Question

以下是我感兴趣的测试： http://www.itl.nist.gov/div898/handbook/eda/section3/eda35h3.htm

如何将此代码调整为接受数值向量的函数，并返回指定要删除哪些数据点的逻辑向量？

我试图在下面这样做，但是我遇到了困难，因为当我对矢量进行排序以返回时，它不会与输入矢量数据对齐。

# input data
y = c(-0.25, 0.68, 0.94, 1.15, 1.20, 1.26, 1.26,
      1.34, 1.38, 1.43, 1.49, 1.49, 1.55, 1.56,
      1.58, 1.65, 1.69, 1.70, 1.76, 1.77, 1.81,
      1.91, 1.94, 1.96, 1.99, 2.06, 2.09, 2.10,
      2.14, 2.15, 2.23, 2.24, 2.26, 2.35, 2.37,
      2.40, 2.47, 2.54, 2.62, 2.64, 2.90, 2.92,
      2.92, 2.93, 3.21, 3.26, 3.30, 3.59, 3.68,
      4.30, 4.64, 5.34, 5.42, 6.01)

## Generate normal probability plot.
qqnorm(y)

removeoutliers = function(dfinputcol) {

  y = as.vector(dfinputcol)

  ## Create function to compute the test statistic.
  rval = function(y){
    ares = abs(y - mean(y))/sd(y)
    df = data.frame(y, ares)
    r = max(df$ares)
    list(r, df)}

  ## Define values and vectors.
  n = length(y)
  alpha = 0.05
  lam = c(1:10)
  R = c(1:10)

  ## Compute test statistic until r=10 values have been
  ## removed from the sample.
  for (i in 1:10){

    if(i==1){
      rt = rval(y)
      R[i] = unlist(rt[1])
      df = data.frame(rt[2])
      newdf = df[df$ares!=max(df$ares),]}

    else if(i!=1){
      rt = rval(newdf$y)
      R[i] = unlist(rt[1])
      df = data.frame(rt[2])
      newdf = df[df$ares!=max(df$ares),]}

    ## Compute critical value.
    p = 1 - alpha/(2*(n-i+1))
    t = qt(p,(n-i-1))
    lam[i] = t*(n-i) / sqrt((n-i-1+t**2)*(n-i+1))

  }
  ## Print results.
  newdf = data.frame(c(1:10),R,lam)
  names(newdf)=c("Outliers","TestStat.", "CriticalVal.")

  # determine how many outliers to remove
  toremove = max(newdf$Outliers[newdf$TestStat. > newdf$CriticalVal.])

  # create vector of same size as input vector
  logicalvectorTifshouldremove = logical(length=length(y))

  # but how to determine which outliers to remove?
  # set largest data points as outliers to remove.. but could be the smallest in some data sets..
  logicalvectorTifshouldremove = replace(logicalvectorTifshouldremove, tail(sort(y), toremove), TRUE)

  return (logicalvectorTifshouldremove)
}

# this should have 3 data points set to TRUE .. but it has 2 and they aren't the correct ones
output = removeoutliers(y)
length(output[output==T])

Answer 1

您可以在库robustHD

上使用winsorize

库（＆＃39; robustHD＆＃39）

set.seed(1234) 
x <- rnorm(10) 
x[1] <- x[1] * 10 
x[2] <- x[2] * 11
x[10] <- x[10] * 10
x
[1] -12.0706575   3.0517217   1.0844412  -2.3456977   0.4291247   0.5060559  -0.5747400  -0.5466319  -0.5644520  -8.9003783
boxplot(x)

enter image description here

y <- winsorize(x)
y
 [1] -4.5609058  3.0517217  1.0844412 -2.3456977  0.4291247  0.5060559 -0.5747400 -0.5466319 -0.5644520 -4.5609058
boxplot(y)

enter image description here

所以如果你有数据框或向量，你可以使用sapply来执行winsorize函数。有关此库的详细信息，请访问此链接http://cran.r-project.org/web/packages/robustHD/index.html

Answer 2

我认为它写在你给出的页面中（不完全是两句话）：

删除最大化| x_i - 意味着（x）|
的 r 观察

因此，只需删除 r 那些超出差异的数据，即可获得没有异常值的数据，使用：

y[abs(y-mean(y)) < sort(abs(y-mean(y)),decreasing=TRUE)[toremove]]

您不需要代码的最后两行。顺便说一下，你可以直接计算：

toremove = which(newdf$TestStat > newdf$CriticalVal)

为简化一点，最终的功能可能是：

# Compute the critical value for ESD Test
esd.critical <- function(alpha, n, i) {
    p = 1 - alpha/(2*(n-i+1))
    t = qt(p,(n-i-1))
    return(t*(n-i) / sqrt((n-i-1+t**2)*(n-i+1)))
}

removeoutliers = function(y) {

    ## Define values and vectors.
    y2 = y
    n = length(y)
    alpha = 0.05
    toremove = 0

    ## Compute test statistic until r=10 values have been
    ## removed from the sample.
    for (i in 1:10){
        if(sd(y2)==0) break
        ares = abs(y2 - mean(y2))/sd(y2)
        Ri = max(ares)
        y2 = y2[ares!=Ri]

        ## Compute critical value.
        if(Ri>esd.critical(alpha,n,i))
            toremove = i
    }

    # Values to keep
    if(toremove>0)
        y = y[abs(y-mean(y)) < sort(abs(y-mean(y)),decreasing=TRUE)[toremove]]

    return (y)
}

返回：

> removeoutliers(y)
 [1] -0.25  0.68  0.94  1.15  1.20  1.26  1.26  1.34  1.38  1.43  1.49
[12]  1.49  1.55  1.56  1.58  1.65  1.69  1.70  1.76  1.77  1.81  1.91
[23]  1.94  1.96  1.99  2.06  2.09  2.10  2.14  2.15  2.23  2.24  2.26
[34]  2.35  2.37  2.40  2.47  2.54  2.62  2.64  2.90  2.92  2.92  2.93
[45]  3.21  3.26  3.30  3.59  3.68  4.30  4.64

用于广义（极端学生偏离）ESD离群值测试的R代码

2 个答案: