How to convert a factor to integer\numeric without loss of information?
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"
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 factorf
to approximately its original numeric values,as.numeric(levels(f))[f]
is recommended and slightly more efficient thanas.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
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
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.