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我目前正在尝试根据所选变量的子集排除异常值，以进行敏感性分析。我已经调整了此处可用的功能：计算 R 中的异常值），但到目前为止一直不成功（我仍然是新手 R 用户）。如果您有任何建议，请告诉我！

df <- data.frame(ID = c(1001, 1002, 1003, 1004, 1005,   1006,   1007,   1008,   1009,   1010,   1011),
                 measure1 = rnorm(11, mean = 8, sd = 4),
                 measure2 = rnorm(11, mean = 40, sd = 5),
                 measure3 = rnorm(11, mean = 20, sd = 2),
                 measure4 = rnorm(11, mean = 9, sd = 3))

vars_of_interest <- c("measure1", "measure3", "measure4")

# define a function to remove outliers
FindOutliers <- function(data) {
  lowerq = quantile(data)[2]
  upperq = quantile(data)[4]
  iqr = upperq - lowerq #Or use IQR(data)
  # we identify extreme outliers
  extreme.threshold.upper = (iqr * 3) + upperq
  extreme.threshold.lower = lowerq - (iqr * 3)
  result <- which(data > extreme.threshold.upper | data < extreme.threshold.lower)
}

# use the function to identify outliers
temp <- FindOutliers(df[vars_of_interest])

# remove the outliers
testData <- testData[-temp]

# show the data with the outliers removed
testData
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1 回答 1

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分离关注点：

  1. 使用 IQR 方法识别数值向量中的异常值。这可以封装在一个采用向量的函数中。
  2. 从 data.frame 的几列中删除异常值。这是一个采用 data.frame 的函数。

我建议返回一个布尔向量而不是索引。这样，返回的值就是数据的大小，这样可以很容易地创建一个新列，例如df$outlier <- is_outlier(df$measure1).

请注意参数名称如何明确预期输入的类型：x是数字向量的标准名称，df显然是 data.frame。cols可能是列名的列表或向量。

我指出只使用base R，但在现实生活中我会使用dplyr包来操作data.frame。

#' Detect outliers using IQR method
#' 
#' @param x A numeric vector
#' @param na.rm Whether to exclude NAs when computing quantiles
#' 
is_outlier <- function(x, na.rm = FALSE) {
  qs = quantile(x, probs = c(0.25, 0.75), na.rm = na.rm)

  lowerq <- qs[1]
  upperq <- qs[2]
  iqr = upperq - lowerq 

  extreme.threshold.upper = (iqr * 3) + upperq
  extreme.threshold.lower = lowerq - (iqr * 3)

  # Return logical vector
  x > extreme.threshold.upper | x < extreme.threshold.lower
}

#' Remove rows with outliers in given columns
#' 
#' Any row with at least 1 outlier will be removed
#' 
#' @param df A data.frame
#' @param cols Names of the columns of interest. Defaults to all columns.
#' 
#' 
remove_outliers <- function(df, cols = names(df)) {
  for (col in cols) {
    cat("Removing outliers in column: ", col, " \n")
    df <- df[!is_outlier(df[[col]]),]
  }
  df
}

有了这两个功能，它变得非常容易：

df <- data.frame(ID = c(1001, 1002, 1003, 1004, 1005,   1006,   1007,   1008,   1009,   1010,   1011),
                 measure1 = rnorm(11, mean = 8, sd = 4),
                 measure2 = rnorm(11, mean = 40, sd = 5),
                 measure3 = rnorm(11, mean = 20, sd = 2),
                 measure4 = rnorm(11, mean = 9, sd = 3))

vars_of_interest <- c("measure1", "measure3", "measure4")


df_filtered <- remove_outliers(df, vars_of_interest)
#> Removing outliers in column:  measure1  
#> Removing outliers in column:  measure3  
#> Removing outliers in column:  measure4

df_filtered
#>      ID  measure1 measure2 measure3   measure4
#> 1  1001  9.127817 40.10590 17.69416  8.6031175
#> 2  1002 18.196182 38.50589 23.65251  7.8630485
#> 3  1003 10.537458 37.97222 21.83248  6.0798316
#> 4  1004  5.590463 46.83458 21.75404  6.9589981
#> 5  1005 14.079801 38.47557 20.93920 -0.6370596
#> 6  1006  3.830089 37.19281 19.56507  6.2165156
#> 7  1007 14.644766 37.09235 19.78774 10.5133674
#> 8  1008  5.462400 41.02952 20.14375 13.5247993
#> 9  1009  5.215756 37.65319 22.23384  7.3131715
#> 10 1010 14.518045 48.97977 20.33128  9.9482211
#> 11 1011  1.594353 44.09224 21.32434 11.1561089

由reprex 包（v0.3.0）于 2020-03-23 创建

于 2020-03-23T13:28:29.163 回答