如何在tensorflow中对图像进行归一化处理？

Question

目前我使用的是tf.image.per_image_standardization(image)，但似乎它需要比下面的方法更长的时间才能收敛：

image = image - image_mean 

我需要知道数据集中的image_mean = [meanR, meanG, meanB]。我做错了什么？

- Ray.R.Chua

Answer 1

该函数执行不同的过程。你只减去平均值，但是tf.image.per_image_standardization()还会除以方差。从API文档中可以看到：

This op computes (x - mean) / adjusted_stddev, where mean is the average of all values in image, and adjusted_stddev = max(stddev, 1.0/sqrt(image.NumElements())).

以下是他们的完整实现，来自这里：

def per_image_standardization(image):
  """Linearly scales `image` to have zero mean and unit norm.
  This op computes `(x - mean) / adjusted_stddev`, where `mean` is the average
  of all values in image, and
  `adjusted_stddev = max(stddev, 1.0/sqrt(image.NumElements()))`.
  `stddev` is the standard deviation of all values in `image`. It is capped
  away from zero to protect against division by 0 when handling uniform images.
  Args:
    image: 3-D tensor of shape `[height, width, channels]`.
  Returns:
    The standardized image with same shape as `image`.
  Raises:
    ValueError: if the shape of 'image' is incompatible with this function.
  """
  image = ops.convert_to_tensor(image, name='image')
  _Check3DImage(image, require_static=False)
  num_pixels = math_ops.reduce_prod(array_ops.shape(image))

  image = math_ops.cast(image, dtype=dtypes.float32)
  image_mean = math_ops.reduce_mean(image)

  variance = (math_ops.reduce_mean(math_ops.square(image)) -
              math_ops.square(image_mean))
  variance = gen_nn_ops.relu(variance)
  stddev = math_ops.sqrt(variance)

  # Apply a minimum normalization that protects us against uniform images.
  min_stddev = math_ops.rsqrt(math_ops.cast(num_pixels, dtypes.float32))
  pixel_value_scale = math_ops.maximum(stddev, min_stddev)
  pixel_value_offset = image_mean

  image = math_ops.subtract(image, pixel_value_offset)
  image = math_ops.div(image, pixel_value_scale)
  return image

- MZHm