Reputation: 5929
I'm trying to clean a sixty column data extract that has been given to me. Part of the data is about thirty columns which have been supplied as "Yes" or "No" values that I would like to convert to logical type. It therefore isn't every column in the data frame, but it is a lot of them. I'm currently doing the equivalent of this:
mtcars %>%
mutate(mpg = as.character(mpg)) %>%
mutate(cyl = as.character(cyl)) %>%
mutate(disp = as.character(disp)) %>%
mutate(hp = as.character(hp))
That is, manually mutating each column in the list. But that feels like it will be prone to error from missing a copy-paste or similar. Is there a function that could do this in one step by being passed a list of field names? I tend to default to tidyverse functions, though base R also works if needed.
Upvotes: 0
Views: 657
Reputation: 389325
This should be a duplicate but cannot find a relevant post right now.
We can use mutate_at
and apply function on selected columns
library(dplyr)
mtcars %>% mutate_at(vars(mpg, cyl, disp, hp), as.character)
Or if we have column names stored in vector called cols
we could do
cols <- c("mpg", "cyl", "disp", "hp")
mtcars %>% mutate_at(cols, as.character)
Upvotes: 1
Reputation: 2009
Perhaps you can use lapply()
?
lapply(mtcars, as.character)
If you would like your data as a data frame:
df = as.data.frame( lapply(mtcars, as.character), stringsAsFactors = F )
> df$mpg
[1] "21" "21" "22.8" "21.4" "18.7" "18.1" "14.3" "24.4" "22.8"
[10] "19.2" "17.8" "16.4" "17.3" "15.2" "10.4" "10.4" "14.7" "32.4"
[19] "30.4" "33.9" "21.5" "15.5" "15.2" "13.3" "19.2" "27.3" "26"
[28] "30.4" "15.8" "19.7" "15" "21.4"
> df$cyl
[1] "6" "6" "4" "6" "8" "6" "8" "4" "4" "6" "6" "8" "8" "8" "8" "8"
[17] "8" "4" "4" "4" "4" "8" "8" "8" "8" "4" "4" "4" "8" "6" "8" "4"
> df$disp
[1] "160" "160" "108" "258" "360" "225" "360" "146.7"
[9] "140.8" "167.6" "167.6" "275.8" "275.8" "275.8" "472" "460"
[17] "440" "78.7" "75.7" "71.1" "120.1" "318" "304" "350"
[25] "400" "79" "120.3" "95.1" "351" "145" "301" "121"
Upvotes: 0