647

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

as.numeric(f)
##  [1] 1 1 3 2 5 5 3 4 4 1 4 2 3 1 3 5 4 5 3 2

as.integer(f)
##  [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:

as.numeric(paste(f))
##  [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?

Jaap
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Adam SO
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    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? – CJB Jan 25 '16 at 09:44
  • If you use a dataframe you can use convert from hablar. `df %>% convert(num(column))`. Or if you have a factor vector you can use `as_reliable_num(factor_vector)` – davsjob Nov 01 '18 at 09:53

11 Answers11

780

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

library(microbenchmark)
microbenchmark(
  as.numeric(levels(f))[f],
  as.numeric(levels(f)[f]),
  as.numeric(as.character(f)),
  paste0(x),
  paste(x),
  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
Jaap
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Joshua Ulrich
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    For timings see this answer: http://stackoverflow.com/questions/6979625/arithmetic-operations-on-r-factors/6980780#6980780 – Ari B. Friedman Aug 08 '11 at 11:27
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    Many thanks for your solution. Can I ask why the as.numeric(levels(f))[f] is more precise and faster? Thanks. – Sam Apr 18 '14 at 00:25
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    @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 Jun 27 '14 at 19:12
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    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 Apr 13 '16 at 21:23
  • @maycca did you overcame this issue? – user08041991 Jan 31 '17 at 12:25
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    @user08041991 I have the same issue as maycca. I suspect this is from gradual changes in R over time (this answer was posted in 2010), and this answer is now outdated – MBorg Dec 13 '20 at 18:27
98

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.

MrLore
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Jealie
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    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 Apr 18 '14 at 12:03
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    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 Apr 18 '14 at 20:11
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    If you find yourself doing that **a lot**, then you should do something upstream to avoid it all-together. – Joshua Ulrich Apr 18 '14 at 22:44
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    as.numeric.factor returns NA? – jO. Aug 08 '14 at 07:56
  • @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 Aug 08 '14 at 14:43
  • `as.numeric(as.character.factor(x))` just did the trick for me – qwerty Feb 25 '21 at 23:44
36

The most easiest way would be to use unfactor function from package varhandle which can accept a factor vector or even a dataframe:

unfactor(your_factor_variable)

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"

library(varhandle)
x <- unfactor(x)
y <- unfactor(y)

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

You can also use it on a dataframe. For example the iris dataset:

sapply(iris, class)

Sepal.Length  Sepal.Width Petal.Length  Petal.Width      Species
   "numeric"    "numeric"    "numeric"    "numeric"     "factor"

# load the package
library("varhandle")
# pass the iris to unfactor
tmp_iris <- unfactor(iris)
# check the classes of the columns
sapply(tmp_iris, class)

Sepal.Length  Sepal.Width Petal.Length  Petal.Width      Species
   "numeric"    "numeric"    "numeric"    "numeric"  "character"

# check if the last column is correctly converted
tmp_iris$Species

  [1] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"    
  [6] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"    
 [11] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"    
 [16] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"    
 [21] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"    
 [26] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"    
 [31] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"
 [36] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"
 [41] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"
 [46] "setosa"     "setosa"     "setosa"     "setosa"     "setosa"
 [51] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [56] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [61] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [66] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [71] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [76] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [81] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [86] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [91] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
 [96] "versicolor" "versicolor" "versicolor" "versicolor" "versicolor"
[101] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[106] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[111] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[116] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[121] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[126] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[131] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[136] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[141] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
[146] "virginica"  "virginica"  "virginica"  "virginica"  "virginica"
Mehrad Mahmoudian
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  • 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. – CJB Jan 25 '16 at 09:32
  • Having said that, the levels of a factor are of character type anyway, so nothing is lost by this approach. – CJB Jan 25 '16 at 09:38
  • The `unfactor` function takes care of things that cannot be converted to numeric. Check the examples in `help("unfactor")` – Mehrad Mahmoudian Jul 25 '16 at 13:15
  • Error: could not find function "unfactor" – Selrac Sep 28 '16 at 16:35
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    @Selrac I've mentioned that this function is available in [varhandle](http://cran.r-project.org/web/packages/varhandle/index.html) package, meaning you should load the package (`library("varhandle")`) first (as I mentioned in the first line of my answer!!) – Mehrad Mahmoudian Sep 29 '16 at 13:06
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    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 Thomas Nov 08 '16 at 20:03
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    @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 Nov 10 '16 at 11:53
27

Note: this particular answer is not for converting numeric-valued factors to numerics, it is for converting categorical factors to their corresponding level numbers.

