我正在尝试使用XGBoost对由不等长的曝光时间段生成的数据的声明频率进行建模，但是无法获得用于正确处理曝光时间的模型.通常，我可以通过将log(exposure)设置为偏移量来做到这一点-您可以在XGBoost中做到这一点吗?
I am trying to use XGBoost to model claims frequency of data generated from unequal length exposure periods, but have been unable to get the model to treat the exposure correctly. I would normally do this by setting log(exposure) as an offset - are you able to do this in XGBoost?
(在此处发布了类似的问题: xgboost，胶印曝光?)
(A similar question was posted here: xgboost, offset exposure?)
为说明问题，下面的R代码使用以下字段生成一些数据:
To illustrate the issue, the R code below generates some data with the fields:
- x1，x2-因子(0或1)
- 暴露-观察数据的保单期限
- 频率-单位曝光的平均索赔数
- 索赔-观察到的索赔数量〜泊松(频率*暴露)
目标是使用x1和x2预测频率-真正的模型是:如果x1 = x2 = 1，则频率= 2，否则，频率= 1.
The goal is to predict frequency using x1 and x2 - the true model is: frequency = 2 if x1 = x2 = 1, frequency = 1 otherwise.
由于一开始就无法得知曝光次数，因此无法用于预测该次数.我们可以使用的唯一方法是说:预期的索赔数量=频率*风险敞口.
Exposure can't be used to predict the frequency as it is not known at the outset of a policy. The only way we can use it is to say: expected number of claims = frequency * exposure.
代码尝试通过以下方式使用XGBoost对此进行预测:
The code tries to predict this using XGBoost by:
在这些下面，我展示了如何处理树(rpart)或gbm的情况.
Below these, I've shown how I would handle the situation for a tree (rpart) or gbm.
set.seed(1) size<-10000 d <- data.frame( x1 = sample(c(0,1),size,replace=T,prob=c(0.5,0.5)), x2 = sample(c(0,1),size,replace=T,prob=c(0.5,0.5)), exposure = runif(size, 1, 10)*0.3 ) d$frequency <- 2^(d$x1==1 & d$x2==1) d$claims <- rpois(size, lambda = d$frequency * d$exposure) #### Try to fit using XGBoost require(xgboost) param0 <- list( "objective" = "count:poisson" , "eval_metric" = "logloss" , "eta" = 1 , "subsample" = 1 , "colsample_bytree" = 1 , "min_child_weight" = 1 , "max_depth" = 2 ) ## 1 - set weight in xgb.Matrix xgtrain = xgb.DMatrix(as.matrix(d[,c("x1","x2")]), label = d$claims, weight = d$exposure) xgb = xgb.train( nrounds = 1 , params = param0 , data = xgtrain ) d$XGB_P_1 <- predict(xgb, xgtrain) ## 2 - set as offset in xgb.Matrix xgtrain.mf <- model.frame(as.formula("claims~x1+x2+offset(log(exposure))"),d) xgtrain.m <- model.matrix(attr(xgtrain.mf,"terms"),data = d) xgtrain <- xgb.DMatrix(xgtrain.m,label = d$claims) xgb = xgb.train( nrounds = 1 , params = param0 , data = xgtrain ) d$XGB_P_2 <- predict(model, xgtrain) #### Fit a tree require(rpart) d[,"tree_response"] <- cbind(d$exposure,d$claims) tree <- rpart(tree_response ~ x1 + x2, data = d, method = "poisson") d$Tree_F <- predict(tree, newdata = d) #### Fit a GBM gbm <- gbm(claims~x1+x2+offset(log(exposure)), data = d, distribution = "poisson", n.trees = 1, shrinkage=1, interaction.depth=2, bag.fraction = 0.5) d$GBM_F <- predict(gbm, newdata = d, n.trees = 1, type="response")推荐答案
我现在已经解决了如何使用setinfo将base_margin属性更改为偏移量(作为线性预测变量)的方法，例如:
I have now worked out how to do this using setinfo to change the base_margin attribute to be the offset (as a linear predictor), ie:
setinfo(xgtrain, "base_margin", log(d$exposure))