else: self.total_N = 1000 self.beta_0 = continuous_beta_0 self.beta_1 = continuous_beta_1 self.cosine_s = 0.008 self.cosine_beta_max = 999. self.cosine_t_max = math.atan(self.cosine_beta_max * (1. + self.cosine_s) / math.pi) * 2. * (1. + self.cosine_s) / math.pi - self.cosine_s self.cosine_log_alpha_0 = math.log(math.cos(self.cosine_s / (1. + self.cosine_s) * math.pi / 2.)) self.schedule = schedule if schedule == 'cosine': # For the cosine schedule, T = 1 will have numerical issues. So we manually set the ending time T. # Note that T = 0.9946 may be not the optimal setting. However, we find it works well. self.T = 0.9946 else: self.T = 1.解析
时间: 2024-02-14 08:18:47 浏览: 22
这段代码是某个类的初始化方法,它设置了该类的一些属性值。其中包括总迭代次数、beta_0、beta_1、cosine_s、cosine_beta_max、cosine_t_max、cosine_log_alpha_0、schedule和T等属性。如果schedule属性的值是'cosine',则设定T属性为0.9946,否则设为1。这段代码的目的是为了初始化该类的属性,为后续的操作做好准备。
相关问题
def setup_layers(self): self.lstm = torch.nn.LSTM( input_size = self.lstm_inputsize, hidden_size = self.lstm_hiddensize, num_layers = self.lstm_layers, batch_first=True, dropout=(0 if self.lstm_layers == 1 else self.lstm_dropout), bidirectional=False )
这段代码使用PyTorch中的`nn.LSTM`构建了一个LSTM层,具体参数如下:
- `input_size`:输入特征的维度。
- `hidden_size`:隐藏状态的维度。
- `num_layers`:LSTM层数。
- `batch_first`:如果为True,则输入和输出张量的形状为(batch_size, seq_length, feature_dims);否则,形状为(seq_length, batch_size, feature_dims)。
- `dropout`:如果`num_layers=1`,则不使用dropout;否则,将应用dropout并指定丢弃概率。
- `bidirectional`:是否使用双向LSTM。
需要注意的是,这段代码构建的是单向LSTM,如果需要构建双向LSTM,需要将`bidirectional`参数设置为True,如下所示:
```python
self.lstm = torch.nn.LSTM(
input_size=self.lstm_inputsize,
hidden_size=self.lstm_hiddensize,
num_layers=self.lstm_layers,
batch_first=True,
dropout=(0 if self.lstm_layers == 1 else self.lstm_dropout),
bidirectional=True
)
```
用python补全代码class Number(): def __init__(self, n1, n2): self.__n1 = n1 self.__n2 = n2 def addition(self): m1 = self.__n1 + self.__n2 print('%d+%d=%d' % (self.__n1,self.__n2,m1)) def subtration(self): m2 = self.__n1 - self.__n2 n1, n2, op = input().split(",") mm = Number(int(n1), int(n2)) if op == 'add': elif op == 'sub': mm.subtration() else: print("error!")
根据题目中的要求,下面是代码的完整实现:
```python
class Number():
def __init__(self, n1, n2):
self.__n1 = n1
self.__n2 = n2
def addition(self):
m1 = self.__n1 + self.__n2
print('%d+%d=%d' % (self.__n1, self.__n2, m1))
def subtration(self):
m2 = self.__n1 - self.__n2
print('%d-%d=%d' % (self.__n1, self.__n2, m2))
n1, n2, op = input().split(",")
mm = Number(int(n1), int(n2))
if op == 'add':
mm.addition()
elif op == 'sub':
mm.subtration()
else:
print("error!")
```
这里定义了一个Number类,包含addition和subtration方法。在创建Number对象mm时,我们从标准输入中读入了n1、n2和op三个变量,并将n1和n2转换为整数类型。如果op为'add',我们调用mm的addition方法,否则我们调用subtration方法。在addition和subtration方法中,我们分别计算出m1和m2,并输出相应的表达式和结果。如果op既不是'add'也不是'sub',我们输出"error!"。