CODEWITHSUNDEEP
rlistdataframe
djc55
djc55

Reputation: 555

How to replace a single value in a list of data frames

I have a list of data frames as follows:

dat_1 <- data.frame(x = c("A", "B", "C", "D") %>% as.factor)
dat_2 <- data.frame(x = c("A", "B", "C", "D") %>% as.character)
dat <- list("dat_1" = dat_1, "dat_2" = dat_2)

Some columns are factors and other are characters. I need to replace any instance of "C" with NA for all data frames in the list. How would I accomplish this?

I know I can accomplish this using the following code, but I want to avoid hardcoding each data frame in the list if possible:

dat[["dat_2"]][dat[["dat_2"]] == "C"] <- NA

Upvotes: 2

Views: 189

Answers (4)

akrun
akrun

Reputation: 887901

We can use na_if

library(dplyr)
library(purrr)
map(dat, ~ .x %>%
             mutate_all(na_if, "C"))
#$dat_1
#     x
#1    A
#2    B
#3 <NA>
#4    D

#$dat_2
#     x
#1    A
#2    B
#3 <NA>
#4    D

Upvotes: 2

Ronak Shah
Ronak Shah

Reputation: 389275

We can use dplyr::recode

library(dplyr)
purrr::map(dat, . %>% mutate_all(~recode(., "C"=NA_character_)))

#$dat_1
#     x
#1    A
#2    B
#3 <NA>
#4    D

#$dat_2
#     x
#1    A
#2    B
#3 <NA>
#4    D

Upvotes: 2

www
www

Reputation: 39174

First of all, the way you created the data frame will cause the columns of the data frames to be all factor. I modified your code as follows to make the first data frame to be factor and the second data frame to be character.

dat_1 <- data.frame(x = c("A", "B", "C", "D"))
dat_2 <- data.frame(x = c("A", "B", "C", "D"), stringsAsFactors = FALSE)
dat <- list("dat_1" = dat_1, "dat_2" = dat_2)

We can use the following code to replace all "C" to be NA.

dat2 <- lapply(dat, function(x){
  x[] <- lapply(x, function(x) replace(x, x %in% "C", NA))
  return(x)
})

dat2
# $dat_1
#      x
# 1    A
# 2    B
# 3 <NA>
# 4    D
# 
# $dat_2
#      x
# 1    A
# 2    B
# 3 <NA>
# 4    D

The code will not change the column types.

lapply(dat2, function(x) sapply(x, class))
# $dat_1
#        x 
# "factor" 
# 
# $dat_2
#           x 
# "character"

Upvotes: 3

David Moore
David Moore

Reputation: 978

lapply(dat, function (x) {sapply(x, function (y) {y <- as.character(y); y[which(y == "C")] <- NA; return (y)})})

Upvotes: 2

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