如何从R输出创建数据帧
我正在尝试从多个操作的输出创建一个数据集。但我不知道如何实现自动化。复制函数很好,但要获得单个新数据点,需要执行多个操作,即调整后的R平方和F统计 R代码:如何从R输出创建数据帧,r,R,我正在尝试从多个操作的输出创建一个数据集。但我不知道如何实现自动化。复制函数很好,但要获得单个新数据点,需要执行多个操作,即调整后的R平方和F统计 R代码: #make dataframe with random data A<-as.integer(round(runif(20, min=1, max=10))) dim(A) <- c(10,2) A<-as.data.frame(A) #extract F-statistic summary(lm(formula=V1~V
#make dataframe with random data
A<-as.integer(round(runif(20, min=1, max=10)))
dim(A) <- c(10,2)
A<-as.data.frame(A)
#extract F-statistic
summary(lm(formula=V1~V2,data=A))$fstatistic[1]
#extract adjusted R squared
summary(lm(formula=V1~V2,data=A))$adj.r.squared
#repeat 100 times and make a dataframe of the unique extracted output, e.g. 2 columns 100 rows
??????????????
#使用随机数据制作数据帧
A只需将其包装在for
循环中即可
df <- as.data.frame(matrix(0, 100, 2))
for (i in 1:100){
A<-as.integer(round(runif(20, min=1, max=10)))
dim(A) <- c(10,2)
A<-as.data.frame(A)
#extract F-statistic
df[i, 1] <- summary(lm(formula=V1~V2,data=A))$fstatistic[1]
#extract adjusted R squared
df[i, 2] <- summary(lm(formula=V1~V2,data=A))$adj.r.squared
}
df复制功能将正常工作。首先,编写一个函数来执行模拟的一次迭代
one.sim <- function() {
A <- matrix(as.integer(runif(20, min=1, max=10)), nrow=10)
A <- as.data.frame(A)
m1.summary <- summary(lm(V1 ~ V2, data=A))
return(c(fstatistic=unname(m1.summary$fstatistic[1]),
adj.r.squared=m1.summary$adj.r.squared))
}
one.sim在5个数据帧上应用线性模型
使用
复制
,它将类似于
> replicate(5, {
A <- data.frame(rnorm(5), rexp(5))
m <- lm(formula = A[,1] ~ A[,2], data = A)
c(f = summary(m)$fstatistic[1], adjR = summary(m)$adj.r.squared)
})
## [,1] [,2] [,3] [,4] [,5]
## f.value 0.4337426 1.3524681 1.17570087 3.8537837 0.04583862
## adjR -0.1649097 0.0809812 0.04207698 0.4163808 -0.31326721
您也可以使用sapply
> sapply(seq(5), function(x){
A <- data.frame(rnorm(5), rexp(5))
m <- lm(formula = A[,1] ~ A[,2], data = A)
c(f = summary(m)$fstatistic[1], adjR = summary(m)$adj.r.squared)
})
## [,1] [,2] [,3] [,4] [,5]
## f.value 0.07245221 0.2076504 0.0003488657 58.5524139 0.92170453
## adjR -0.30189169 -0.2470187 -0.3331783000 0.9350147 -0.01996465
>sapply(序号(5),函数(x){
A您就快到了…replicate
可以正常工作。只需将两个摘要输出包装在c()
replicate(100,{我想做的事情;c(summary(lm(formula=V1~V2,data=A))$fstatistic[1],summary(lm(formula=V1~V2,data=A))$adj.r.squared)}
> do.call(rbind, lapply(seq(5), function(x){
A <- data.frame(rnorm(5), rexp(5))
m <- lm(formula = A[,1] ~ A[,2], data = A)
c(f = summary(m)$fstatistic[1], adjR = summary(m)$adj.r.squared)
}))
## f.value adjR
## [1,] 1.9820243 0.19711351
## [2,] 21.6698543 0.83785879
## [3,] 4.4484639 0.46297652
## [4,] 0.9084373 -0.02342693
## [5,] 0.0388510 -0.31628698
> sapply(seq(5), function(x){
A <- data.frame(rnorm(5), rexp(5))
m <- lm(formula = A[,1] ~ A[,2], data = A)
c(f = summary(m)$fstatistic[1], adjR = summary(m)$adj.r.squared)
})
## [,1] [,2] [,3] [,4] [,5]
## f.value 0.07245221 0.2076504 0.0003488657 58.5524139 0.92170453
## adjR -0.30189169 -0.2470187 -0.3331783000 0.9350147 -0.01996465