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我正在尝试从以下数据框中一次删除一组数据框中的异常值：

set.seed(1234)
library('mvoutlier')
x <- rnorm(10)     # standard normal
x[1] <- x[1] * 10  # introduce outlier

y <- rnorm(10)     # standard normal
y[4] <- y[4] * 10  # introduce outlier

w <- rnorm(10)     # standard normal
w[9] <- w[9] * 10  # introduce outlier

grp = c(rep('a',3), rep('b',4), rep('c',3)) #Introduce groups
df = data.frame(grp, x,y,w)

数据框如下所示：

> df
   grp           x          y          w
1     a -12.0706575 -0.4771927  0.1340882
2     a   0.2774292 -0.9983864 -0.4906859
3     a   1.0844412 -0.7762539 -0.4405479
4     b  -2.3456977  0.6445882  0.4595894
5     b   0.4291247  0.9594941 -0.6937202
6     b   0.5060559 -0.1102855 -1.4482049
7     b  -0.5747400 -0.5110095  0.5747557
8     c  -0.5466319 -0.9111954 -1.0236557
9     c  -0.5644520 -0.8371717 -0.1513830
10    c  -0.8900378  2.4158352 -0.9359486

我编写了以下函数来从数据框中删除异常值：

removeOutliers = function(data)
  {
  print("inside")
  print(dim(data))
  z = sign2(data[, -which(colnames(data)=="grp")],makeplot=FALSE) 
  idx = which(z$wfinal01==0)  #Get the index of outliers
  return(data[-idx,]) #Return the remaining rows
  }

我想为每个组分别删除异常行（即a, b, ans c）。我需要将具有 group 的子数据帧传递a给上述函数并收集结果并对 groupb和c.

我知道aggregate可以在这里使用该功能，但不确定如何实现。

aggregate( . ~ grp, data=df, removeOutliers)

任何帮助appriciated。谢谢

4

2 回答 2

1

这是一种快速的方法。 .SD表示除by变量（grp在此示例中）之外的所有变量。

#Set data as data.table object
require(data.table)
setDT(df)

#Apply function and extract rows where wfinal is 0
tokeep <- df[ , sign2(.SD), by=grp][wfinal01==0,which=TRUE]

#Get rid of outliers
df[-tokeep]

没有异常值的结果数据集：

   grp          x          y          w
1:   a  0.2774292 -0.9983864 -0.4906859
2:   a  1.0844412 -0.7762539 -0.4405479
3:   b  0.4291247  0.9594941 -0.6937202
4:   b  0.5060559 -0.1102855 -1.4482049
5:   b -0.5747400 -0.5110095  0.5747557
6:   c -0.5466319 -0.9111954 -1.0236557
7:   c -0.5644520 -0.8371717 -0.1513830

如果你想要异常值：

df[tokeep]

   grp           x          y          w
1:   a -12.0706575 -0.4771927  0.1340882
2:   b  -2.3456977  0.6445882  0.4595894
3:   c  -0.8900378  2.4158352 -0.9359486
于 2014-08-22T02:00:06.003 回答
0

尝试：

for(i in unique(df$grp)) print(df[grp==i,])
  grp           x          y          w
1   a -12.0706575 -0.4771927  0.1340882
2   a   0.2774292 -0.9983864 -0.4906859
3   a   1.0844412 -0.7762539 -0.4405479
  grp          x          y          w
4   b -2.3456977  0.6445882  0.4595894
5   b  0.4291247  0.9594941 -0.6937202
6   b  0.5060559 -0.1102855 -1.4482049
7   b -0.5747400 -0.5110095  0.5747557
   grp          x          y          w
8    c -0.5466319 -0.9111954 -1.0236557
9    c -0.5644520 -0.8371717 -0.1513830
10   c -0.8900378  2.4158352 -0.9359486

for(i in unique(df$grp)) removeOutliers(df[grp==i,])
[1] "inside"
[1] 3 4
[1] "inside"
[1] 4 4
[1] "inside"
[1] 3 4
于 2014-08-22T01:25:43.633 回答