通常，输出大小可以计算为

[(W−K + 2P / S] + 1

在哪里

W is the input volume - in your case you have not given us this
K is the Kernel size - in your case 2 == "filter" 
P is the padding - in your case 2
S is the stride - in your case 3

另一个更漂亮的说法是:

你可以通过使用:model.layers[index].output轻松获得任何层的输出

对于所有层使用这个:

from keras import backend as K

inp = model.input                                           # input placeholder
outputs = [layer.output for layer in model.layers]          # all layer outputs
functors = [K.function([inp, K.learning_phase()], [out]) for out in outputs]    # evaluation functions

# Testing
test = np.random.random(input_shape)[np.newaxis,...]
layer_outs = [func([test, 1.]) for func in functors]
print layer_outs

注意:要模拟Dropout，请使用learning_phase作为1。在layer_outs中，否则使用0。

编辑:(根据评论)

K.function创建了ano/tensorflow张量函数，之后用于从给定输入的符号图中获得输出。

现在需要K.learning_phase()作为输入，因为许多Keras层(如Dropout/Batchnomalization)依赖它来改变训练和测试期间的行为。

所以如果你在你的代码中删除dropout层，你可以简单地使用:

from keras import backend as K

inp = model.input                                           # input placeholder
outputs = [layer.output for layer in model.layers]          # all layer outputs
functors = [K.function([inp], [out]) for out in outputs]    # evaluation functions

# Testing
test = np.random.random(input_shape)[np.newaxis,...]
layer_outs = [func([test]) for func in functors]
print layer_outs

编辑2:更优化

我刚刚意识到，前面的答案并没有优化，因为对于每个函数的计算，数据将被CPU->GPU内存传输，而且张量计算需要对较低的层进行n-over。

相反，这是一种更好的方式，因为你不需要多个函数，而是一个函数给你所有输出的列表:

from keras import backend as K

inp = model.input                                           # input placeholder
outputs = [layer.output for layer in model.layers]          # all layer outputs
functor = K.function([inp, K.learning_phase()], outputs )   # evaluation function

# Testing
test = np.random.random(input_shape)[np.newaxis,...]
layer_outs = functor([test, 1.])
print layer_outs

希望将此作为评论(但没有足够高的代表)添加到@indraforyou的回答中，以纠正@mathtick的评论中提到的问题。为了避免InvalidArgumentError: input_X:Y被提供和获取。异常，只需替换行输出=[层。模型中层的输出。输出= [layer.]模型中层的输出。层][1:]。

调整indraforyou的最小工作示例:

from keras import backend as K 
inp = model.input                                           # input placeholder
outputs = [layer.output for layer in model.layers][1:]        # all layer outputs except first (input) layer
functor = K.function([inp, K.learning_phase()], outputs )   # evaluation function

# Testing
test = np.random.random(input_shape)[np.newaxis,...]
layer_outs = functor([test, 1.])
print layer_outs

附注:我尝试输出=[层。模型中层的输出。Layers[1:]]不起作用。

如果你有以下情况之一:

InvalidArgumentError: input_X:Y既被提供也被获取 多输入情况

您需要做以下更改:

为输出变量中的输入层添加过滤 函子循环的微小变化

最小的例子:

from keras.engine.input_layer import InputLayer
inp = model.input
outputs = [layer.output for layer in model.layers if not isinstance(layer, InputLayer)]
functors = [K.function(inp + [K.learning_phase()], [x]) for x in outputs]
layer_outputs = [fun([x1, x2, xn, 1]) for fun in functors]

我为自己写了这个函数(在Jupyter)，它的灵感来自indraforyou的答案。它会自动绘制所有的层输出。您的图像必须具有(x, y, 1)形状，其中1代表1个通道。您只需调用plot_layer_outputs(…)来绘图。

%matplotlib inline
import matplotlib.pyplot as plt
from keras import backend as K

def get_layer_outputs():
    test_image = YOUR IMAGE GOES HERE!!!
    outputs    = [layer.output for layer in model.layers]          # all layer outputs
    comp_graph = [K.function([model.input]+ [K.learning_phase()], [output]) for output in outputs]  # evaluation functions

    # Testing
    layer_outputs_list = [op([test_image, 1.]) for op in comp_graph]
    layer_outputs = []

    for layer_output in layer_outputs_list:
        print(layer_output[0][0].shape, end='\n-------------------\n')
        layer_outputs.append(layer_output[0][0])

    return layer_outputs

def plot_layer_outputs(layer_number):    
    layer_outputs = get_layer_outputs()

    x_max = layer_outputs[layer_number].shape[0]
    y_max = layer_outputs[layer_number].shape[1]
    n     = layer_outputs[layer_number].shape[2]

    L = []
    for i in range(n):
        L.append(np.zeros((x_max, y_max)))

    for i in range(n):
        for x in range(x_max):
            for y in range(y_max):
                L[i][x][y] = layer_outputs[layer_number][x][y][i]


    for img in L:
        plt.figure()
        plt.imshow(img, interpolation='nearest')

通常，输出大小可以计算为

[(W−K + 2P / S] + 1

在哪里

W is the input volume - in your case you have not given us this
K is the Kernel size - in your case 2 == "filter" 
P is the padding - in your case 2
S is the stride - in your case 3

另一个更漂亮的说法是: