关于keras多任务多loss回传的思考

如果有一个多任务多loss的网络，那么在训练时，loss是如何工作的呢？

比如下面：

model = Model(inputs = input, outputs = [y1, y2])
l1 = 0.5
l2 = 0.3
model.compile(loss = [loss1, loss2], loss_weights=[l1, l2], ...)

其实我们最终得到的loss为

final_loss = l1 * loss1 + l2 * loss2

我们最终的优化效果是最小化final_loss。

问题来了，在训练过程中，是否loss2只更新得到y2的网络通路，还是loss2会更新所有的网络层呢？

此问题的关键在梯度回传上，即反向传播算法。

所以loss1只对x1和x2有影响，而loss2只对x1和x3有影响。

补充：keras 多个LOSS总和定义

用字典形式，名字是模型中输出那一层的名字，这里的loss可以是自己定义的，也可是自带的

补充：keras实战-多类别分割loss实现

本文样例均为3d数据的onehot标签形式，即y_true(batch_size,x,y,z,class_num)

1、dice loss

def dice_coef_fun(smooth=1):
    def dice_coef(y_true, y_pred):
        #求得每个sample的每个类的dice
        intersection = K.sum(y_true * y_pred, axis=(1,2,3))
        union = K.sum(y_true, axis=(1,2,3)) + K.sum(y_pred, axis=(1,2,3))
        sample_dices=(2. * intersection + smooth) / (union + smooth) #一维数组 为各个类别的dice
        #求得每个类的dice
        dices=K.mean(sample_dices,axis=0)
        return K.mean(dices) #所有类别dice求平均的dice
    return dice_coef

def dice_coef_loss_fun(smooth=0):
    def dice_coef_loss(y_true,y_pred):
        return 1-1-dice_coef_fun(smooth=smooth)(y_true=y_true,y_pred=y_pred)
    return dice_coef_loss

2、generalized dice loss

def generalized_dice_coef_fun(smooth=0):
    def generalized_dice(y_true, y_pred):
        # Compute weights: "the contribution of each label is corrected by the inverse of its volume"
        w = K.sum(y_true, axis=(0, 1, 2, 3))
        w = 1 / (w ** 2 + 0.00001)
        # w为各个类别的权重，占比越大，权重越小
        # Compute gen dice coef:
        numerator = y_true * y_pred
        numerator = w * K.sum(numerator, axis=(0, 1, 2, 3))
        numerator = K.sum(numerator)

        denominator = y_true + y_pred
        denominator = w * K.sum(denominator, axis=(0, 1, 2, 3))
        denominator = K.sum(denominator)

        gen_dice_coef = numerator / denominator

        return  2 * gen_dice_coef
    return generalized_dice

def generalized_dice_loss_fun(smooth=0):
    def generalized_dice_loss(y_true,y_pred):
        return 1 - generalized_dice_coef_fun(smooth=smooth)(y_true=y_true,y_pred=y_pred)
    return generalized_dice_loss

3、tversky coefficient loss

# Ref: salehi17, "Twersky loss function for image segmentation using 3D FCDN"
# -> the score is computed for each class separately and then summed
# alpha=beta=0.5 : dice coefficient
# alpha=beta=1   : tanimoto coefficient (also known as jaccard)
# alpha+beta=1   : produces set of F*-scores
# implemented by E. Moebel, 06/04/18
def tversky_coef_fun(alpha,beta):
    def tversky_coef(y_true, y_pred):
        p0 = y_pred  # proba that voxels are class i
        p1 = 1 - y_pred  # proba that voxels are not class i
        g0 = y_true
        g1 = 1 - y_true

        # 求得每个sample的每个类的dice
        num = K.sum(p0 * g0, axis=( 1, 2, 3))
        den = num + alpha * K.sum(p0 * g1,axis= ( 1, 2, 3)) + beta * K.sum(p1 * g0, axis=( 1, 2, 3))
        T = num / den  #[batch_size,class_num]

        # 求得每个类的dice
        dices=K.mean(T,axis=0) #[class_num]

        return K.mean(dices)
    return tversky_coef

def tversky_coef_loss_fun(alpha,beta):
    def tversky_coef_loss(y_true,y_pred):
        return 1-tversky_coef_fun(alpha=alpha,beta=beta)(y_true=y_true,y_pred=y_pred)
    return tversky_coef_loss

4、IoU loss

def IoU_fun(eps=1e-6):
    def IoU(y_true, y_pred):
        # if np.max(y_true) == 0.0:
        #     return IoU(1-y_true, 1-y_pred) ## empty image; calc IoU of zeros
        intersection = K.sum(y_true * y_pred, axis=[1,2,3])
        union = K.sum(y_true, axis=[1,2,3]) + K.sum(y_pred, axis=[1,2,3]) - intersection
        #
        ious=K.mean((intersection + eps) / (union + eps),axis=0)
        return K.mean(ious)
    return IoU

def IoU_loss_fun(eps=1e-6):
    def IoU_loss(y_true,y_pred):
        return 1-IoU_fun(eps=eps)(y_true=y_true,y_pred=y_pred)
    return IoU_loss

以上为个人经验，希望能给大家一个参考，也希望大家多多支持我们。