Replace values outside range with NA using replace_with_na function

Question

I have the following dataset

structure(list(a = c(2, 1, 9, 2, 9, 8), b = c(4, 5, 1, 9, 12, 
NA), c = c(50, 34, 77, 88, 33, 60)), class = "data.frame", row.names = c(NA, 
-6L))

  a  b  c
1 2  4 50
2 1  5 34
3 9  1 77
4 2  9 88
5 9 12 33
6 8 NA 60

From column b I only want values between 4-9. Column c between 50-80. Replacing the values outside the range with NA, resulting in

structure(list(a = c(2, 1, 9, 2, 9, 8), b = c(4, 5, NA, 9, NA, 
NA), c = c(50, NA, 77, NA, NA, 60)), class = "data.frame", row.names = c(NA, 
-6L))

  a  b  c
1 2  4 50
2 1  5 NA
3 9 NA 77
4 2  9 NA
5 9 NA NA
6 8 NA 60

I've tried several things with replace_with_na_at function where this seemed most logical:

test <- replace_with_na_at(data = test, .vars="c",
                          condition = ~.x < 2 & ~.x > 2)

However, nothing I tried works. Does somebody know why? Thanks in advance! :)

`replace_with_na_at()` comes from what package? You should include the relevant library call in your reproducible example. — s_baldur, Jun 02 '20 at 10:33

score 4 · Answer 1 · answered Jun 02 '20 at 10:41

Yet another base R solution, this time with function is.na<-

is.na(test$b) <- with(test, b < 4 | b > 9)
is.na(test$c) <- with(test, c < 50 | c > 80)

A package naniar solution with a pipe could be

library(naniar)
library(magrittr)

test %>%
  replace_with_na_at(
    .vars = 'b',
    condition = ~(.x < 4 | .x > 9)
  ) %>%
  replace_with_na_at(
    .vars = 'c',
    condition = ~(.x < 50 | .x > 80)
  )

score 3 · Answer 2 · answered Jun 02 '20 at 10:30

You simply could use Map to replace your values with NA.

dat[2:3] <- Map(function(x, y) {x[!x %in% y] <- NA;x}, dat[2:3], list(4:9, 50:80))
dat
#   a  b  c
# 1 2  4 50
# 2 1  5 NA
# 3 9 NA 77
# 4 2  9 NA
# 5 9 NA NA
# 6 8 NA 60

Data:

dat <- structure(list(a = c(2, 1, 9, 2, 9, 8), b = c(4, 5, 1, 9, 12, 
NA), c = c(50, 34, 77, 88, 33, 60)), class = "data.frame", row.names = c(NA, 
-6L))

score 3 · Answer 3 · answered Jun 02 '20 at 10:36

You can subset with a logical vector testing your conditions.

x$b[x$b < 4 | x$b > 9] <- NA
x$c[x$c < 50 | x$c > 80] <- NA
x
#  a  b  c
#1 2  4 50
#2 1  5 NA
#3 9 NA 77
#4 2  9 NA
#5 9 NA NA
#6 8 NA 60

Data:

x <- structure(list(a = c(2, 1, 9, 2, 9, 8), b = c(4, 5, 1, 9, 12, 
NA), c = c(50, 34, 77, 88, 33, 60)), class = "data.frame", row.names = c(NA, 
-6L))

score 3 · Answer 4 · answered Jun 02 '20 at 19:42

3

We can use map2

library(purrr)
library(dplyr)
df1[c('b', 'c')] <- map2(df1 %>% 
       select(b, c), list(c(4, 9), c(50,80)), ~ 
           replace(.x, .x < .y[1]|.x > .y[2], NA))

answered Jun 02 '20 at 19:42

akrun

874,273
37
540
662

shs · Accepted Answer · 2020-06-03T15:15:16.487

You should mention the packages you are using. From googling, i'm guessing you are using naniar. The problem appears to be that you did not properly specify the condition, but the following should work:

library(naniar)

test <- structure(list(a = c(2, 1, 9, 2, 9, 8), 
                    b = c(4, 5, 1, 9, 12, NA),
                    c = c(50, 34, 77, 88, 33, 60)), 
               class = "data.frame", 
               row.names = c(NA, -6L)) 

replace_with_na_at(test, "c", ~.x < 50 | .x > 80)
#>   a  b  c
#> 1 2  4 50
#> 2 1  5 NA
#> 3 9  1 77
#> 4 2  9 NA
#> 5 9 12 NA
#> 6 8 NA 60

^{Created on 2020-06-02 by the reprex package (v0.3.0)}

Replace values outside range with NA using replace_with_na function

5 Answers5