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

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

What worked for me is this - 

as.integer(y2)
# [1] 1 2 3 4 1
Axeman
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Indi
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  • Are you sure you had a factor? Look at this example.`y% as.numeric` This returns 4,1,3,2, not 5,15,20,2. This seems like incorrect information. – MrFlick Feb 22 '17 at 19:19
  • Ok, this is similar to what I was trying to do today :- y2% as.numeric gave me the results that I needed. – Indi Feb 22 '17 at 19:34
  • Let me update my scenario in the answer that I had provided – Indi Feb 22 '17 at 19:36
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    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 Feb 22 '17 at 19:37
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    Well, I really hope it helps someone who was in a hurry like me and read just the title ! – Indi Feb 22 '17 at 19:45
  • @jogo `%>%` is from the `magrittr` package. – Phil May 09 '17 at 16:39
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    If you have characters representing the integers as factors, this is the one I would recommend. this is the only one that worked for me. – aimme Dec 12 '19 at 16:14
9

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)
$levels
[1] "A" "B" "C" "D"

$class
[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.

djhurio
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3

You can use hablar::convert if you have a data frame. The syntax is easy:

Sample df

library(hablar)
library(dplyr)

df <- dplyr::tibble(a = as.factor(c("7", "3")),
                    b = as.factor(c("1.5", "6.3")))

Solution

df %>% 
  convert(num(a, b))

gives you:

# A tibble: 2 x 2
      a     b
  <dbl> <dbl>
1    7.  1.50
2    3.  6.30

Or if you want one column to be integer and one numeric:

df %>% 
  convert(int(a),
          num(b))

results in:

# A tibble: 2 x 2
      a     b
  <int> <dbl>
1     7  1.50
2     3  6.30
davsjob
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2

late to the game, accidently, I found trimws() can convert factor(3:5) to c("3","4","5"). Then you can call as.numeric(). That is:

as.numeric(trimws(x_factor_var))
Jerry T
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    Is there a reason you would recommend using `trimws` over `as.character` as described in the accepted answer? It seems to me like unless you actually had whitespace you needed to remove, `trimws` is just going to do a bunch of unnecessary regular expression work to return the same result. – MrFlick Nov 13 '18 at 18:54
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    as.numeric(levels(f))[f] is might be a bit confusing and hard to remember for beginners. trimws does no harm. – Jerry T Feb 22 '19 at 18:54
2

type.convert(f) on a factor whose levels are completely numeric is another base option.

Performance-wise it's about equivalent to as.numeric(as.character(f)) but not nearly as quick as as.numeric(levels(f))[f].

identical(type.convert(f), as.numeric(levels(f))[f])

[1] TRUE

That said, if the reason the vector was created as a factor in the first instance has not been addressed (i.e. it likely contained some characters that could not be coerced to numeric) then this approach won't work and it will return a factor.

levels(f)[1] <- "some character level"
identical(type.convert(f), as.numeric(levels(f))[f])

[1] FALSE
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0

strtoi() works if your factor levels are integers.

Robert Bray
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-1

From the many answers I could read, the only given way was to expand the number of variables according to the number of factors. If you have a variable "pet" with levels "dog" and "cat", you would end up with pet_dog and pet_cat.

In my case I wanted to stay with the same number of variables, by just translating the factor variable to a numeric one, in a way that can applied to many variables with many levels, so that cat=1 and dog=0 for instance.

Please find the corresponding solution below:

crime <- data.frame(city = c("SF", "SF", "NYC"),
                    year = c(1990, 2000, 1990),
                    crime = 1:3)

indx <- sapply(crime, is.factor)

crime[indx] <- lapply(crime[indx], function(x){ 
  listOri <- unique(x)
  listMod <- seq_along(listOri)
  res <- factor(x, levels=listOri)
  res <- as.numeric(res)
  return(res)
}
)
Xavier Prudent
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-1

Looks like the solution as.numeric(levels(f))[f] no longer work with R 4.0.

Alternative solution:

factor2number <- function(x){
    data.frame(levels(x), 1:length(levels(x)), row.names = 1)[x, 1]
}

factor2number(yourFactor)
Life_Searching_Steps
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