Keras：将前一层的一部分送到下一层，在CNN中

Question

我想将前一层的各个内核输出馈入一个新的转换过滤器，以获得下一层。为此，我尝试通过Conv2D传递每个内核输出，并通过索引来调用它们。我使用的功能是：

def modification(weights_path=None, classes=2): 

    ########### 
    ## Input ## 
    ########### 

    ### 224x224x3 sized RGB Input 
    inputs = Input(shape=(224,224,3)) 

    ################################# 
    ## Conv2D Layer with 5 kernels ## 
    ################################# 

    k = 5 
    x = Conv2D(k, (3,3), data_format='channels_last', padding='same', name='block1_conv1')(inputs) 
    y = np.empty(k, dtype=object) 
    for i in range(0,k): 
     y[i] = Conv2D(1, (3,3), data_format='channels_last', padding='same')(np.asarray([x[i]])) 
    y = keras.layers.concatenate([y[i] for i in range (0,k)], axis=3, name='block1_conv1_loc') 
    out = Activation('relu')(y) 
    print ('Output shape is, ' +str(out.get_shape())) 

    ### Maxpooling(2,2) with a stride of (2,2) 
    out = MaxPooling2D((2,2), strides=(2,2), data_format='channels_last')(out) 

    ############################################ 
    ## Top layer, with fully connected layers ## 
    ############################################ 

    out = Flatten(name='flatten')(out) 
    out = Dense(4096, activation='relu', name='fc1')(out) 
    out = Dropout(0.5)(out) 
    out = Dense(4096, activation='relu', name='fc2')(out) 
    out = Dropout(0.5)(out) 
    out = Dense(classes, activation='softmax', name='predictions')(out) 

    if weights_path: 
     model.load_weights(weights_path) 

    model = Model(inputs, out, name='modification') 

    return model 

但这不是工作，并抛出以下错误：

Traceback (most recent call last): 
    File "sim-conn-edit.py", line 137, in <module> 
    model = modification() 
    File "sim-conn-edit.py", line 38, in modification 
    y[i] = Conv2D(1, (3,3), data_format='channels_last', padding='same')(np.asarray([x[i]])) 
    File "/home/yx96/anaconda2/lib/python2.7/site-packages/keras/engine/topology.py", line 511, in __call__ 
    self.assert_input_compatibility(inputs) 
    File "/home/yx96/anaconda2/lib/python2.7/site-packages/keras/engine/topology.py", line 408, in assert_input_compatibil 
ity 
    if K.ndim(x) != spec.ndim: 
    File "/home/yx96/anaconda2/lib/python2.7/site-packages/keras/backend/tensorflow_backend.py", line 437, in ndim 
    dims = x.get_shape()._dims 
AttributeError: 'numpy.ndarray' object has no attribute 'get_shape' 

我送进层x[i]作为[ x[i] ]满足Conv2D层的渔政船的要求。任何帮助解决这个问题将深受赞赏！

Answer 1

在StackOverflow发布this和this问题以及一些个人探索之后，我想出了一个解决方案。可以用Lambda层做到这一点;通过调用Lambda图层来提取前一图层的子部分。例如，如果Lambda函数被定义为，

def layer_slice(x,i): 
    return x[:,:,:,i:i+1] 

然后，称为，

k = 5 
x = Conv2D(k, (3,3), data_format='channels_last', padding='same', name='block1_conv1')(inputs) 
y = np.empty(k, dtype=object) 
for i in range(0,k): 
    y[i] = Lambda(layer_slice, arguments={'i':i})(x) 
    y[i] = Conv2D(1,(3,3), data_format='channels_last', padding='same')(y[i]) 
y = keras.layers.concatenate([y[i] for i in range (0,k)], axis=3, name='block1_conv1_loc') 
out = Activation('relu')(y) 
print ('Output shape is, ' +str(out.get_shape())) 

它应该在各个内核输出有效供给到一个新Conv2D层。从model.summary()获得的层次形状和相应的可训练参数数量符合预期。感谢Daniel指出Lambda图层不能有可训练的权重。

Answer 2

Prabaha。我知道你已经解决了你的问题，但是现在我看到了你的答案，你可以在不使用lambda层的情况下做到这一点，只需分割第一个Conv2D。一层具有k滤波器等效于k层，一个过滤器：

for i in range(0,k): 
    y[i] = Conv2D(1, (3,3), ... , name='block1_conv'+str(i))(inputs)  
    y[i] = Conv2D(1,(3,3), ...)(y[i]) 
y = Concatenate()([y[i] for i in range (0,k)]) 
out = Activation('relu')(y) 

你可以在你的答案，在这个答案比较计算总参数。