When I convert a factor to a numeric or integer, I get the underlying level codes, not the values as numbers.

f <- factor(sample(runif(5), 20, replace = TRUE))
##  [1] 0.0248644019011408 0.0248644019011408 0.179684827337041 
##  [4] 0.0284090070053935 0.363644931698218  0.363644931698218 
##  [7] 0.179684827337041  0.249704354675487  0.249704354675487 
## [10] 0.0248644019011408 0.249704354675487  0.0284090070053935
## [13] 0.179684827337041  0.0248644019011408 0.179684827337041 
## [16] 0.363644931698218  0.249704354675487  0.363644931698218 
## [19] 0.179684827337041  0.0284090070053935
## 5 Levels: 0.0248644019011408 0.0284090070053935 ... 0.363644931698218

##  [1] 1 1 3 2 5 5 3 4 4 1 4 2 3 1 3 5 4 5 3 2

##  [1] 1 1 3 2 5 5 3 4 4 1 4 2 3 1 3 5 4 5 3 2

I have to resort to paste to get the real values:

##  [1] 0.02486440 0.02486440 0.17968483 0.02840901 0.36364493 0.36364493
##  [7] 0.17968483 0.24970435 0.24970435 0.02486440 0.24970435 0.02840901
## [13] 0.17968483 0.02486440 0.17968483 0.36364493 0.24970435 0.36364493
## [19] 0.17968483 0.02840901

Is there a better way to convert a factor to numeric?

The levels of a factor are stored as character data type anyway (attributes(f)), so I don't think there is anything wrong with as.numeric(paste(f)). Perhaps it would be better to think why (in the specific context) you are getting a factor in the first place, and try to stop that. E.g., is the dec argument in read.table set correctly? – Bazz

5 Answers 11

up vote 499 down vote accepted

See the Warning section of ?factor:

In particular, as.numeric applied to a factor is meaningless, and may happen by implicit coercion. To transform a factor f to approximately its original numeric values, as.numeric(levels(f))[f] is recommended and slightly more efficient than as.numeric(as.character(f)).

The FAQ on R has similar advice.

Why is as.numeric(levels(f))[f] more efficent than as.numeric(as.character(f))?

as.numeric(as.character(f)) is effectively as.numeric(levels(f)[f]), so you are performing the conversion to numeric on length(x) values, rather than on nlevels(x) values. The speed difference will be most apparent for long vectors with few levels. If the values are mostly unique, there won't be much difference in speed. However you do the conversion, this operation is unlikely to be the bottleneck in your code, so don't worry too much about it.

Some timings

  times = 1e5
## Unit: microseconds
##                         expr   min    lq      mean median     uq      max neval
##     as.numeric(levels(f))[f] 3.982 5.120  6.088624  5.405  5.974 1981.418 1e+05
##     as.numeric(levels(f)[f]) 5.973 7.111  8.352032  7.396  8.250 4256.380 1e+05
##  as.numeric(as.character(f)) 6.827 8.249  9.628264  8.534  9.671 1983.694 1e+05
##                    paste0(x) 7.964 9.387 11.026351  9.956 10.810 2911.257 1e+05
##                     paste(x) 7.965 9.387 11.127308  9.956 11.093 2419.458 1e+05
1 upvote
1 upvote
Many thanks for your solution. Can I ask why the as.numeric(levels(f))[f] is more precise and faster? Thanks. – Sam
6 upvote
@Sam as.character(f) requires a "primitive lookup" to find the function as.character.factor(), which is defined as as.numeric(levels(f))[f]. – Jonathan
4 upvote
when apply as.numeric(levels(f))[f] OR as.numeric(as.character(f)), I have an warning msg: Warning message:NAs introduced by coercion. Do you know where the problem could be? thank you ! – maycca
@maycca did you overcame this issue? – user08041991
@user08041991, no sorry, I don't think so... – maycca

R has a number of (undocumented) convenience functions for converting factors:

  • as.character.factor
  • as.data.frame.factor
  • as.Date.factor
  • as.list.factor
  • as.vector.factor
  • ...

But annoyingly, there is nothing to handle the factor -> numeric conversion. As an extension of Joshua Ulrich's answer, I would suggest to overcome this omission with the definition of your own idiomatic function:

as.numeric.factor <- function(x) {as.numeric(levels(x))[x]}

that you can store at the beginning of your script, or even better in your .Rprofile file.

