我目前正在关注以下link中的幻灯片。我在幻灯片121/128上，我想知道如何复制AUC。作者没有解释如何做（也与幻灯片124相同）。其次，在幻灯片125上产生以下代码；
bestRound = which.max(as.matrix(cv.res)[,3]-as.matrix(cv.res)[,4])
bestRound
我收到以下错误；
as.matrix（cv.res）[，2]中的错误：下标超出范围
可以从here下载以下代码的数据，我提供了以下代码供您参考。
问题：我如何以作者的身份制作AUC，为什么下标超出范围？
-----代码------
# Kaggle Winning Solutions
train <- 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 other
new.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')
在此处绘制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
#bestRound: - subscript out of bounds
bestRound <- which.max(as.matrix(cv.res)[,3]-as.matrix(cv.res)[,4])
bestRound
cv.res
cv.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)
最佳答案
代码的许多部分对我来说意义不大，但这是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
prediction = T时指定xgb.cv。为了获得最佳迭代：
it = which.max(cv.res$evaluation_log$test_auc_mean)
best.iter = cv.res$evaluation_log$iter[it]
在交叉验证结果上绘制ROC曲线：
library(pROC)
plot(pROC::roc(response = y,
predictor = cv.res$pred,
levels=c(0, 1)),
lwd=1.5)
要获得一个混淆矩阵（假设阈值为0.5概率）：
library(caret)
confusionMatrix(ifelse(cv.res$pred <= 0.5, 0, 1), y)
#output
Reference
Prediction 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
话虽这么说，但应该旨在通过交叉验证来调整超参数，例如eta，gamma，lambda，子样本，colsample_bytree，colsample_bylevel等。
最简单的方法是构造一个网格搜索，在该搜索中，您可以对所有超参数组合使用
expand.grid，并在网格上使用lccly（其中xgb.cv作为自定义函数的一部分）。如果您需要更多详细信息，请发表评论。关于r - 从R中的xgboost模型绘制AUC，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/46736934/