'Apply an `as.character()` function to a list of dataframes
So essentially I have a list of dataframes that I want to apply as.character() to.
To obtain the list of dataframes I have a list of files that I read in using a map() function and a read funtion that I created. I can't use map_df() because there are columns that are being read in as different data types. All of the files are the same and I know that I could hard code the data types in the read function if I wanted, but I want to avoid that if I can.
At this point I throw the list of dataframes in a for loop and apply another map() function to apply the as.character() function. This final list of dataframes is then compressed using bind_rows().
All in all, this seems like an extremely convoluted process, see code below.
audits <- list.files()
my_reader <- function(x) {
my_file <- read_xlsx(x)
}
audits <- map(audits, my_reader)
for (i in 1:length(audits)) {
audits[[i]] <- map_df(audits[[i]], as.character)
}
audits <- bind_rows(audits)
Does anybody have any ideas on how I can improve this? Ideally to the point where I can do everything in a single vectorised map() function?
For reproducibility you can use two iris datasets with one of the columns datatypes changed.
iris2 <- iris
iris2[1] <- as.character(iris2[1])
my_list <- list(iris, iris2)
Solution 1:[1]
as.character works on vector whereas data.frame is a list of vectors. An option is to use across if we want only a single use of map
library(dplyr)
library(purrr)
map_dfr(my_list, ~ .x %>%
mutate(across(everything(), as.character)))
Solution 2:[2]
I wanted to show a base R solution just incase if it helps anyone else. You can use rapply to recursively go through the list and apply a function. you can specify class and if you want to replace or unlist/list the returned object:
iris2 <- iris
iris2[1] <- as.character(iris2[1])
my_list <- list(iris, iris2)
mylist2 <- rapply(my_list, class = "ANY", f = as.character, how = "replace")
bigdf <- do.call(rbind, mylist2)
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | Mike |