9 upvote
There's nothing to handle the factor-to-integer (or numeric) conversion because it's expected that as.integer(factor) returns the underlying integer codes (as shown in the examples section of ?factor). It's probably okay to define this function in your global environment, but you might cause problems if you actually register it as an S3 method. – Joshua Ulrich
That's a good point and I agree: a complete redefinition of the factor->numeric conversion is likely to mess a lot of things. I found myself writing the cumbersome factor->numeric conversion a lot before realizing that it is in fact a shortcoming of R: some convenience function should be available... Calling it as.numeric.factor makes sense to me, but YMMV. – Jealie
2 upvote
If you find yourself doing that a lot, then you should do something upstream to avoid it all-together. – Joshua Ulrich
as.numeric.factor returns NA? – jO.
@jO.: in the cases where you used something like v=NA;as.numeric.factor(v) or v='something';as.numeric.factor(v), then it should, otherwise you have a weird thing going on somewhere. – Jealie

It is possible only in the case when the factor labels match the original values. I will explain it with an example.

Assume the data is vector x:

x <- c(20, 10, 30, 20, 10, 40, 10, 40)

Now I will create a factor with four labels:

f <- factor(x, levels = c(10, 20, 30, 40), labels = c("A", "B", "C", "D"))

1) x is with type double, f is with type integer. This is the first unavoidable loss of information. Factors are always stored as integers.

> typeof(x)
[1] "double"
> typeof(f)
[1] "integer"

2) It is not possible to revert back to the original values (10, 20, 30, 40) having only f available. We can see that f holds only integer values 1, 2, 3, 4 and two attributes - the list of labels ("A", "B", "C", "D") and the class attribute "factor". Nothing more.

> str(f)
 Factor w/ 4 levels "A","B","C","D": 2 1 3 2 1 4 1 4
> attributes(f)
[1] "A" "B" "C" "D"

[1] "factor"

To revert back to the original values we have to know the values of levels used in creating the factor. In this case c(10, 20, 30, 40). If we know the original levels (in correct order), we can revert back to the original values.

> orig_levels <- c(10, 20, 30, 40)
> x1 <- orig_levels[f]
> all.equal(x, x1)
[1] TRUE

And this will work only in case when labels have been defined for all possible values in the original data.

So if you will need the original values, you have to keep them. Otherwise there is a high chance it will not be possible to get back to them only from a factor.

The most easiest way would be to use unfactor function from package varhandle


This example can be a quick start:

x <- rep(c("a", "b", "c"), 20)
y <- rep(c(1, 1, 0), 20)

class(x)  # -> "character"
class(y)  # -> "numeric"

x <- factor(x)
y <- factor(y)

class(x)  # -> "factor"
class(y)  # -> "factor"

x <- unfactor(x)
y <- unfactor(y)

class(x)  # -> "character"
class(y)  # -> "numeric"
The unfactor function converts to character data type first and then converts back to numeric. Type unfactor at the console and you can see it in the middle of the function. Therefore it doesn't really give a better solution than what the asker already had. – Bazz
Having said that, the levels of a factor are of character type anyway, so nothing is lost by this approach. – Bazz
The unfactor function takes care of things that cannot be converted to numeric. Check the examples in help("unfactor") – Mehrad Mahmoudian
Error: could not find function "unfactor" – Selrac
1 upvote
@Selrac I've mentioned that this function is available in varhandle package, meaning you should load the package (library("varhandle")) first (as I mentioned in the first line of my answer!!) – Mehrad Mahmoudian
You are right. Sorry I miss this. – Selrac
I appreciate that your package probably has some other nice functions too, but installing a new package (and adding an external dependency to your code) isn't as nice or easy as typing as.character(as.numeric()). – Gregor
@Gregor adding a light dependency does not harm usually and of course if you are looking for the most efficient way, writing the code your self might perform faster. but as you can also see in your comment this is not trivial since you also put the as.numeric() and as.character() in a wrong order ;) What your code chunk does is to turn the factor's level index into a character matrix, so what you will have at the and is a character vector that contains some numbers that has been once assigned to certain level of your factor. Functions in that package are there to prevent these confusions – Mehrad Mahmoudian

Every answer in this post failed to generate results for me , NAs were getting generated.

[1] NA NA NA NA NA Warning message: NAs introduced by coercion

What worked for me is this -

unclass(y2) %>% as.numeric 
  [1] 1 2 3 4 1
Are you sure you had a factor? Look at this example.y<-factor(c("5","15","20","2")); unclass(y) %>% as.numeric This returns 4,1,3,2, not 5,15,20,2. This seems like incorrect information. – MrFlick
Ok, this is similar to what I was trying to do today :- y2<-factor(c("A","B","C","D","A")); as.numeric(levels(y2))[y2] [1] NA NA NA NA NA Warning message: NAs introduced by coercion whereas unclass(y2) %>% as.numeric gave me the results that I needed. – Indi
Let me update my scenario in the answer that I had provided – Indi
OK, well that's not the question that was asked above. In this question the factor levels are all "numeric". In your case , as.numeric(y) should have worked just fine, no need for the unclass(). But again, that's not what this question was about. This answer isn't appropriate here. – MrFlick
3 upvote
Well, I really hope it helps someone who was in a hurry like me and read just the title ! – Indi
@jogo %>% is from the magrittr package. – Phil

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