Tensorflow 2，cpp 中的服务预测问题

Question

我正在尝试使用 cpp 中的最终

saved_model.pb

发送

text as input

并接收

predicted class id and score

。 但我收到这个错误：

W tensorflow/core/framework/op_kernel.cc:1767] OP_REQUIRES failed at strided_slice_op.cc:108 : Invalid argument: slice index 0 of dimension 0 out of bounds.

我的模型是使用 python 训练的。现在我在 TensorFlow 2.5 中使用它。

使用saved_model_cli的更多信息：

MetaGraphDef with tag-set: 'serve' contains the following SignatureDefs:

signature_def['classification']:
  The given SavedModel SignatureDef contains the following input(s):
    inputs['inputs'] tensor_info:
        dtype: DT_STRING
        shape: (-1)
        name: Placeholder:0
  The given SavedModel SignatureDef contains the following output(s):
    outputs['classes'] tensor_info:
        dtype: DT_STRING
        shape: (-1, 74)
        name: head/Tile:0
    outputs['scores'] tensor_info:
        dtype: DT_FLOAT
        shape: (-1, 74)
        name: head/predictions/probabilities:0
  Method name is: tensorflow/serving/classify

signature_def['predict']:
  The given SavedModel SignatureDef contains the following input(s):
    inputs['content'] tensor_info:
        dtype: DT_STRING
        shape: (-1)
        name: Placeholder:0
  The given SavedModel SignatureDef contains the following output(s):
    outputs['all_class_ids'] tensor_info:
        dtype: DT_INT32
        shape: (-1, 74)
        name: head/predictions/Tile:0
    outputs['all_classes'] tensor_info:
        dtype: DT_STRING
        shape: (-1, 74)
        name: head/predictions/Tile_1:0
    outputs['class_ids'] tensor_info:
        dtype: DT_INT64
        shape: (-1, 1)
        name: head/predictions/ExpandDims:0
    outputs['classes'] tensor_info:
        dtype: DT_STRING
        shape: (-1, 1)
        name: head/predictions/hash_table_Lookup/LookupTableFindV2:0
    outputs['logits'] tensor_info:
        dtype: DT_FLOAT
        shape: (-1, 74)
        name: add:0
    outputs['probabilities'] tensor_info:
        dtype: DT_FLOAT
        shape: (-1, 74)
        name: head/predictions/probabilities:0
  Method name is: tensorflow/serving/predict

signature_def['serving_default']:
  The given SavedModel SignatureDef contains the following input(s):
    inputs['inputs'] tensor_info:
        dtype: DT_STRING
        shape: (-1)
        name: Placeholder:0
  The given SavedModel SignatureDef contains the following output(s):
    outputs['classes'] tensor_info:
        dtype: DT_STRING
        shape: (-1, 74)
        name: head/Tile:0
    outputs['scores'] tensor_info:
        dtype: DT_FLOAT
        shape: (-1, 74)
        name: head/predictions/probabilities:0
  Method name is: tensorflow/serving/classify

我的源代码：

// run the model
std::string test(const char *modelPath, const char *text) {
    tensorflow::SavedModelBundle model;
    tensorflow::Status status = tensorflow::LoadSavedModel(
            tensorflow::SessionOptions(),
            tensorflow::RunOptions(),
            modelPath,
            {"serve"},
            &model);

    TF_CHECK_OK(status);

    // Provide input data.
    tensorflow::Tensor tensor(tensorflow::DT_STRING, tensorflow::TensorShape());
    tensor.scalar<tensorflow::tstring>()() = tensorflow::tstring(text);

    // Link the data with some tags so tensorflow know where to put those data entries.
    std::vector<std::pair<std::string, tensorflow::Tensor>> feedInputs = {{"Placeholder:0", tensor}};
    std::vector<std::string> fetches = {"head/Tile:0", "head/predictions/probabilities:0"};

    // We need to store the results somewhere.
    std::vector<tensorflow::Tensor> outputs;

    // Let's run the model...
    status = model.GetSession()->Run(feedInputs, fetches, {}, &outputs);
    TF_CHECK_OK(status);

    for (const auto& output : outputs) {
        // TODO::
    }

    return "";
}

我尝试更改 cpp 中的输入和输出名称，但似乎正确的名称是此源代码中使用的名称。 我对 ternsorflow cpp api 不太熟悉。 谢谢。

Answer 1

问题出在

tensorflow::Tensor tensor(tensorflow::DT_STRING, tensorflow::TensorShape());

应该是

tensorflow::Tensor tensor(tensorflow::DT_STRING, tensorflow::TensorShape({1}));

错误告诉您

Invalid argument: slice index 0 of dimension 0 out of bounds.

看起来张量对于 1 个字符串没有额外的维度（一维）。