在keras中添加以前的图层？ - Conv2D' 对象没有属性 'is_placeholder'

Question

我在 keras 中附加图层时似乎遇到了一些问题。

示例：

import keras
from keras.layers.merge import Concatenate
from keras.models import Model
from keras.layers import Input, Dense
from keras.layers import Dropout
from keras.layers.core import Dense, Activation, Lambda, Reshape,Flatten
from keras.layers import Conv2D, MaxPooling2D, Reshape, ZeroPadding2D

input_img = Input(shape=(3, 6, 3))

conv2d_1_1 = Conv2D(filters = 32, kernel_size = (3,3) , padding = "same" , activation = 'relu' , name = "conv2d_1_1" )(input_img)
conv2d_2_1 = Conv2D(filters = 64, kernel_size = (3,3) , padding = "same" , activation = 'relu' )(conv2d_1_1)
conv2d_3_1 = Conv2D(filters = 64, kernel_size = (3,3) , padding = "same" , activation = 'relu' )(conv2d_2_1)
conv2d_4_1 = Conv2D(filters = 32, kernel_size = (1,1) , padding = "same" , activation = 'relu' )(conv2d_3_1)
conv2d_4_1_flatten = Flatten()(conv2d_4_1)

conv2d_1_2 = Conv2D(filters = 32, kernel_size = (3,3) , padding = "same" , activation = 'relu' , name = "conv2d_1_2")(input_img)
conv2d_2_2 = Conv2D(filters = 64, kernel_size = (3,3) , padding = "same" , activation = 'relu' )(conv2d_1_2)
conv2d_3_2 = Conv2D(filters = 64, kernel_size = (3,3) , padding = "same" , activation = 'relu' )(conv2d_2_2)
conv2d_4_2 = Conv2D(filters = 32, kernel_size = (1,1) , padding = "same" , activation = 'relu' )(conv2d_3_2)
conv2d_4_2_flatten = Flatten()(conv2d_4_2)


merge = keras.layers.concatenate([conv2d_4_1_flatten, conv2d_4_2_flatten])

dense1 = Dense(100, activation = 'relu')(merge)
dense2 = Dense(50,activation = 'relu')(dense1)
dense3 = Dense(1 ,activation = 'softmax')(dense2)


model = Model(inputs = [conv2d_1_1 , conv2d_1_2] , outputs = dense3)
model.compile(loss="crossentropy", optimizer="adam")

print model.summary()

为什么我不能像这样附加图层？ 输入是我手动分离成 (3,6,3) 形状的图像。

Answer 1

你输入的不对，你自己说的，输入的是你的图片。更改创建模型的方式：

model = Model(inputs = input_img , outputs = dense3)

这应该有效。