以前的解决方案对我不起作用。我对这个问题的处理如下所示。

layer_outputs = []
for i in range(1, len(model.layers)):
    tmp_model = Model(model.layers[0].input, model.layers[i].output)
    tmp_output = tmp_model.predict(img)[0]
    layer_outputs.append(tmp_output)

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')

以下对我来说很简单:

model.layers[idx].output

上面是一个张量对象，所以你可以使用应用于张量对象的操作来修改它。

例如，要获取形状model.layers[idx].output.get_shape()

Idx是该层的索引，你可以从model.summary()中找到它

以前的解决方案对我不起作用。我对这个问题的处理如下所示。

layer_outputs = []
for i in range(1, len(model.layers)):
    tmp_model = Model(model.layers[0].input, model.layers[i].output)
    tmp_output = tmp_model.predict(img)[0]
    layer_outputs.append(tmp_output)

来自:https://github.com/philipperemy/keras-visualize-activations/blob/master/read_activations.py

import keras.backend as K

def get_activations(model, model_inputs, print_shape_only=False, layer_name=None):
    print('----- activations -----')
    activations = []
    inp = model.input

    model_multi_inputs_cond = True
    if not isinstance(inp, list):
        # only one input! let's wrap it in a list.
        inp = [inp]
        model_multi_inputs_cond = False

    outputs = [layer.output for layer in model.layers if
               layer.name == layer_name or layer_name is None]  # all layer outputs

    funcs = [K.function(inp + [K.learning_phase()], [out]) for out in outputs]  # evaluation functions

    if model_multi_inputs_cond:
        list_inputs = []
        list_inputs.extend(model_inputs)
        list_inputs.append(0.)
    else:
        list_inputs = [model_inputs, 0.]

    # Learning phase. 0 = Test mode (no dropout or batch normalization)
    # layer_outputs = [func([model_inputs, 0.])[0] for func in funcs]
    layer_outputs = [func(list_inputs)[0] for func in funcs]
    for layer_activations in layer_outputs:
        activations.append(layer_activations)
        if print_shape_only:
            print(layer_activations.shape)
        else:
            print(layer_activations)
    return activations