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

Question

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?

Answer 1

The collapse package includes a wrapper around as.numeric(levels(f))[f] and as.character(levels(f))[f] in as_numeric_factor and as_character_factor.

library(collapse)
set.seed(1)
f <- factor(sample(runif(5), 5, replace = TRUE))

as_numeric_factor(f)
# [1] 0.2016819 0.5728534 0.3721239 0.5728534 0.5728534

as_character_factor(f)
# [1] "0.201681931037456" "0.572853363351896" "0.37212389963679" "0.572853363351896" "0.572853363351896"

It gives similar performances compared to as.numeric(levels(f))[f].

# Unit: milliseconds
#                      expr      min        lq       mean    median        uq      max neval
#  as.numeric(levels(f))[f]   2.6026   3.01305   5.834900   3.54310   8.57450  66.3497   100
#  as.numeric(levels(f)[f]) 317.2509 336.78690 350.215388 349.85620 361.57980 401.1002   100
#      as_numeric_factor(f)   2.5793   2.92970   5.383223   3.23355   4.29355  68.4460   100

Code:

set.seed(1)
f <- factor(sample(runif(5), 1e6, replace = TRUE))
library(microbenchmark)
microbenchmark(
  as.numeric(levels(f))[f],
  as.numeric(levels(f)[f]),
  as_numeric_factor(f),
  times = 100
)

Answer 2

I found as.numeric(levels(f))[f] difficult to apply across a list of column names using tidyverse syntax. Converting to a character first then an integer gave me the original numeric values without having to add additional packages. Perhaps not the most performant/elegant solution but kept things simple and readable.

library(tidyverse)

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

cols <- c("a", "b")

tbl_df %>%
  mutate(across(all_of(cols), as.character)) %>% 
  mutate(across(all_of(cols), as.numeric))

Answer 3

If you have many factor columns to convert to numeric,

df <- rapply(df, function(x) as.numeric(levels(x))[x], "factor", how =  "replace")

This solution is robust for data.frames containing mixed types, provided all factor levels are numbers.

Answer 4

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.

Answer 5

strtoi() works if your factor levels are integers.

Answer 6

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"

Answer 7

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

Answer 8

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)

Answer 9

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)
}
)

Answer 10

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

Answer 11

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))

Answer 12

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

Answer 13

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

Answer 14

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.