我不知道您的Stata命令应该再现什么，但为了回答有关层次结构问题的第二部分：使用list很容易做到这一点。例如，您可以定义如下结构：

tree = list(
      "var1" = list(
         "panel" = list(type ='p',mean = 1,sd=0)
         ,"cluster" = list(type = 'c',value = c(5,8,10)))
      ,"var2" = list(
          "panel" = list(type ='p',mean = 2,sd=0.5)
         ,"cluster" = list(type="c",value =c(1,2)))
)

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要创建此lapply，可以方便地使用list

tree <- lapply(list('var1','var2'),function(x){ 
  ll <- list(panel= list(type ='p',mean = rnorm(1),sd=0), ## I use symbol here not name
             cluster= list(type = 'c',value = rnorm(3)))  ## R prefer symbols
})
names(tree) <-c('var1','var2')

型
可以使用str查看结构

str(tree)
List of 2
 $ var1:List of 2
  ..$ panel  :List of 3
  .. ..$ type: chr "p"
  .. ..$ mean: num 0.284
  .. ..$ sd  : num 0
  ..$ cluster:List of 2
  .. ..$ type : chr "c"
  .. ..$ value: num [1:3] 0.0722 -0.9413 0.6649
 $ var2:List of 2
  ..$ panel  :List of 3
  .. ..$ type: chr "p"
  .. ..$ mean: num -0.144
  .. ..$ sd  : num 0
  ..$ cluster:List of 2
  .. ..$ type : chr "c"
  .. ..$ value: num [1:3] -0.595 -1.795 -0.439

型

在OP澄清后编辑

我认为reshape2包是您想要。我将在这里演示。
为了进行多层次分析，我们需要对数据进行整形。
首先，将变量分为两组：标识符变量和测量变量。

library(reshape2)
dat.m <- melt(dat,id.vars=c('son_id','mom_id')) ## other columns are measured 

str(dat.m)
'data.frame':   21 obs. of  4 variables:
 $ son_id  : Factor w/ 3 levels "1","2","3": 1 2 3 1 2 1 2 1 2 3 ...
 $ mom_id  : Factor w/ 3 levels "1","2","3": 1 1 1 2 2 3 3 1 1 1 ...
 $ variable: Factor w/ 3 levels "hispanic","mom_smoke",..: 1 1 1 1 1 1 1 2 2 2 ...
 $ value   : num  1 1 1 0 0 0 0 1 0 0 ..

型
一旦你有了“moten”形式的数据，你就可以“cast”把它重新排列成你想要的形状：

# mom1 means for all variable
 acast(dat.m,variable~mom_id,mean)
                           1    2      3
hispanic           1.0000000    0    0.0
mom_smoke          0.3333333    1    0.5
son_birthweigth 3943.3333333 4160 2977.5
# Within-mother variance for birthweigth

acast(dat.m,variable~mom_id,function(x) sum((x-mean(x))^2))
                           1    2    3
hispanic           0.0000000    0  0.0
mom_smoke          0.6666667    0  0.5
son_birthweigth 5066.6666667 3200 12.5

## overall mean of each variable
acast(dat.m,variable~.,mean)
[,1]
hispanic           0.4285714
mom_smoke          0.5714286
son_birthweigth 3729.2857143

型

我知道这个问题已经有四年的历史了，但是最近我想在R中做同样的事情，并提出了以下函数。它取决于dplyr和tibble。其中：df是嵌套框，columns是嵌套框的子集的数值向量，individuals是包含individuals的列。

xtsumR<-function(df,columns,individuals){
  df<-dplyr::arrange_(df,individuals)
  panel<-tibble::tibble()
  for (i in columns){
    v<-df %>% dplyr::group_by_() %>%
      dplyr::summarize_(
        mean=mean(df[[i]]),
        sd=sd(df[[i]]),
        min=min(df[[i]]),
        max=max(df[[i]])
      )
    v<-tibble::add_column(v,variacao="overal",.before=-1)
    v2<-aggregate(df[[i]],list(df[[individuals]]),"mean")[[2]]
    sdB<-sd(v2)
    varW<-df[[i]]-rep(v2,each=12) #
    varW<-varW+mean(df[[i]])
    sdW<-sd(varW)
    minB<-min(v2)
    maxB<-max(v2)
    minW<-min(varW)
    maxW<-max(varW)
    v<-rbind(v,c("between",NA,sdB,minB,maxB),c("within",NA,sdW,minW,maxW))
    panel<-rbind(panel,v)
  }
  var<-rep(names(df)[columns])
  n1<-rep(NA,length(columns))
  n2<-rep(NA,length(columns))
  var<-c(rbind(var,n1,n1))
  panel$var<-var
  panel<-panel[c(6,1:5)]
  names(panel)<-c("variable","variation","mean","standard.deviation","min","max")
  panel[3:6]<-as.numeric(unlist(panel[3:6]))
  panel[3:6]<-round(unlist(panel[3:6]),2)
  return(panel)
}

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