如何在 20 列中用 0 替换 NA？

Question

I want to substitute NA by 0 in 20 columns. I found this approach for 2 columns, however I guess it's not optimal if the number of columns is 20. Is there any alternative and more compact solution?

mydata[,c("a", "c")] <-
        apply(mydata[,c("a","c")], 2, function(x){replace(x, is.na(x), 0)})

UPDATE: For simplicity lets take this data with 8 columns and substitute NAs in columns b, c, e, f and d

a  b  c  d  e  f  g  d
1  NA NA 2  3  4  7  6
2  g  3  NA 4  5  4  Y
3  r  4  4  NA t  5  5

The result must be this one:

a  b  c  d  e  f  g  d
1  0  0  2  3  4  7  6
2  g  3  NA 4  5  4  Y
3  r  4  4  0  t  5  5

Answer 1

The replace_na function from tidyr can be applied over a vector as well as a dataframe ( http://tidyr.tidyverse.org/reference/replace_na.html ).

Use it with a mutate_at variation from dplyr to apply it to multiple columns at the same time:

my_data %>% mutate_at(vars(b,c,e,f), replace_na, 0)

or

my_data %>% mutate_at(c('b','c','e','f'), replace_na, 0)

Answer 2

Another option:

library(tidyr)
v <- c('b', 'c', 'e', 'f')
replace_na(df, as.list(setNames(rep(0, length(v)), v)))

Which gives:

#  a b c  d e f g d.1
#1 1 0 0  2 3 4 7   6
#2 2 g 3 NA 4 5 4   Y
#3 3 r 4  4 0 t 5   5

Answer 3

Here is a tidyverse way to replace NA with different values based on the data type of the column.

library(tidyverse)

dataset %>% mutate_if(is.numeric, replace_na, 0) %>%  
    mutate_if(is.character, replace_na, "")

Answer 4

We can use NAer from qdap to convert the NA to 0. If there are multiple column, loop using lapply .

library(qdap)
nm1 <- c('b', 'c', 'e', 'f')
mydata[nm1] <- lapply(mydata[nm1], NAer)
mydata
#  a b c  d e f g d.1
#1 1 0 0  2 3 4 7   6
#2 2 g 3 NA 4 5 4   Y
#3 3 r 4  4 0 t 5   5

---

Or using dplyr

library(dplyr)
mydata %>% 
   mutate_each_(funs(replace(., which(is.na(.)), 0)), nm1)
#  a b c  d e f g d.1
#1 1 0 0  2 3 4 7   6
#2 2 g 3 NA 4 5 4   Y
#3 3 r 4  4 0 t 5   5

Answer 5

Another strategy using tidyr::replace_na()

library(tidyverse)

df <- read.table(header = T, text = 'a  b  c  d  e  f  g  h
1  NA NA 2  3  4  7  6
2  g  3  NA 4  5  4  Y
3  r  4  4  NA t  5  5')

df %>%
  mutate(across(everything(), ~replace_na(., 0)))
#>   a b c d e f g h
#> 1 1 0 0 2 3 4 7 6
#> 2 2 g 3 0 4 5 4 Y
#> 3 3 r 4 4 0 t 5 5

^{Created on 2021-08-22 by the reprex package (v2.0.0)}

Answer 6

Knowing that replace_na() accepts a named list for the replace argument, using purrr::map() is a good option here to reduce the amount of code. It is also possible to replace different values in each column using 'map2()'.

code:

library(tidyverse)

tbl <-read_table("a  b  c  d  e  f  g  d
1  NA NA 2  3  4  7  6
2  g  3  NA 4  5  4  Y
3  r  4  4  NA t  5  5")
#> Warning: Duplicated column names deduplicated: 'd' => 'd_1' [8]

replace_na(tbl, map(tbl, ~0))
#> # A tibble: 3 × 8
#>       a b         c     d     e f         g d_1  
#>   <dbl> <chr> <dbl> <dbl> <dbl> <chr> <dbl> <chr>
#> 1     1 0         0     2     3 4         7 6    
#> 2     2 g         3     0     4 5         4 Y    
#> 3     3 r         4     4     0 t         5 5

#alternative

tbl %>%
  replace_na(map(., ~0))
#> # A tibble: 3 × 8
#>       a b         c     d     e f         g d_1  
#>   <dbl> <chr> <dbl> <dbl> <dbl> <chr> <dbl> <chr>
#> 1     1 0         0     2     3 4         7 6    
#> 2     2 g         3     0     4 5         4 Y    
#> 3     3 r         4     4     0 t         5 5

^{Created on 2021-09-11 by the reprex package (v2.0.0)}