如何检测主要是统计问题。您可以使用的一种方法是 Hampel 过滤器(其优缺点不在此答案的范围内)。
它将median ± 3*(median absolute deviation) 之外的值视为异常值。
假设我们将使用这个标准。您可以通过data.table 的by 参数在测试内部和测试之间进行。
将数据转换为长格式更好吗?
这会使分析更容易,因此我通过melt进行了转换
my_data <- data.frame(Company = c("A","B","C","D"), CITY = c("Paris", "Paris", "Quimper", "Nice"), year_creation = c("2010", "2009", "2008", "2009"), revenue_2008 = c(NA, NA, 10, NA),
revenue_2009 = c(NA,10, 20, 15000), revenue_2010 = c(02, 10, 2500, 20000), revenue_2011 = c(14, 16, 10, 30000),
size = c(2, 3, 5, 1))
library(data.table)
my_data <- as.data.table(my_data)
my_data <- melt(my_data, id.vars = c("Company", "CITY", "year_creation", "size"))
hampel_filter <- function(x){
x_med <- median(x, na.rm = TRUE)
x_mad <- mad(x, na.rm = TRUE)
(x > x_med + 3*x_mad | x < x_med - 3*x_mad)
}
my_data[, between_out := hampel_filter(value), by = variable]
my_data[, within_out := hampel_filter(value), by = Company]
> my_data
Company CITY year_creation size variable value between_out within_out
1: A Paris 2010 2 revenue_2008 NA NA NA
2: B Paris 2009 3 revenue_2008 NA NA NA
3: C Quimper 2008 5 revenue_2008 10 FALSE FALSE
4: D Nice 2009 1 revenue_2008 NA NA NA
5: A Paris 2010 2 revenue_2009 NA NA NA
6: B Paris 2009 3 revenue_2009 10 FALSE FALSE
7: C Quimper 2008 5 revenue_2009 20 FALSE FALSE
8: D Nice 2009 1 revenue_2009 15000 TRUE FALSE
9: A Paris 2010 2 revenue_2010 2 FALSE FALSE
10: B Paris 2009 3 revenue_2010 10 FALSE FALSE
11: C Quimper 2008 5 revenue_2010 2500 FALSE TRUE
12: D Nice 2009 1 revenue_2010 20000 TRUE FALSE
13: A Paris 2010 2 revenue_2011 14 FALSE FALSE
14: B Paris 2009 3 revenue_2011 16 FALSE TRUE
15: C Quimper 2008 5 revenue_2011 10 FALSE FALSE
16: D Nice 2009 1 revenue_2011 30000 TRUE FALSE
您还可以使用来自DescTools 的Winsorize() 同时检测和处理异常值。查看详情:https://en.wikipedia.org/wiki/Winsorizing