Keras如何自定义IOU-创新互联
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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
文章名称:Keras如何自定义IOU-创新互联
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