骰子系数大于1

时间:2020-02-06 08:14:18

标签: keras deep-learning image-segmentation

当我训练UNET时,骰子系数和iou有时会大于1和iou > dice,然后在几批处理之后它们又会恢复正常。 如picture所示。

我对它们的定义如下:

def dice_coef(y_true, y_pred, smooth=1):
    y_true_f = K.flatten(y_true)
    y_pred_f = K.flatten(y_pred)
    intersection = K.sum(y_true_f * y_pred_f)
    return (2. * intersection + smooth) / (K.sum(y_true_f) + K.sum(y_pred_f) + smooth)

def iou(y_true, y_pred, smooth=1):
    y_true_f = K.flatten(y_true)
    y_pred_f = K.flatten(y_pred)
    intersection = K.sum(y_true_f * y_pred_f)
    union = K.sum(y_true_f) + K.sum(y_pred_f) - intersection
    return (intersection + smooth) / (union + smooth)

def dice_loss(y_true, y_pred):
    return 1. - dice_coef(y_true, y_pred)

我曾尝试在y_pred中添加K.abs(),但这会导致性能下降。我觉得既然输出是S型激活的,是否添加K.abs()应该会得到相同的结果?另外,如您所见,我的准确性很奇怪,我一直依靠骰子来判断我的模型性能,如果有人指出这个问题,那就更好了。

1 个答案:

答案 0 :(得分:2)

我相信您的y_true图像可能不在0到1范围内。...您确定它们不在0到255之间吗?还是他们只有一个频道(而不是3个频道?)

这不是原因,但您使用的是批次骰子,则应使用图像骰子:

def dice_coef(y_true, y_pred, smooth=1):
    y_true_f = K.batch_flatten(y_true)
    y_pred_f = K.batch_flatten(y_pred)

    intersection = K.sum(y_true_f * y_pred_f, axis=-1)
    sums = K.sum(y_true_f, axis=-1) + K.sum(y_pred_f, axis=-1)

    return (2. * intersection + smooth) / (sums + smooth)

通常,我使用K.epsilon()来表示“平滑”(很小的东西)。

iou也是如此:

def iou(y_true, y_pred, smooth=1):
    y_true_f = K.batch_flatten(y_true)
    y_pred_f = K.batch_flatten(y_pred)

    intersection = K.sum(y_true_f * y_pred_f, axis=-1)
    union = K.sum(y_true_f, axis=-1) + K.sum(y_pred_f, axis=-1) - intersection
    return (intersection + smooth) / (union + smooth)

频道骰子的示例:

#considering shape (batch, classes, image_size, image_size)
def dice_coef(y_true, y_pred, smooth=1):

    intersection = K.sum(y_true * y_pred, axis=[2,3])
    sums = K.sum(y_true, axis=[2,3]) + K.sum(y_pred, axis=[2,3])

    dice = (2. * intersection + smooth) / (sums + smooth)
    return K.mean(dice, axis=-1)