How to convert a factor to integer\numeric without loss of information?

r casting r-faq

1,081,311

Solution 1

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

Solution 2

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.double.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.

Solution 3

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

Solution 4

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"

Solution 5

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.

View more solutions

1,081,311

Author by

Adam SO

Experimental Psychology, vision, taste, smell and multisensory integration.

Updated on July 16, 2022

Comments

Adam SO almost 2 years

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?

CJB over 8 years

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?
davsjob over 5 years

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)
Denis Cousineau almost 2 years

Thank good for this question. This is SO MUCH frustrating to see numbers get transformed into other numbers pretty much randomly.

Ari B. Friedman over 12 years

For timings see this answer: stackoverflow.com/questions/6979625/…
Sam about 10 years

Many thanks for your solution. Can I ask why the as.numeric(levels(f))[f] is more precise and faster? Thanks.
Joshua Ulrich about 10 years

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.
Jealie about 10 years

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.
Joshua Ulrich about 10 years

If you find yourself doing that a lot, then you should do something upstream to avoid it all-together.
Jonathan almost 10 years

@Sam as.character(f) requires a "primitive lookup" to find the function as.character.factor(), which is defined as as.numeric(levels(f))[f].
jO. over 9 years

as.numeric.factor returns NA?
Jealie over 9 years

@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.
CJB over 8 years

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 over 8 years

Having said that, the levels of a factor are of character type anyway, so nothing is lost by this approach.
maycca about 8 years

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 !
Mehrad Mahmoudian almost 8 years

The unfactor function takes care of things that cannot be converted to numeric. Check the examples in help("unfactor")
Selrac over 7 years

Error: could not find function "unfactor"
Mehrad Mahmoudian over 7 years

@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!!)
Gregor Thomas over 7 years

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()).
Mehrad Mahmoudian over 7 years

@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
user08041991 over 7 years

@maycca did you overcame this issue?
MrFlick about 7 years

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.
Indi about 7 years

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 about 7 years

Let me update my scenario in the answer that I had provided
MrFlick about 7 years

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.
Indi about 7 years

Well, I really hope it helps someone who was in a hurry like me and read just the title !
Phil almost 7 years

@jogo %>% is from the magrittr package.
MrFlick over 5 years

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.
Jerry T about 5 years

as.numeric(levels(f))[f] is might be a bit confusing and hard to remember for beginners. trimws does no harm.
aimme over 4 years

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.
MBorg over 3 years

@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
qwerty about 3 years

as.numeric(as.character.factor(x)) just did the trick for me
Phil almost 3 years

Nice simple solution, as fast as other solutions too.
luchonacho over 2 years

This is the answer so many of us are after and the first hit in Google. I can't find a similar question.
Jealie over 2 years

@rui-barradas comment = as a historical anomaly, R has two types for floating point vectors: numeric and double. According to the documentation, it is better to write code for the double type, thus as.double.factor seems like a more proper name. Link to documentation: stat.ethz.ch/R-manual/R-devel/library/base/html/numeric.html . Thanks @rui-barradas !
Denis Cousineau almost 2 years

?? On R 4.1, it does work.
Denis Cousineau almost 2 years

However, loading another package just for that single operation is not parcimonious