这篇教程关于keras多任务多loss回传的思考写得很实用，希望能帮到您。
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 = 0.5l2 = 0.3model.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 lossdef 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 lossdef 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, 06/04/18def 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 lossdef 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51zixue.net。
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