我目前正在按照以下链接中的幻灯片进行学习。我在第121/128页，我想知道如何复制AUC。作者没有解释如何做到这一点（第124页也是如此）。其次，在第125页上，出现了以下代码；
bestRound = which.max(as.matrix(cv.res)[,3]-as.matrix(cv.res)[,4])bestRound我收到了以下错误；
Error in as.matrix(cv.res)[, 2] : subscript out of bounds
以下代码的数据可以从这里下载，我已经为您提供了下面的代码参考。
问题：我如何像作者一样生成AUC，以及为什么会出现下标越界错误？
—– 代码 ——
# Kaggle Winning Solutionstrain <- read.csv('train.csv', header = TRUE)test <- read.csv('test.csv', header = TRUE)y <- train[, 1]train <- as.matrix(train[, -1])test <- as.matrix(test)train[1, ]#We want to determin who is more influencial than the othernew.train <- cbind(train[, 12:22], train[, 1:11])train = rbind(train, new.train)y <- c(y, 1 - y)x <- rbind(train, test)(dat[,i]+lambda)/(dat[,j]+lambda)A.follow.ratio = calcRatio(x,1,2)A.mention.ratio = calcRatio(x,4,6)A.retweet.ratio = calcRatio(x,5,7)A.follow.post = calcRatio(x,1,8)A.mention.post = calcRatio(x,4,8)A.retweet.post = calcRatio(x,5,8)B.follow.ratio = calcRatio(x,12,13)B.mention.ratio = calcRatio(x,15,17)B.retweet.ratio = calcRatio(x,16,18)B.follow.post = calcRatio(x,12,19)B.mention.post = calcRatio(x,15,19)B.retweet.post = calcRatio(x,16,19)x = cbind(x[,1:11], A.follow.ratio,A.mention.ratio,A.retweet.ratio, A.follow.post,A.mention.post,A.retweet.post, x[,12:22], B.follow.ratio,B.mention.ratio,B.retweet.ratio, B.follow.post,B.mention.post,B.retweet.post)AB.diff = x[,1:17]-x[,18:34]x = cbind(x,AB.diff)train = x[1:nrow(train),]test = x[-(1:nrow(train)),]set.seed(1024)cv.res <- xgb.cv(data = train, nfold = 3, label = y, nrounds = 100, verbose = FALSE, objective = 'binary:logistic', eval_metric = 'auc')在这里绘制AUC图
set.seed(1024)cv.res = xgb.cv(data = train, nfold = 3, label = y, nrounds = 3000, objective='binary:logistic', eval_metric = 'auc', eta = 0.005, gamma = 1,lambda = 3, nthread = 8, max_depth = 4, min_child_weight = 1, verbose = F, subsample = 0.8,colsample_bytree = 0.8)这是我遇到的代码中断点
#bestRound: - subscript out of boundsbestRound <- which.max(as.matrix(cv.res)[,3]-as.matrix(cv.res)[,4])bestRoundcv.rescv.res[bestRound,]set.seed(1024) bst <- xgboost(data = train, label = y, nrounds = 3000, objective='binary:logistic', eval_metric = 'auc', eta = 0.005, gamma = 1,lambda = 3, nthread = 8, max_depth = 4, min_child_weight = 1, subsample = 0.8,colsample_bytree = 0.8)preds <- predict(bst,test,ntreelimit = bestRound)result <- data.frame(Id = 1:nrow(test), Choice = preds)write.csv(result,'submission.csv',quote=FALSE,row.names=FALSE)回答：
代码的许多部分对我来说意义不大，但这里是一个使用提供的数据构建模型的最小示例：
数据：
train <- read.csv('train.csv', header = TRUE)y <- train[, 1]train <- as.matrix(train[, -1])模型：
library(xgboost)cv.res <- xgb.cv(data = train, nfold = 3, label = y, nrounds = 100, verbose = FALSE, objective = 'binary:logistic', eval_metric = 'auc', prediction = T)要获得交叉验证预测，必须在调用xgb.cv时指定prediction = T。
要获得最佳迭代次数：
it = which.max(cv.res$evaluation_log$test_auc_mean)best.iter = cv.res$evaluation_log$iter[it]要在交叉验证结果上绘制ROC曲线：
library(pROC)plot(pROC::roc(response = y, predictor = cv.res$pred, levels=c(0, 1)), lwd=1.5) 
要获得混淆矩阵（假设0.5的概率是阈值）：
library(caret)confusionMatrix(ifelse(cv.res$pred <= 0.5, 0, 1), y)#output ReferencePrediction 0 1 0 2020 638 1 678 2164 Accuracy : 0.7607 95% CI : (0.7492, 0.772) No Information Rate : 0.5095 P-Value [Acc > NIR] : <2e-16 Kappa : 0.5212 Mcnemar's Test P-Value : 0.2823 Sensitivity : 0.7487 Specificity : 0.7723 Pos Pred Value : 0.7600 Neg Pred Value : 0.7614 Prevalence : 0.4905 Detection Rate : 0.3673 Detection Prevalence : 0.4833 Balanced Accuracy : 0.7605 'Positive' Class : 0 尽管如此，应该通过交叉验证来调整超参数，如eta、gamma、lambda、subsample、colsample_bytree、colsample_bylevel等。
最简单的方法是构建一个网格搜索，您可以对所有超参数组合使用expand.grid，并在网格上使用lapply，将xgb.cv作为自定义函数的一部分。如果您需要更多详细信息，请评论。
