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我就废话不多说了，大家还是直接看代码吧！

def iou(y_true, y_pred, label: int):
  """
  Return the Intersection over Union (IoU) for a given label.
  Args:
    y_true: the expected y values as a one-hot
    y_pred: the predicted y values as a one-hot or softmax output
    label: the label to return the IoU for
  Returns:
    the IoU for the given label
  """
  # extract the label values using the argmax operator then
  # calculate equality of the predictions and truths to the label
  y_true = K.cast(K.equal(K.argmax(y_true), label), K.floatx())
  y_pred = K.cast(K.equal(K.argmax(y_pred), label), K.floatx())
  # calculate the |intersection| (AND) of the labels
  intersection = K.sum(y_true * y_pred)
  # calculate the |union| (OR) of the labels
  union = K.sum(y_true) + K.sum(y_pred) - intersection
  # avoid divide by zero - if the union is zero, return 1
  # otherwise, return the intersection over union
  return K.switch(K.equal(union, 0), 1.0, intersection / union)
 
def mean_iou(y_true, y_pred):
  """
  Return the Intersection over Union (IoU) score.
  Args:
    y_true: the expected y values as a one-hot
    y_pred: the predicted y values as a one-hot or softmax output
  Returns:
    the scalar IoU value (mean over all labels)
  """
  # get number of labels to calculate IoU for
  num_labels = K.int_shape(y_pred)[-1] - 1
  # initialize a variable to store total IoU in
  mean_iou = K.variable(0)
  
  # iterate over labels to calculate IoU for
  for label in range(num_labels):
    mean_iou = mean_iou + iou(y_true, y_pred, label)
    
  # divide total IoU by number of labels to get mean IoU
  return mean_iou / num_labels

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