TensorFlow：冻结图形后显着的精度损失？

2 回答

• 1

  我认为问题与冻结模型无关 . 相反，它与您预处理图像的方式有关 .

  我建议您使用InceptionResnet V2中的默认预处理功能 .

  下面，我将发布一个采用图像路径（JPG或PNG）并返回预处理图像的代码 . 您可以修改它以使其接收一批图像 . 它不是专业代码 . 它需要一些优化 . 但是，它运作良好 .

  一，加载图片：

  def load_img(path_img):
      """
      Load an image to tensorflow
      :param path_img: image path on the disk
      :return: 3D tensorflow image
      """
      filename_queue = tf.train.string_input_producer([path_img])  # list of files to read
  
      reader = tf.WholeFileReader()
      key, value = reader.read(filename_queue)
  
      my_img = tf.image.decode_image(value)  # use png or jpg decoder based on your files.
  
      init_op = tf.global_variables_initializer()
      with tf.Session() as sess:
          sess.run(init_op)
  
          # Start populating the filename queue.
  
          coord = tf.train.Coordinator()
          threads = tf.train.start_queue_runners(coord=coord)
  
          for i in range(1):  # length of your filename list
              image = my_img.eval()  # here is your image Tensor :)
  
          print(image.shape)
          # Image.fromarray(np.asarray(image)).show()
  
          coord.request_stop()
          coord.join(threads)
  
          return image
  

  然后，预处理代码：

  def preprocess(image, height, width,
                 central_fraction=0.875, scope=None):
      """Prepare one image for evaluation.
  
      If height and width are specified it would output an image with that size by
      applying resize_bilinear.
  
      If central_fraction is specified it would cropt the central fraction of the
      input image.
  
      Args:
        image: 3-D Tensor of image. If dtype is tf.float32 then the range should be
          [0, 1], otherwise it would converted to tf.float32 assuming that the range
          is [0, MAX], where MAX is largest positive representable number for
          int(8/16/32) data type (see `tf.image.convert_image_dtype` for details)
        height: integer
        width: integer
        central_fraction: Optional Float, fraction of the image to crop.
        scope: Optional scope for name_scope.
      Returns:
        3-D float Tensor of prepared image.
      """
  
      image = tf.image.convert_image_dtype(image, dtype=tf.float32)
      # Crop the central region of the image with an area containing 87.5% of
      # the original image.
      if central_fraction:
          image = tf.image.central_crop(image, central_fraction=central_fraction)
  
      if height and width:
          # Resize the image to the specified height and width.
          image = tf.expand_dims(image, 0)
          image = tf.image.resize_bilinear(image, [height, width],
                                           align_corners=False)
          image = tf.squeeze(image, [0])
      image = tf.subtract(image, 0.5)
      image = tf.multiply(image, 2.0)
      return image
  

  最后，对于我的情况，我不得不将处理后的张量转换为numpy数组：

  image = tf.Session().run(image)
  

  因此，该图像可以馈送到冻结模型

  persistent_sess = tf.Session(graph=graph)  # , config=sess_config)
  
      input_node = graph.get_tensor_by_name('prefix/batch:0')
      output_node = graph.get_tensor_by_name('prefix/InceptionResnetV2/Logits/Predictions:0')
  
      predictions = persistent_sess.run(output_node, feed_dict={input_node: [image]})
      print(predictions)
      label_predicted = np.argmax(predictions[0])
      print(label_predicted)
  

  回复于 2024-05-02T11:46:40+08:00
• 1

  我有类似的问题，使用冷冻模型时准确度降低了1.5％ . 问题在于冻结模型的代码中的保护程序对象 . 您需要将移动平均值衰减作为参数传递给保护程序 . 我使用Inception模型中的代码，这就是我在冻结脚本中创建保护程序的方法：

  variable_averages = tf.train.ExponentialMovingAverage(0.9997)
  variables_to_restore = variable_averages.variables_to_restore()
  saver = tf.train.Saver(variables_to_restore)
  

  对我来说，它解决了这个问题 .

  回复于 2024-05-02T11:46:40+08:00