目前我正在使用 tf.image.per_image_standardization(image) 但它似乎比使用以下方法需要更长的时间来收敛:
image = image - image_mean
数据集的 image_mean = [meanR, meanG, meanB]。我做错了什么？
请您参考如下方法:
该函数执行不同的过程。您只需减去平均值，但 tf.image.per_image_standardization() 还会除以方差。来自API docs :
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())).
这是来自 here 的完整实现:
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
