#    Copyright 2020 Division of Medical Image Computing, German Cancer Research Center (DKFZ), Heidelberg, Germany
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#    you may not use this file except in compliance with the License.
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#        http://www.apache.org/licenses/LICENSE-2.0
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import numpy as np
from medpy.metric.binary import __surface_distances


def normalized_surface_dice(a: np.ndarray, b: np.ndarray, threshold: float, spacing: tuple = None, connectivity=1):
    """
    This implementation differs from the official surface dice implementation! These two are not comparable!!!!!

    The normalized surface dice is symmetric, so it should not matter whether a or b is the reference image

    This implementation natively supports 2D and 3D images. Whether other dimensions are supported depends on the
    __surface_distances implementation in medpy

    :param a: image 1, must have the same shape as b
    :param b: image 2, must have the same shape as a
    :param threshold: distances below this threshold will be counted as true positives. Threshold is in mm, not voxels!
    (if spacing = (1, 1(, 1)) then one voxel=1mm so the threshold is effectively in voxels)
    must be a tuple of len dimension(a)
    :param spacing: how many mm is one voxel in reality? Can be left at None, we then assume an isotropic spacing of 1mm
    :param connectivity: see scipy.ndimage.generate_binary_structure for more information. I suggest you leave that
    one alone
    :return:
    """
    assert all([i == j for i, j in zip(a.shape, b.shape)]), "a and b must have the same shape. a.shape= %s, " \
                                                            "b.shape= %s" % (str(a.shape), str(b.shape))
    if spacing is None:
        spacing = tuple([1 for _ in range(len(a.shape))])
    a_to_b = __surface_distances(a, b, spacing, connectivity)
    b_to_a = __surface_distances(b, a, spacing, connectivity)

    numel_a = len(a_to_b)
    numel_b = len(b_to_a)

    tp_a = np.sum(a_to_b <= threshold) / numel_a
    tp_b = np.sum(b_to_a <= threshold) / numel_b

    fp = np.sum(a_to_b > threshold) / numel_a
    fn = np.sum(b_to_a > threshold) / numel_b

    dc = (tp_a + tp_b) / (tp_a + tp_b + fp + fn + 1e-8)  # 1e-8 just so that we don't get div by 0
    return dc