这是我的问题,我想在 TimeDistributed 层中使用预训练 CNN 网络之一。但是我在实施它时遇到了一些问题。
这是我的模型:
def bnn_model(max_len):
# sequence length and resnet input size
x = Input(shape=(maxlen, 224, 224, 3))
base_model = ResNet50.ResNet50(weights='imagenet', include_top=False)
for layer in base_model.layers:
layer.trainable = False
som = TimeDistributed(base_model)(x)
#the ouput of the model is [1, 1, 2048], need to squeeze
som = Lambda(lambda x: K.squeeze(K.squeeze(x,2),2))(som)
bnn = Bidirectional(LSTM(300))(som)
bnn = Dropout(0.5)(bnn)
pred = Dense(1, activation='sigmoid')(bnn)
model = Model(input=x, output=pred)
model.compile(optimizer=Adam(lr=1.0e-5), loss="mse", metrics=["accuracy"])
return model
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编译模型时我没有错误。但是当我开始训练时,我收到以下错误:
tensorflow/core/framework/op_kernel.cc:975] Invalid argument: You must feed a value …Run Code Online (Sandbox Code Playgroud) 所以我的问题是要知道是否有一种方法可以将值直接从vector(但我们也可以考虑array)传递给a tensorflow::tensor?
我知道的唯一方法是一个一个地复制每个值。
示例(2D矢量):
tensorflow::Tensor input(tensorflow::DT_FLOAT, tensorflow::TensorShape({50, 20}));
auto input_map = input.tensor<float, 2>();
for (int b = 0; b < 50; b++) {
for (int c = 0; c < 20; c++) {
input_map(b, c) = (vector_name)[b][c];
}
}
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有更方便的方法吗?
例如array到vector:
int x[3] = {1, 2, 3};
std::vector<int> v(x, x + sizeof x / sizeof x[0]);
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