R中的标准差之间/之内

Ste*_*rdi 6 r hierarchical-data multi-level stata

在处理分层/多级/面板数据集时,采用一个返回可用变量的组内和组间标准差的包可能非常有用。

Stata通过以下命令可以轻松完成以下数据操作

xtsum, i(momid)
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我进行了研究,但找不到任何R可以做到这一点的软件包。

编辑:

为了解决问题,分层数据集的示例可能是这样的:

son_id       mom_id      hispanic     mom_smoke     son_birthweigth

  1            1            1            1              3950
  2            1            1            0              3890
  3            1            1            0              3990
  1            2            0            1              4200
  2            2            0            1              4120
  1            3            0            0              2975
  2            3            0            1              2980
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每个母亲(较高级别)有两个或更多儿子(较低级别)的事实给出了“多级”结构。因此,每个母亲都定义了一组观察结果。

因此,每个数据集变量可以在母亲之间和母亲之间或仅在母亲之间变化。birtweigth母亲之间会有所不同,但同一位母亲之间也会有所不同。而是hispanic固定为同一个母亲。

例如,母体内部方差son_birthweigth为:

# mom1 means
    bwt_mean1 <- (3950+3890+3990)/3
    bwt_mean2 <- (4200+4120)/2
    bwt_mean3 <- (2975+2980)/2

# Within-mother variance for birthweigth
    ((3950-bwt_mean1)^2 + (3890-bwt_mean1)^2 + (3990-bwt_mean1)^2 + 
    (4200-bwt_mean2)^2 + (4120-bwt_mean2)^2 + 
    (2975-bwt_mean3)^2 + (2980-bwt_mean3)^2)/(7-1)
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而母亲之间的差异是:

# overall mean of birthweigth:
# mean <- sum(data$son_birthweigth)/length(data$son_birthweigth)
    mean <- (3950+3890+3990+4200+4120+2975+2980)/7

# within variance:
    ((bwt_mean1-mean)^2 + (bwt_mean2-mean)^2 + (bwt_mean3-mean)^2)/(3-1)
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ags*_*udy 1

我不知道你的 Stata 命令应该重现什么,但要回答有关层次结构问题的第二部分:使用list. 例如,您定义如下结构:

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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要创建它,lapply可以很方便地使用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')
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您可以使用以下命令查看结构str

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
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OP澄清后编辑

我想那个包reshape2就是你想要的。我在这里演示一下。

为了进行多级分析,我们需要重塑数据。

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 ..
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一旦你有了“moten”形式的数据,你就可以“cast”将其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
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