medpy.metric.binary.ravd#

medpy.metric.binary.ravd(result, reference)[source]#

Relative absolute volume difference.

Compute the relative absolute volume difference between the (joined) binary objects in the two images.

Parameters:

resultarray_like: Input data containing objects. Can be any type but will be converted into binary: background where 0, object everywhere else.
referencearray_like: Input data containing objects. Can be any type but will be converted into binary: background where 0, object everywhere else.

Returns:

ravdfloat: The relative absolute volume difference between the object(s) in result and the object(s) in reference. This is a percentage value in the range \([-1.0, +inf]\) for which a \(0\) denotes an ideal score.

Raises:

RuntimeError: If the reference object is empty.

See also

dc
precision
recall

Notes

This is not a real metric, as it is directed. Negative values denote a smaller and positive values a larger volume than the reference. This implementation does not check, whether the two supplied arrays are of the same size.

Examples

Considering the following inputs

>>> import numpy
>>> arr1 = numpy.asarray([[0,1,0],[1,1,1],[0,1,0]])
>>> arr1
array([[0, 1, 0],
       [1, 1, 1],
       [0, 1, 0]])
>>> arr2 = numpy.asarray([[0,1,0],[1,0,1],[0,1,0]])
>>> arr2
array([[0, 1, 0],
       [1, 0, 1],
       [0, 1, 0]])

comparing arr1 to arr2 we get

>>> ravd(arr1, arr2)
-0.2

and reversing the inputs the directivness of the metric becomes evident

>>> ravd(arr2, arr1)
0.25

It is important to keep in mind that a perfect score of 0 does not mean that the binary objects fit exactely, as only the volumes are compared:

>>> arr1 = numpy.asarray([1,0,0])
>>> arr2 = numpy.asarray([0,0,1])
>>> ravd(arr1, arr2)
0.0