Can Dense() and Conv2d() can act as same layer function?-CodePudding

Lets suppose that I have a input layer with shape (h,w,f) = (1 x 1 x 256 )

And let me make two sequence

case 1 :

input = keras.models.Input((1,1,256))
x = keras.layers.Conv2d(f= 32, k=(1,1),s = 1)(input)
x = keras.layers.ReLU()(x) 
x = keras.layers.Conv2d(f= 256, k=(1,1),s = 1)(x)

case 2 :

input = keras.models.Input((1,1,256))
x = keras.layers.Flatten()(input)
x = keras.layers.Dense(32)(x)
x = keras.layers.ReLU()(x) 
x = keras.layers.Dense(256)(x)
x = keras.layers.reshape((1,1,256))(x)

In this 2 cases are the output x are same?

I am making an SE-Net-like attention module but not same.

CodePudding user response：

Yes, and you do not need to apply Flatten() and Reshape() in code 2. Dense will be applied on the last channel automatically.