Nsryjdtyk

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?

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

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.

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

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"