def affine_forward(x, w, b): out = None # Reshape x into rows N = x.shape[0] x_row = x.reshape(N, -1) # (N,784) out = np.dot(x_row, w) + b # (N,M) cache = (x, w, b) return out, cacheshapes (40,6272) and (8192,100) not aligned: 6272 (dim 1) != 8192 (dim 0)报错
时间: 2023-11-16 09:06:53 浏览: 33
根据你提供的代码,错误可能在于你的权重矩阵和偏置项矩阵的维度不匹配。在这个具体的例子中,权重矩阵 w 的维度为 (6272, 8192),而偏置项矩阵 b 的维度为 (100,),这意味着你需要将偏置项矩阵 b 扩展为一个维度为 (1, 100) 的矩阵,然后将其加到 np.dot(x_row, w) 的结果上。你可以尝试以下代码:
```python
def affine_forward(x, w, b):
out = None
# Reshape x into rows
N = x.shape[0]
x_row = x.reshape(N, -1) # (N, 6272)
out = np.dot(x_row, w) + b.reshape(1, -1) # (N, 100)
cache = (x, w, b)
return out, cache
```
这应该可以解决报错问题。
相关问题
def affine_forward(x, w, b): out = None N = x.shape[0] x_row = x.reshape(N, -1) out = np.dot(x_row, w) + b.reshape(1, -1) cache = (x, w, b) return out, cache,W为超参数会导致程序错误,应该如何修改
如果W是超参数,则应该在函数外部定义并传递给函数。你可以修改函数定义,使其接受额外的W参数,如下所示:
```
def affine_forward(x, w, b, W):
out = None
N = x.shape[0]
x_row = x.reshape(N, -1)
out = np.dot(x_row, W) + b.reshape(1, -1)
cache = (x, W, b)
return out, cache
```
在调用函数时,你需要将W作为参数传递给函数。例如:
```
W = np.random.randn(784, 100)
out, cache = affine_forward(x, w, b, W)
```
这样,你就可以在函数内部使用超参数W,而不会出现错误。
class Affine: def __init__(self, W, b): self.W =W self.b = b self.x = None self.original_x_shape = None # 权重和偏置参数的导数 self.dW = None self.db = None def forward(self, x): # 对应张量 self.original_x_shape = x.shape x = x.reshape(x.shape[0], -1) self.x = x out = np.dot(self.x, self.W) + self.b return out def backward(self, dout): dx = np.dot(dout, self.W.T) self.dW = np.dot(self.x.T, dout) self.db = np.sum(dout, axis=0) dx = dx.reshape(*self.original_x_shape) # 还原输入数据的形状(对应张量) return dx
这是一个全连接层的类实现,也称为Affine层。在类的初始化函数中,初始化该层的权重W和偏置b,并且初始化该层的输入x以及原始输入x的形状original_x_shape。在前向传播函数中,将输入x展开为二维矩阵,然后与权重W相乘并加上偏置b,得到该层的输出out。在反向传播函数中,首先计算该层的输入梯度dx,即损失函数关于该层输入的梯度,然后计算该层的权重W和偏置b的梯度dW和db。最后将输入梯度dx还原为原始输入x的形状,从而得到该层的输入梯度dx。最终返回该层的输入梯度dx。