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All Levels of a Factor in a Model Matrix in R

I have a data.frame consisting of numeric and factor variables as seen below.

testFrame <- data.frame(First=sample(1:10, 20, replace=T),
           Second=sample(1:20, 20, replace=T), Third=sample(1:10, 20, replace=T),
           Fourth=rep(c("Alice","Bob","Charlie","David"), 5),
           Fifth=rep(c("Edward","Frank","Georgia","Hank","Isaac"),4))

I want to build out a matrix that assigns dummy variables to the factor and leaves the numeric variables alone.

model.matrix(~ First + Second + Third + Fourth + Fifth, data=testFrame)

As expected when running lm this leaves out one level of each factor as the reference level. However, I want to build out a matrix with a dummy/indicator variable for every level of all the factors. I am building this matrix for glmnet so I am not worried about multicollinearity.

Is there a way to have model.matrix create the dummy for every level of the factor?

like image 476
Jared Avatar asked Dec 30 '10 06:12

Jared


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10 Answers

(Trying to redeem myself...) In response to Jared's comment on @Fabians answer about automating it, note that all you need to supply is a named list of contrast matrices. contrasts() takes a vector/factor and produces the contrasts matrix from it. For this then we can use lapply() to run contrasts() on each factor in our data set, e.g. for the testFrame example provided:

> lapply(testFrame[,4:5], contrasts, contrasts = FALSE)
$Fourth
        Alice Bob Charlie David
Alice       1   0       0     0
Bob         0   1       0     0
Charlie     0   0       1     0
David       0   0       0     1

$Fifth
        Edward Frank Georgia Hank Isaac
Edward       1     0       0    0     0
Frank        0     1       0    0     0
Georgia      0     0       1    0     0
Hank         0     0       0    1     0
Isaac        0     0       0    0     1

Which slots nicely into @fabians answer:

model.matrix(~ ., data=testFrame, 
             contrasts.arg = lapply(testFrame[,4:5], contrasts, contrasts=FALSE))
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Gavin Simpson Avatar answered Oct 14 '22 22:10

Gavin Simpson


You need to reset the contrasts for the factor variables:

model.matrix(~ Fourth + Fifth, data=testFrame, 
        contrasts.arg=list(Fourth=contrasts(testFrame$Fourth, contrasts=F), 
                Fifth=contrasts(testFrame$Fifth, contrasts=F)))

or, with a little less typing and without the proper names:

model.matrix(~ Fourth + Fifth, data=testFrame, 
    contrasts.arg=list(Fourth=diag(nlevels(testFrame$Fourth)), 
            Fifth=diag(nlevels(testFrame$Fifth))))
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fabians Avatar answered Oct 14 '22 23:10

fabians


caret implemented a nice function dummyVars to achieve this with 2 lines:

library(caret) dmy <- dummyVars(" ~ .", data = testFrame) testFrame2 <- data.frame(predict(dmy, newdata = testFrame))

Checking the final columns:

colnames(testFrame2)

"First"  "Second"         "Third"          "Fourth.Alice"   "Fourth.Bob"     "Fourth.Charlie" "Fourth.David"   "Fifth.Edward"   "Fifth.Frank"   "Fifth.Georgia"  "Fifth.Hank"     "Fifth.Isaac"   

The nicest point here is you get the original data frame, plus the dummy variables having excluded the original ones used for the transformation.

More info: http://amunategui.github.io/dummyVar-Walkthrough/

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Pablo Casas Avatar answered Oct 14 '22 23:10

Pablo Casas


dummyVars from caret could also be used. http://caret.r-forge.r-project.org/preprocess.html

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Sagar Jauhari Avatar answered Oct 14 '22 22:10

Sagar Jauhari


Ok. Just reading the above and putting it all together. Suppose you wanted the matrix e.g. 'X.factors' that multiplies by your coefficient vector to get your linear predictor. There are still a couple extra steps:

X.factors = 
  model.matrix( ~ ., data=X, contrasts.arg = 
    lapply(data.frame(X[,sapply(data.frame(X), is.factor)]),
                                             contrasts, contrasts = FALSE))

(Note that you need to turn X[*] back into a data frame in case you have only one factor column.)

