如果有一個(gè)多任務(wù)多l(xiāng)oss的網(wǎng)絡(luò)，那么在訓(xùn)練時(shí)，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], ...)
其實(shí)我們最終得到的loss為
final_loss = l1 * loss1 + l2 * loss2
我們最終的優(yōu)化效果是最小化final_loss。
問題來了，在訓(xùn)練過程中，是否loss2只更新得到y(tǒng)2的網(wǎng)絡(luò)通路，還是loss2會更新所有的網(wǎng)絡(luò)層呢？
此問題的關(guān)鍵在梯度回傳上，即反向傳播算法。
所以loss1只對x1和x2有影響，而loss2只對x1和x3有影響。
補(bǔ)充：keras 多個(gè)LOSS總和定義
用字典形式，名字是模型中輸出那一層的名字，這里的loss可以是自己定義的，也可是自帶的
補(bǔ)充：keras實(shí)戰(zhàn)-多類別分割loss實(shí)現(xiàn)
本文樣例均為3d數(shù)據(jù)的onehot標(biāo)簽形式，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):
#求得每個(gè)sample的每個(gè)類的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) #一維數(shù)組 為各個(gè)類別的dice
#求得每個(gè)類的dice
dices=K.mean(sample_dices,axis=0)
return K.mean(dices) #所有類別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為各個(gè)類別的權(quán)重，占比越大，權(quán)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
# 求得每個(gè)sample的每個(gè)類的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]
# 求得每個(gè)類的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
以上為個(gè)人經(jīng)驗(yàn)，希望能給大家一個(gè)參考，也希望大家多多支持腳本之家。
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