner(data, Model3(), metrics=accuracy)
earn.fit_one_cycle(20, 3e-3)
现在我们获得的准确性比以前更高了！(h.detach()防止累积梯度回传)
nn.RNN
class Model4(nn.Module):def __init__(self):super().__init__()self.i_h = nn.Embedding(nv,nh)self.rnn = nn.RNN(nh,nh, batch_first=True)self.h_o = nn.Linear(nh,nv)self.bn = BatchNorm1dFlat(nh)self.h = torch.zeros(1, bs, nh).cuda()def forward(self, x):res,h = self.rnn(self.i_h(x), self.h)self.h = h.detach()return self.h_o(self.bn(res))
learn = Learner(data, Model4(), metrics=accuracy)
learn.fit_one_cycle(20, 3e-3)
2-layer GRU
当你拥有较长的时间尺度和较深的网络时，这些就变得无法训练。解决这个问题的一种方法是添加 mini-NN 来决定保留多少绿色箭头和多少橙色箭头。这些 mini-NN 可以是 GRU 或 LSTM。我们将在后面的课程中介绍更多细节。
class Model5(nn.Module):def __init__(self):super().__init__()self.i_h = nn.Embedding(nv,nh)self.rnn = nn.GRU(nh, nh, 2, batch_first=True)self.h_o = nn.Linear(nh,nv)self.bn = BatchNorm1dFlat(nh)self.h = torch.zeros(2, bs, nh).cuda()def forward(self, x):res,h = self.rnn(self.i_h(x), self.h)self.h = h.detach()return self.h_o(self.bn(res))
learn = Learner(data, Model5(), metrics=accuracy)
learn.fit_one_cycle(10, 1e-2)
Connection to ULMFit
在上一课中，我们基本上用分类器替换了 self.h_o 来对文本进行分类。
结尾
RNN 只是一个重构的全连接神经网络。
你可以使用相同的方法处理任何序列标记任务（词性、分类材料是否敏感等）
