I have mutliple data frames (different size of rows, however, same columns). I want to create a factor in each of them using a loop.
list.dfs <- list(d1, d2, d3, d4, d5, d6, d7, d8)
for (i in 1:length(list.dfs)){
d[i]$Gender <- factor(d[i]$Gender,
levels = c(1, 2, 3),
labels = c("female", "male", "divers")
)
}
This is not working
Call your factor instruction in a lapply loop and assign the result back to list.dfs.
list.dfs <- list(structure(list(Gender = c(1, 2, 3)),
class = "data.frame", row.names = c(NA, -3L)),
structure(list(Gender = c(1, 2, 3)),
class = "data.frame", row.names = c(NA, -3L)))
list.dfs <- lapply(list.dfs, \(x) {
x$Gender <- factor(x$Gender, levels = c(1, 2, 3), labels = c("female", "male", "divers"))
x
})
list.dfs
#> [[1]]
#> Gender
#> 1 female
#> 2 male
#> 3 divers
#>
#> [[2]]
#> Gender
#> 1 female
#> 2 male
#> 3 divers
Created on 2022-05-24 by the reprex package (v2.0.1)
R would not read d[i] as the object d1 if i = 1. You can access items of a list using [[i]]. Note that class(list.dfs[1]) is list while class(list.dfs[[1]]) is data.frame.
As an example
#example data
list.dfs <- list(structure(list(gender = c(1, 2, 3)), class = "data.frame", row.names = c(NA,
-3L)), structure(list(gender = c(1, 2, 3)), class = "data.frame", row.names = c(NA,
-3L)))
#check first item
list.dfs[[1]]
gender
1 1
2 2
3 3
#use for loop to access all items of the list, apply function
for(i in 1:length(list.dfs)){
list.dfs[[i]]$gender <- factor(list.dfs[[i]]$gender, levels = c(1, 2, 3),
labels = c("female", "male", "diverse"))
}
You also might want to read into lapply which applies a function on every object in the list.
#example using lapply
lapply(list.dfs, FUN = function(x) x$gender <- factor(x$gender, levels = c(1,2,3), labels = c("female", "male", "diverse")))
Regarding your comment (assign objects in the list to global environment).
#say we used that lapply function before and end up with this list
list.dfs <- list(structure(1:3, .Label = c("female", "male", "diverse"), class = "factor"),
structure(1:3, .Label = c("female", "male", "diverse"), class = "factor"))
Then we can add names to the list according to the order the items were put in the list and assign list objects to global environment.
#add names (will be the object names later)
names(list.dfs) <- c("d1", "d2")
#assign to global environment
for(i in 1:length(list.dfs)){
assign(names(list.dfs)[i], list.dfs[[i]])
}
Note, I am not a fan of assigning objects like that. Personally I would probably write a function and use that function 8 times. E.g.
#create function
gender_fun <- function(x){
#read data.frame
dt1 <- x
#set gender
dt1$gender <- factor(dt1$gender, levels = c(1,2,3), labels = c("female", "male", "diverse"))
#return
return(dt1)
}
#apply function on the data.frames
dt1 <- gender_fun(dt1)
dt2 <- gender_fun(dt2)
#etc...
assign but can consider writing a function and use that function for each data.frame individually. Or keep working with the list so you can use lapply like functions. I personally avoid using assign.
list.dfs[[i]]$Gender.dput(d1). Or, if it is too big with the output ofdput(head(d1, 20)). And say in the question whether alld1throughd8have the same structure.