R中的蒙特卡罗模拟、自举和回归
我已经使用SAS很长时间了,现在我想用R翻译我的代码。我需要帮助来完成以下工作:R中的蒙特卡罗模拟、自举和回归,r,simulation,apply,resampling,replicate,R,Simulation,Apply,Resampling,Replicate,我已经使用SAS很长时间了,现在我想用R翻译我的代码。我需要帮助来完成以下工作: 生成几个引导样本 对每个样本运行线性回归模型 通过复制样本将参数存储在新的数据集中 为了更清晰,我编辑了这段代码。 我使用了很多for循环,我知道这并不总是被推荐的。这个过程非常缓慢 是否有一个代码/包(例如,apply family functions,“caret”包)可以使这个非常干净高效/快速,特别是当samplesize*bootsample>1000万时 任何帮助都将不胜感激 samplesize &l
samplesize <- 200
bootsize<- 500
myseed <- 123
#generating a fake dataset
id=1:n
set.seed(myseed)
x <- rnorm(samplesize, 5, 5)
y <- rnorm(samplesize, 2 + 0.4*x, 0.5)
data <- data.frame(id, x, y)
head(data)
id x y
1 1 2.197622 3.978454
2 2 3.849113 4.195852
3 3 12.793542 6.984844
4 4 5.352542 4.412614
5 5 5.646439 4.051405
6 6 13.575325 7.192007
# generate bootstrap samples
bootstrap <- function(nbootsamples, data, seed) {
bootdata <- data.frame() #to initialize it
set.seed(seed)
for (i in 1:nbootsamples) {
replicate <- i
bootstrapIndex <- sample(1:nrow(data), replace = TRUE)
datatemp <- data[bootstrapIndex, ]
tempall <- cbind(replicate, datatemp)
bootdata <- rbind(bootdata, tempall)
}
return(bootdata)
}
bootdata <- bootstrap(nbootsamples=bootsize, data=data, seed=myseed)
bootdata <- dplyr::arrange(bootdata, replicate, id)
head(bootdata)
#The data should look like this
replicate id x y
1 1 1 2.197622 3.978454
2 1 3 12.793542 6.984844
3 1 5 5.646439 4.051405
4 1 9 1.565736 3.451748
5 1 10 2.771690 3.081662
6 1 10 2.771690 3.081662
#Model-fitting and saving coefficient and means
modelFitting <- function(y, x, data) {
modeltemp <- glm(y ~ x,
data = data,
family = gaussian('identity'))
Inty <- coef(modeltemp)["(Intercept)"]
betaX <- coef(modeltemp)["x"]
sdy <- sd(residuals.glm(modeltemp))
data.frame(Inty, betaX, sdy, row.names = NULL)
}
saveParameters <- function(nbootsamples, data, seed) {
parameters <- data.frame() #to initialize it
for (i in 1:length(unique(data$replicate))) {
replicate <- i
datai <- data[ which(data$replicate==i),]
datatemp <- modelFitting(y, x,data=datai)
meandata <- data.frame(Pr_X=mean(datai$x))
tempall <- cbind(replicate, datatemp, meandata)
parameters <- rbind(parameters, tempall)
}
return(parameters)
}
parameters <- saveParameters(nbootsamples=bootsize, data=bootdata, seed=myseed)
head(parameters)
#Ultimately all I want is my final dataset to look like the following
replicate Inty betaX sdy Pr_X
1 1 2.135529 0.3851757 0.5162728 4.995836
2 2 1.957152 0.4094682 0.5071635 4.835884
3 3 2.044257 0.3989742 0.4734178 5.111185
4 4 2.093452 0.3861861 0.4921470 4.741299
5 5 2.017825 0.4037699 0.5240363 4.931793
6 6 2.026952 0.3979731 0.4898346 5.502320
samplesize使用该软件包可以轻松完成带重采样的回归。给定示例数据,通过广义线性模型运行200个引导样本的代码如下所示
library(caret)
x = round(rnorm(200, 5, 5))
y= rnorm(200, 2 + 0.4*x, 0.5)
theData <- data.frame(id=1:200,x, y)
# configure caret training parameters to 200 bootstrap samples
fitControl <- trainControl(method = "boot",
number = 200)
fit <- train(y ~ x, method="glm",data=theData,
trControl = fitControl)
# print output object
fit
# print first 10 resamples
fit$resample[1:10,]
有关如何使用插入符号(包括结果模型对象的内容)的详细信息(例如,访问各个模型,以便您可以使用模拟的predict()
函数生成预测),请访问
插入符号还支持并行处理。有关如何使用带插入符号的并行处理的示例,请阅读
此外,蒙特卡罗模拟通过R中的包在R中得到支持。对于连续因变量,family=gaussian(“恒等式”)不应该吗?谢谢Len Greski。我会修好的谢谢你,伦格雷斯基。为了更清晰,我编辑了问题和代码。我尝试了插入符号包
,但我不知道如何获得想要的结果。有什么建议吗?谢谢
> fit
Generalized Linear Model
200 samples
1 predictor
No pre-processing
Resampling: Bootstrapped (200 reps)
Summary of sample sizes: 200, 200, 200, 200, 200, 200, ...
Resampling results:
RMSE Rsquared MAE
0.4739306 0.9438834 0.3772199
> fit$resample[1:10,]
RMSE Rsquared MAE Resample
1 0.5069606 0.9520896 0.3872257 Resample001
2 0.4636029 0.9460214 0.3711900 Resample002
3 0.4446103 0.9549866 0.3435148 Resample003
4 0.4464119 0.9443726 0.3636947 Resample004
5 0.5193685 0.9191259 0.4010104 Resample005
6 0.4995917 0.9451417 0.4044659 Resample006
7 0.4347831 0.9494606 0.3383224 Resample007
8 0.4725041 0.9483434 0.3716319 Resample008
9 0.5295650 0.9458453 0.4241543 Resample009
10 0.4796985 0.9514595 0.3927207 Resample010
>