使用lightgbm Tweedie目标将SHAP值从原始单位转换为原始单位吗？

如何解决使用lightgbm Tweedie目标将SHAP值从原始单位转换为原始单位吗？

编辑：

Shapley附加说明（SHAP值）的实用程序是为了了解每个特征如何有助于模型的预测。对于某些目标，例如以RMSE作为目标函数的回归，SHAP值以标签值的本机单位表示。例如，如果估算住房成本，SHAP值可以表示为美元。正如您将在下面看到的，并非所有目标函数都如此。特别是，Tweedie回归目标不会以本机单位生成SHAP值。这是一个解释上的问题，因为我们想知道房屋成本如何受+/-美元影响。

鉴于此信息，我的问题是：在解释具有Tweedie回归目标的模型时，如何将每个单个要素的SHAP值转换到目标标签的数据空间中？

我不知道当前有任何实现此类转换的软件包。 This remains unresolved in the package put out by the shap authors themselves.

我在下面用lightgbm的R实现说明了这个问题的精妙之处：

library(tweedie)
library(lightgbm)

set.seed(123)

tweedie_variance_power <- 1.2

labels <- rtweedie(1000,mu = 1,phi = 1,power = tweedie_variance_power)
hist(labels)

feat1 <- labels + rnorm(1000) #good signal for label with some noise
feat2 <-rnorm(1000) #garbage feature 
feat3 <-rnorm(1000) #garbage feature 

features <- cbind(feat1,feat2,feat3)

dTrain <- lgb.Dataset(data = features,label = labels)

params <- c(objective = 'tweedie',tweedie_variance_power = tweedie_variance_power)

mod <- lgb.train(data = dTrain,params = params,nrounds = 100)

#Predictions in the native units of the labels
predsNative <- predict(mod,features,rawscore = FALSE)
#Predictions in the raw format
predsRaw <- predict(mod,rawscore = TRUE)

#We do not expect these values to be equal
all.equal(predsTrans,predsRaw)
"Mean relative difference: 1.503072"

#We expect values to be equal if raw scores are exponentiated
all.equal(predsTrans,exp(predsRaw))
"TRUE" #... our expectations are correct

#SHAP values 
shapNative <- predict(mod,rawscore = FALSE,predcontrib = TRUE)
shapRaw <- predict(mod,rawscore = TRUE,predcontrib = TRUE )

#Are there differences between shap values when rawscore is TRUE or FALSE?
all.equal(shapNative,shapRaw)
"TRUE" #outputs are identical,that is surprising!

#So are the shap values in raw or native formats?
#To anwser this question we can sum them

#testing raw the raw case first
all.equal(rowSums(shapRaw),predsRaw)
"TRUE" 

#from this we can conclude that shap values are not in native units,#regardless of whether rawscore is TRUE or FALSE

#Test native scores just to prove point
all.equal(rowSums(shapNative),predsNative)
"Mean relative difference: 1.636892" # reaffirms that shap values are not in native units

#However,we can perform this operation on the raw shap scores
#to get the prediction in the native value
all.equal(exp(rowSums(shapRaw)),predsNative)
'TRUE'

#reversing the operations does not yield the same result
all.equal(rowSums(exp(shapRaw)),predsNative)
"Mean relative difference: 0.7662481"

#The last line is relevant because it implies 
#The relationship between native predictions
#and exponentiated shap values is not linear

#So,given the point of SHAP is to understand how each 
#feature impacts the prediction in its native units
#the raw shap values are not as useful as they Could be

#Thus,how how would we convert 
#each of these four raw shap value elements to native units,#thus understanding their contributions to their predictions
#in currency of native units?
shapRaw[1,]
-0.15429227  0.04858757 -0.27715359 -0.48454457

原始帖子和编辑

我对SHAP值的理解是，进行回归时，它们以标签/响应的本机单位为单位，并且SHAP值的总和近似于模型的预测。

我正在尝试使用Tweedie回归目标在LightGBM包中提取SHAP值，但是发现SHAP值不在标签的本机单位中，并且它们不等于预测值。

似乎必须对它们求幂，这是正确的吗？

旁注：我了解SHAP值矩阵的最后一列代表基本预测，必须添加。

可复制的示例：

library(tweedie)
library(caret)
library(lightgbm)

set.seed(123)

tweedie_variance_power <- 1.2

labels <- rtweedie(1000,nrounds = 100)

preds <- predict(mod,features)

plot(preds,labels,main = paste('RMSE =',RMSE(pred = preds,obs = labels)))

#shap values are summing to negative values?
shap_vals <- predict(mod,predcontrib = TRUE,rawscore = FALSE)
shaps_sum <- rowSums(shap_vals)
plot(shaps_sum,RMSE(pred = shaps_sum,obs = labels)))

#maybe we need to exponentiate?
shap_vals_exp <- exp(shap_vals)
shap_vals_exp_sum <- rowSums(shap_vals_exp)
#still looks a little weird,overpredicting 
plot(shap_vals_exp_sum,RMSE(pred = shap_vals_exp_sum,obs = labels)))

编辑

操作的顺序是先求和，然后取幂SHAP值，这将使您以本机单位进行预测。尽管我仍然不清楚如何将要素级别值转换为本地响应单位。

shap_vals_sum_exp <- exp(shaps_sum)
plot(shap_vals_sum_exp,RMSE(pred = shap_vals_sum_exp,obs = labels)))

解决方法

我将展示如何在原始分数和原始单位中调和Python中的shap值和模型预测。希望它将帮助您了解您在R中的位置。

步骤1。生成数据集

# pip install tweedie
import tweedie
y = tweedie.tweedie(1.2,1,1).rvs(size=1000)
X = np.random.randn(1000,3)

第2步。拟合模型

from lightgbm.sklearn import LGBMRegressor
lgb = LGBMRegressor(objective = 'tweedie')
lgb.fit(X,y)

第3步。了解什么是shap值。

第0个数据点的Shap值

shap_values = lgb.predict(X,pred_contrib=True)
shap_values[0]
array([ 0.36841812,-0.15985678,0.28910617,-0.27317984])

前3个是模型对基线的贡献，即shap值本身：

shap_values[0,:3].sum()
0.4976675073764354

第四是原始分数的基线：

shap_values[0,3]
-0.2731798364061747

它们的总和加到原始分数的模型预测中：

shap_values[0,:3].sum() + shap_values[0,3]
0.22448767097026068

让我们检查一下原始模型的预测：

preds = lgb.predict(X,raw_score=True)
preds[0]
0.2244876709702609

编辑。原始分数和原始utut之间的转换

要在Tweedie（以及Poisson和Gamma）分布的原始得分和原始单位之间进行转换，您需要了解以下两个事实：

1. 原始物是exp的原始物
exp的
2. sum是product的{​​{1}}

演示：

1. 第0个原始单位的预测：

exps

1. 原始分数空间中第0行的Shap值：

lgb.predict([X[0,:]])
array([0.39394102])

1. 将shap值转换为原始单位（指数的乘积）：

shap_values = lgb.predict(X,pred_contrib=True,raw_score=True)
shap_values[0]
array([-0.77194274,-0.08343294,0.22740536,-0.30358374])

再次看起来类似于我。