Question

我已经将自定义训练的图像分割TensorFlow模型转换为TF Lite格式，现在正尝试使用CPP API在该模型上进行推断。

我正在研究输入和输出张量，并遵循以下link进行此操作。奇怪的是，我的输入张量的形状大于需要馈入模型的图像。图像的形状为480x640x3，但是下面的代码片段中的dims->data包含的信息远不止于此（请参见控制台的screenshot）。怎么会来？

#include <cstdio>
#include <iostream>
#include <iomanip>
#include "tensorflow/lite/interpreter.h"
#include "tensorflow/lite/kernels/register.h"
#include "tensorflow/lite/model.h"
#include "tensorflow/lite/optional_debug_tools.h"
#include <time.h>


using namespace tflite;
using namespace std;

#define TFLITE_MINIMAL_CHECK(x)                              \
  if (!(x)) {                                                \
    fprintf(stderr, "Error at %s:%d\n", __FILE__, __LINE__); \
    exit(1);                                                 \
  }

int main(int argc, char* argv[]) {
  if (argc != 2) {
    fprintf(stderr, "minimal <tflite model>\n");
    return 1;
  }
  const char* filename = argv[1];
  int startt, endd;
  startt = clock();

  // Load model
  // Load model
  std::unique_ptr<tflite::FlatBufferModel> model =
      tflite::FlatBufferModel::BuildFromFile(filename);
  TFLITE_MINIMAL_CHECK(model != nullptr);

  // Build the interpreter
  tflite::ops::builtin::BuiltinOpResolver resolver;
  std::unique_ptr<Interpreter> interpreter(new Interpreter());
  InterpreterBuilder builder(*model, resolver);
  std::cout << "allocating a number of threads" << std::endl;
  int numthreads=4;
  interpreter->SetNumThreads(numthreads);
  std::cout << "threads allocated" << std::endl;

  builder(&interpreter);
  TFLITE_MINIMAL_CHECK(interpreter != nullptr);
  endd = clock();

  // Allocate tensor buffers.
  TFLITE_MINIMAL_CHECK(interpreter->AllocateTensors() == kTfLiteOk);
  printf("=== Pre-invoke Interpreter State ===\n");
  tflite::PrintInterpreterState(interpreter.get());

  double time_taken = double(endd-startt)/double(CLOCKS_PER_SEC);
  std::cout << "a number of threads should be selected" << std::endl;
  std::cout << "Time taken to load in the model in tflite using CPP API :" << fixed << time_taken << std::setprecision(5);
  std::cout << " sec " << std::endl;


  // Fill input buffers
  std::cout << "inputs : " << interpreter->inputs().size() << "\n";
  std::cout << "inputs(0) name : " << interpreter->GetInputName(0) << "\n";
  std::cout << "tensors size: " << interpreter->tensors_size() << "\n";
  std::cout << "nodes size: " << interpreter->nodes_size() << "\n";
  int startinput = clock();
  int input = interpreter->inputs()[0];
  std::cout << "input.1 : " << input <<"\n";
  const std::vector<int> inputs = interpreter->inputs();
  const std::vector<int> outputs = interpreter->outputs();
  std::cout << "number of inputs: " <<inputs.size() << "\n";
  std::cout << "number of outputs: " <<outputs.size() << "\n";

  TfLiteIntArray* dims = interpreter->tensor(input)->dims;
  int test0 = dims->data[0];
  int wanted_channels = dims->data[1];
  int wanted_height = dims->data[2];
  int wanted_width = dims->data[3];
  int test4 = dims->data[4];
  int test5 = dims->data[5];
  int test6 = dims->data[6];
  std::cout << "type of input tensor: " << interpreter->tensor(input)->type << std::endl;
  std::cout << "height, width, channels of input : " << wanted_height << " " << wanted_width << " "<< wanted_channels <<  " " << test0 << " " << test4 << " " << test5 << " " << test6 << std::endl;
}

为什么输入张量包含比输入形状更多的尺寸？

0 个答案: