r - 如何在没有by子句的data.table中使用lapply

Question

我正在尝试使用 data.table、lapply 和函数调用对同一变量运行多个回归。我想得到一个简单的表格作为输出，显示每个变量和每个变量的决定系数。

我正在使用 Rstudio 1.2.1335，data.table 1.12.2
我使用的数据集是“http://users.stat.ufl.edu/~rrandles/sta4210/Rclassnotes/data/textdatasets/KutnerData/Appendix%20C%20Data%20Sets/APPENC02.txt”

cnames<-c("ID","County","State","Area","Pop","Young","Old","Phys","Beds","Crime","HighSchool","BA","Poverty","Unemploy","PerCapitaIncome","TotalIncome","Region")
df62<-fread("APPENC02.txt", col.names=cnames)
df62[,c("ID", "County","State","Region"):=NULL]
variability<-function(y){
     model<-eval(substitute(lm(Phys~y, data=df62)))
     anova<-anova(model)
     SSR<- anova$`Sum Sq`[1]
     SSE<- anova$`Sum Sq`[2]
     SSTO<-SSR+SSE
     R2<-SSR/SSTO
     return(R2)
}
df62[ , lapply(.SD, variability)]

如果最后一行是:

df62[ , lapply(.SD, Variability), by=Phys]

Error Message when I omit the 'by' clause: "Error in (function(x, i, exact) if (is.matrix(i)) as.matrix(x)[[i]] else .subset2(x, : object 'i' not found"

如果我按变量“Phys”分组，我会得到正确的结果，但我会不必要地重复每个结果。

Answer 1

我们可以用 reformulate 创建表达式.在这里，我们可以传递两个参数，'data' 和 'y'，y 将把列名作为参数。

variability<-function(data, y){
     model<- lm(reformulate(y, "Phys"), data=data)
     anova<-anova(model)
     SSR<- anova$`Sum Sq`[1]
     SSE<- anova$`Sum Sq`[2]
     SSTO<-SSR+SSE
     R2<-SSR/SSTO
     return(R2)
}

选择感兴趣的列名

nm1 <- setdiff(names(df62), "Phys")

循环遍历它们，应用函数，而 data是 .SD

setnames(df62[, lapply(nm1, variability, data = .SD)], nm1)[]
#    Area       Pop      Young          Old      Beds     Crime   HighSchool         BA     Poverty    Unemploy PerCapitaIncome TotalIncome
#1: 0.006095652 0.8840674 0.01432791 9.788323e-06 0.9033826 0.6731538 1.804622e-05 0.05605789 0.004113459 0.002551878       0.0999411   0.8989137

数据

cnames<-c("ID","County","State","Area","Pop","Young","Old","Phys","Beds","Crime","HighSchool","BA","Poverty","Unemploy","PerCapitaIncome","TotalIncome","Region")

df62 <- fread("http://users.stat.ufl.edu/~rrandles/sta4210/Rclassnotes/data/textdatasets/KutnerData/Appendix%20C%20Data%20Sets/APPENC02.txt", col.names = cnames)
df62[,c("ID", "County","State","Region"):=NULL]