deleting outlier in r with account of nominal var

我与影子孤独终老i 提交于 2020-01-02 20:09:14

问题


Say, i have three columns

x <- c(-10, 1:6, 50)
x1<- c(-20, 1:6, 60)
z<- c(1,2,3,4,5,6,7,8)

check outliers for x

bx <- boxplot(x)
bx$out

check outliers for x1

bx1 <- boxplot(x1)
bx1$out

now we must delete outliers

x <- x[!(x %in% bx$out)]
x

x1 <- x1[!(x1 %in% bx1$out)]
x1

but we have variable Z(nominal) and we must remove observations, which correspond to the outlier of variables x and x1, in our case it is 1 and 8 obs. of Z

How to do it? in output we must have

x   x1  z
Na  Na  Na
1   1   2
2   2   3
3   3   4
4   4   5
5   5   6
6   6   7
Na  Na  Na

回答1:


Try this solution:

x_to_remove<-which(x %in% bx$out)
x <- x[!(x %in% bx$out)]

x1_to_remove<-which(x1 %in% bx1$out)
x1 <- x1[!(x1 %in% bx1$out)]

z<-z[-unique(c(x_to_remove,x1_to_remove))]
z    
[1] 2 3 4 5 6 7

Before delete values in x and x1 you have to save the positions (x_to_remove and x1_to_remove) and than use to clean z.

Your output:

data.frame(cbind(x,x1,z))
  x x1 z
1 1  1 2
2 2  2 3
3 3  3 4
4 4  4 5
5 5  5 6
6 6  6 7



回答2:


If you have a dataframe as

x  <- c(-10, 1:6, 50)
x1 <- c(-20, 1:6, 60)
z  <- c(1,2,3,4,5,6,7,8)
df <- data.frame(x = x, x1 = x1, z = z)

You can do this to remove rows with outliers in x or x1

is.outlier <- sapply(df[c('x', 'x1')], function(x) x %in% boxplot(x)$out)
df[!rowSums(is.outlier),]

#   x x1 z
# 2 1  1 2
# 3 2  2 3
# 4 3  3 4
# 5 4  4 5
# 6 5  5 6
# 7 6  6 7

In tidyverse (same result)

library(tidyverse)
df %>% 
  filter(map(list(x, x1), ~!.x %in% boxplot(.x)$out) %>% pmap_lgl(`&`))



回答3:


You can try

z[!((x1 %in% bx1$out) | (x %in% bx$out))]

Or a tidyverse

library(tidyverse)
data.frame(x, x1, z) %>% 
  select(starts_with("x")) %>%  
  map_dfr(~.x %in% boxplot(.x, plot = F)$out) %>%  
  with(.,!rowSums(.)) %>% 
  filter(df, .)
   x x1 z
1 50  1 2
2  1  2 3
3  2  3 4
4  3  4 5
5  4  5 6
6  5  6 7


来源:https://stackoverflow.com/questions/50569314/deleting-outlier-in-r-with-account-of-nominal-var

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