How to substitute NA by 0 in 20 columns?

问题

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

回答1:

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

回答2:

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)

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

回答3:

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

来源：https://stackoverflow.com/questions/33067547/how-to-substitute-na-by-0-in-20-columns

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