Question

为什么aggregate（）在这里不起作用？

> r = aggregate(data.frame(var1 = 1:10, var2 = 101:110), by=list(range=cut(1:10, breaks=c(2,4,8,10))), FUN = function(x) { c(obs=length(x), avg=mean(x), sd=sd(x)) })
> class(r)
[1] "data.frame"
> dim(r)
[1] 3 3
> r[,1]
[1] (2,4]  (4,8]  (8,10]
Levels: (2,4] (4,8] (8,10]
> r[,2]
     obs avg       sd
[1,]   2 3.5 0.707107
[2,]   4 6.5 1.290994
[3,]   2 9.5 0.707107
> r[,3]
     obs   avg       sd
[1,]   2 103.5 0.707107
[2,]   4 106.5 1.290994
[3,]   2 109.5 0.707107
> class(r[,2])
[1] "matrix"
> class(r[,3])
[1] "matrix"

更新

运行正确版本后返回的aggregate（）值：

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Answer 1

那是因为aggregate没有将data.frames传递给其FUN=参数。它传递了观察向量。此外，[, "name"]索引不适用于矩阵。确保传入data.frame而不是示例中的矩阵。也许你想要by函数

by(data.frame(var1 = 1:10, var2 = 101:110), 
    list(range=cut(1:10, breaks=c(2,4,8,10))), 
    FUN = function(x) { c(obs=length(x[, "var2"]), avg=mean(x[, "var2"]), sd=sd(x[, "var2"])) })

Answer 2

提供数据框并理解聚合只传递列向量，因此使用x[ , "colname"]注定要失败，因为＆＃34; x＆＃34;不是数据帧：

 aggregate(data.frame(var1 = 1:10, var2 = 101:110), 
       by=list(range=cut(1:10, breaks=c(2,4,8,10))), 
       FUN = function(x) 
         { 
         c(obs=length(x), avg=mean(x), sd=sd(x)) 
         })
#------------
   range  var1.obs  var1.avg   var1.sd    var2.obs    var2.avg     var2.sd
1  (2,4] 2.0000000 3.5000000 0.7071068   2.0000000 103.5000000   0.7071068
2  (4,8] 4.0000000 6.5000000 1.2909944   4.0000000 106.5000000   1.2909944
3 (8,10] 2.0000000 9.5000000 0.7071068   2.0000000 109.5000000   0.7071068

在data.frame对象上使用aggregate（）

2 个答案: