change argument object in R
0
I want to mutate a data.frame object within a function. The following does not do what I intended:
# function to change factors to characters using dplyr
# x: a data.frame
fa_clean <- function(x,...) {
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
print(x)
return(x)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test) # applying the function
str(test) # letter is still a factor ???
I am obviously missing something. Edit: And I am aware of:
test <- fa_clean(test)
But I would like to run it without this assignment.
r function
asked Nov 23 '18 at 8:30
Krisselack
10010
add a comment |
0
I want to mutate a data.frame object within a function. The following does not do what I intended:
# function to change factors to characters using dplyr
# x: a data.frame
fa_clean <- function(x,...) {
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
print(x)
return(x)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test) # applying the function
str(test) # letter is still a factor ???
I am obviously missing something. Edit: And I am aware of:
test <- fa_clean(test)
But I would like to run it without this assignment.
r function
asked Nov 23 '18 at 8:30
Krisselack
10010
R is a functional language. It's a feature that you have to explicitly assign the result to something.
– Hong Ooi
Nov 23 '18 at 8:54
Also,printing something to the console is not the same as returning a result
– Hong Ooi
Nov 23 '18 at 8:55
1
Wouldmagrittr's inplace pipe work for your needs?test %<>% mutate_if(is.factor, as.character)will updatetestin place, although if you use that inside a function with argumentx, it won't updatetestoutside of the function.
– andrew_reece
Nov 23 '18 at 8:57
add a comment |
0
0
0
I want to mutate a data.frame object within a function. The following does not do what I intended:
# function to change factors to characters using dplyr
# x: a data.frame
fa_clean <- function(x,...) {
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
print(x)
return(x)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test) # applying the function
str(test) # letter is still a factor ???
I am obviously missing something. Edit: And I am aware of:
test <- fa_clean(test)
But I would like to run it without this assignment.
r function
asked Nov 23 '18 at 8:30
Krisselack
10010
I want to mutate a data.frame object within a function. The following does not do what I intended:
# function to change factors to characters using dplyr
# x: a data.frame
fa_clean <- function(x,...) {
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
print(x)
return(x)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test) # applying the function
str(test) # letter is still a factor ???
I am obviously missing something. Edit: And I am aware of:
test <- fa_clean(test)
But I would like to run it without this assignment.
r function
r function
asked Nov 23 '18 at 8:30
Krisselack
10010
asked Nov 23 '18 at 8:30
Krisselack
10010
edited Nov 23 '18 at 10:40
asked Nov 23 '18 at 8:30
Krisselack
10010
asked Nov 23 '18 at 8:30
Krisselack
10010
asked Nov 23 '18 at 8:30
Krisselack
10010
10010
R is a functional language. It's a feature that you have to explicitly assign the result to something.
– Hong Ooi
Nov 23 '18 at 8:54
Also,printing something to the console is not the same as returning a result
– Hong Ooi
Nov 23 '18 at 8:55
1
Wouldmagrittr's inplace pipe work for your needs?test %<>% mutate_if(is.factor, as.character)will updatetestin place, although if you use that inside a function with argumentx, it won't updatetestoutside of the function.
– andrew_reece
Nov 23 '18 at 8:57
add a comment |
R is a functional language. It's a feature that you have to explicitly assign the result to something.
– Hong Ooi
Nov 23 '18 at 8:54
Also,printing something to the console is not the same as returning a result
– Hong Ooi
Nov 23 '18 at 8:55
1
Wouldmagrittr's inplace pipe work for your needs?test %<>% mutate_if(is.factor, as.character)will updatetestin place, although if you use that inside a function with argumentx, it won't updatetestoutside of the function.
– andrew_reece
Nov 23 '18 at 8:57
R is a functional language. It's a feature that you have to explicitly assign the result to something.
– Hong Ooi
Nov 23 '18 at 8:54
R is a functional language. It's a feature that you have to explicitly assign the result to something.
– Hong Ooi
Nov 23 '18 at 8:54
Also,
printing something to the console is not the same as returning a result– Hong Ooi
Nov 23 '18 at 8:55
Also,
printing something to the console is not the same as returning a result– Hong Ooi
Nov 23 '18 at 8:55
1
1
Would
magrittr's inplace pipe work for your needs? test %<>% mutate_if(is.factor, as.character) will update test in place, although if you use that inside a function with argument x, it won't update test outside of the function.– andrew_reece
Nov 23 '18 at 8:57
Would
magrittr's inplace pipe work for your needs? test %<>% mutate_if(is.factor, as.character) will update test in place, although if you use that inside a function with argument x, it won't update test outside of the function.– andrew_reece
Nov 23 '18 at 8:57
add a comment |
1 Answer
1
active
oldest
votes
1
Here is your code with the necessary modification to make it work:
fa_clean <- function(x) {
varname <- deparse(substitute(x))
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
assign(varname, x, envir = .GlobalEnv)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test)
str(test) # letter is no longer a factor
'data.frame': 10 obs. of 2 variables:
$ number: int 1 2 3 4 5 6 7 8 9 10
$ letter: chr "a" "b" "c" "d" ...
answered Nov 23 '18 at 9:34
snoram
6,402831
Thanks, so basically removing the piping did the trick. Your first version, which I also tried before returns a factor.
– Krisselack
Nov 23 '18 at 10:32
fa_clean <- function(x,...) { require(dplyr) x <- x %>% mutate_if(is.factor, as.character) x # don't print, just return } test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10])) fa_clean(test) str(test) # returns a factor
– Krisselack
Nov 23 '18 at 10:35
1
This is exactly what I wanted. I thought that I would need environments, I just fooled around with <<-, but your solution taught me a lot. Thank you very much.
– Krisselack
Nov 23 '18 at 10:51
add a comment |
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1 Answer
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votes
1 Answer
1
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votes
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1
Here is your code with the necessary modification to make it work:
fa_clean <- function(x) {
varname <- deparse(substitute(x))
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
assign(varname, x, envir = .GlobalEnv)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test)
str(test) # letter is no longer a factor
'data.frame': 10 obs. of 2 variables:
$ number: int 1 2 3 4 5 6 7 8 9 10
$ letter: chr "a" "b" "c" "d" ...
answered Nov 23 '18 at 9:34
snoram
6,402831
Thanks, so basically removing the piping did the trick. Your first version, which I also tried before returns a factor.
– Krisselack
Nov 23 '18 at 10:32
fa_clean <- function(x,...) { require(dplyr) x <- x %>% mutate_if(is.factor, as.character) x # don't print, just return } test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10])) fa_clean(test) str(test) # returns a factor
– Krisselack
Nov 23 '18 at 10:35
1
This is exactly what I wanted. I thought that I would need environments, I just fooled around with <<-, but your solution taught me a lot. Thank you very much.
– Krisselack
Nov 23 '18 at 10:51
add a comment |
1
Here is your code with the necessary modification to make it work:
fa_clean <- function(x) {
varname <- deparse(substitute(x))
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
assign(varname, x, envir = .GlobalEnv)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test)
str(test) # letter is no longer a factor
'data.frame': 10 obs. of 2 variables:
$ number: int 1 2 3 4 5 6 7 8 9 10
$ letter: chr "a" "b" "c" "d" ...
answered Nov 23 '18 at 9:34
snoram
6,402831
Thanks, so basically removing the piping did the trick. Your first version, which I also tried before returns a factor.
– Krisselack
Nov 23 '18 at 10:32
fa_clean <- function(x,...) { require(dplyr) x <- x %>% mutate_if(is.factor, as.character) x # don't print, just return } test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10])) fa_clean(test) str(test) # returns a factor
– Krisselack
Nov 23 '18 at 10:35
1
This is exactly what I wanted. I thought that I would need environments, I just fooled around with <<-, but your solution taught me a lot. Thank you very much.
– Krisselack
Nov 23 '18 at 10:51
add a comment |
1
1
1
Here is your code with the necessary modification to make it work:
fa_clean <- function(x) {
varname <- deparse(substitute(x))
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
assign(varname, x, envir = .GlobalEnv)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test)
str(test) # letter is no longer a factor
'data.frame': 10 obs. of 2 variables:
$ number: int 1 2 3 4 5 6 7 8 9 10
$ letter: chr "a" "b" "c" "d" ...
answered Nov 23 '18 at 9:34
snoram
6,402831
Here is your code with the necessary modification to make it work:
fa_clean <- function(x) {
varname <- deparse(substitute(x))
require(dplyr)
x <- x %>% mutate_if(is.factor, as.character)
assign(varname, x, envir = .GlobalEnv)
}
# example set
test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10]))
fa_clean(test)
str(test) # letter is no longer a factor
'data.frame': 10 obs. of 2 variables:
$ number: int 1 2 3 4 5 6 7 8 9 10
$ letter: chr "a" "b" "c" "d" ...
answered Nov 23 '18 at 9:34
snoram
6,402831
edited Dec 13 '18 at 10:05
answered Nov 23 '18 at 9:34
snoram
6,402831
answered Nov 23 '18 at 9:34
snoram
6,402831
answered Nov 23 '18 at 9:34
snoram
6,402831
6,402831
Thanks, so basically removing the piping did the trick. Your first version, which I also tried before returns a factor.
– Krisselack
Nov 23 '18 at 10:32
fa_clean <- function(x,...) { require(dplyr) x <- x %>% mutate_if(is.factor, as.character) x # don't print, just return } test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10])) fa_clean(test) str(test) # returns a factor
– Krisselack
Nov 23 '18 at 10:35
1
This is exactly what I wanted. I thought that I would need environments, I just fooled around with <<-, but your solution taught me a lot. Thank you very much.
– Krisselack
Nov 23 '18 at 10:51
add a comment |
Thanks, so basically removing the piping did the trick. Your first version, which I also tried before returns a factor.
– Krisselack
Nov 23 '18 at 10:32
fa_clean <- function(x,...) { require(dplyr) x <- x %>% mutate_if(is.factor, as.character) x # don't print, just return } test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10])) fa_clean(test) str(test) # returns a factor
– Krisselack
Nov 23 '18 at 10:35
1
This is exactly what I wanted. I thought that I would need environments, I just fooled around with <<-, but your solution taught me a lot. Thank you very much.
– Krisselack
Nov 23 '18 at 10:51
Thanks, so basically removing the piping did the trick. Your first version, which I also tried before returns a factor.
– Krisselack
Nov 23 '18 at 10:32
Thanks, so basically removing the piping did the trick. Your first version, which I also tried before returns a factor.
– Krisselack
Nov 23 '18 at 10:32
fa_clean <- function(x,...) { require(dplyr) x <- x %>% mutate_if(is.factor, as.character) x # don't print, just return } test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10])) fa_clean(test) str(test) # returns a factor
– Krisselack
Nov 23 '18 at 10:35
fa_clean <- function(x,...) { require(dplyr) x <- x %>% mutate_if(is.factor, as.character) x # don't print, just return } test <- data.frame(number=c(1:10),letter=as.factor(letters[1:10])) fa_clean(test) str(test) # returns a factor
– Krisselack
Nov 23 '18 at 10:35
1
1
This is exactly what I wanted. I thought that I would need environments, I just fooled around with <<-, but your solution taught me a lot. Thank you very much.
– Krisselack
Nov 23 '18 at 10:51
This is exactly what I wanted. I thought that I would need environments, I just fooled around with <<-, but your solution taught me a lot. Thank you very much.
– Krisselack
Nov 23 '18 at 10:51
add a comment |
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R is a functional language. It's a feature that you have to explicitly assign the result to something.
– Hong Ooi
Nov 23 '18 at 8:54
Also,
printing something to the console is not the same as returning a result– Hong Ooi
Nov 23 '18 at 8:55
1
Would
magrittr's inplace pipe work for your needs?test %<>% mutate_if(is.factor, as.character)will updatetestin place, although if you use that inside a function with argumentx, it won't updatetestoutside of the function.– andrew_reece
Nov 23 '18 at 8:57