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:
- 原始物是
exp
的原始物 -
sum
是product
的{{1}}
exp
的演示:
- 第0个原始单位的预测:
exps
- 原始分数空间中第0行的Shap值:
lgb.predict([X[0,:]])
array([0.39394102])
- 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。
版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。