Then say you get something like this:

attr(X.factors,"assign")
[1]  0  1  **2**  2  **3**  3  3  **4**  4  4  5  6  7  8  9 10 #emphasis added

We want to get rid of the **'d reference levels of each factor

att = attr(X.factors,"assign")
factor.columns = unique(att[duplicated(att)])
unwanted.columns = match(factor.columns,att)
X.factors = X.factors[,-unwanted.columns]
X.factors = (data.matrix(X.factors))
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user36302 Avatar answered Oct 14 '22 23:10

user36302


A tidyverse answer:

library(dplyr)
library(tidyr)
result <- testFrame %>% 
    mutate(one = 1) %>% spread(Fourth, one, fill = 0, sep = "") %>% 
    mutate(one = 1) %>% spread(Fifth, one, fill = 0, sep = "")

yields the desired result (same as @Gavin Simpson's answer):

> head(result, 6)
  First Second Third FourthAlice FourthBob FourthCharlie FourthDavid FifthEdward FifthFrank FifthGeorgia FifthHank FifthIsaac
1     1      5     4           0         0             1           0           0          1            0         0          0
2     1     14    10           0         0             0           1           0          0            1         0          0
3     2      2     9           0         1             0           0           1          0            0         0          0
4     2      5     4           0         0             0           1           0          1            0         0          0
5     2     13     5           0         0             1           0           1          0            0         0          0
6     2     15     7           1         0             0           0           1          0            0         0          0
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shosaco Avatar answered Oct 14 '22 22:10

shosaco


Using the R package 'CatEncoders'

library(CatEncoders)
testFrame <- data.frame(First=sample(1:10, 20, replace=T),
           Second=sample(1:20, 20, replace=T), Third=sample(1:10, 20, replace=T),
           Fourth=rep(c("Alice","Bob","Charlie","David"), 5),
           Fifth=rep(c("Edward","Frank","Georgia","Hank","Isaac"),4))

fit <- OneHotEncoder.fit(testFrame)

z <- transform(fit,testFrame,sparse=TRUE) # give the sparse output
z <- transform(fit,testFrame,sparse=FALSE) # give the dense output
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asdf123 Avatar answered Oct 14 '22 22:10

asdf123


I am currently learning Lasso model and glmnet::cv.glmnet(), model.matrix() and Matrix::sparse.model.matrix()(for high dimensions matrix, using model.matrix will killing our time as suggested by the author of glmnet.).

Just sharing there has a tidy coding to get the same answer as @fabians and @Gavin's answer. Meanwhile, @asdf123 introduced another package library('CatEncoders') as well.

> require('useful')
> # always use all levels
> build.x(First ~ Second + Fourth + Fifth, data = testFrame, contrasts = FALSE)
> 
> # just use all levels for Fourth
> build.x(First ~ Second + Fourth + Fifth, data = testFrame, contrasts = c(Fourth = FALSE, Fifth = TRUE))

Source : R for Everyone: Advanced Analytics and Graphics (page273)

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Rγσ ξηg Lιαη Ημ Avatar answered Oct 15 '22 00:10

Rγσ ξηg Lιαη Ημ


I write a package called ModelMatrixModel to improve the functionality of model.matrix(). The ModelMatrixModel() function in the package in default return a class containing a sparse matrix with all levels of dummy variables which is suitable for input in cv.glmnet() in glmnet package. Importantly, returned class also stores transforming parameters such as the factor level information, which can then be applied to new data. The function can hand most items in r formula like poly() and interaction. It also gives several other options like handle invalid factor levels , and scale output.

#devtools::install_github("xinyongtian/R_ModelMatrixModel")
library(ModelMatrixModel)
testFrame <- data.frame(First=sample(1:10, 20, replace=T),
                        Second=sample(1:20, 20, replace=T), Third=sample(1:10, 20, replace=T),
                        Fourth=rep(c("Alice","Bob","Charlie","David"), 5))
newdata=data.frame(First=sample(1:10, 2, replace=T),
                   Second=sample(1:20, 2, replace=T), Third=sample(1:10, 2, replace=T),
                   Fourth=c("Bob","Charlie"))
mm=ModelMatrixModel(~First+Second+Fourth, data = testFrame)
class(mm)
## [1] "ModelMatrixModel"
class(mm$x) #default output is sparse matrix
## [1] "dgCMatrix"
## attr(,"package")
## [1] "Matrix"
data.frame(as.matrix(head(mm$x,2)))
##   First Second FourthAlice FourthBob FourthCharlie FourthDavid
## 1     7     17           1         0             0           0
## 2     9      7           0         1             0           0

#apply the same transformation to new data, note the dummy variables for 'Fourth' includes the levels not appearing in new data     
mm_new=predict(mm,newdata)
data.frame(as.matrix(head(mm_new$x,2))) 
##   First Second FourthAlice FourthBob FourthCharlie FourthDavid
## 1     6      3           0         1             0           0
## 2     2     12           0         0             1           0
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Ben2018 Avatar answered Oct 14 '22 23:10

Ben2018


You can use tidyverse to achieve this without specifying each column manually.

The trick is to make a "long" dataframe.

Then, munge a few things, and spread it back to wide to create the indicators/dummy variables.

Code:

library(tidyverse)

## add index variable for pivoting
testFrame$id <- 1:nrow(testFrame)

testFrame %>%
    ## pivot to "long" format
    gather(feature, value, -id) %>%
    ## add indicator value
    mutate(indicator=1) %>%
    ## create feature name that unites a feature and its value
    unite(feature, value, col="feature_value", sep="_") %>%
    ## convert to wide format, filling missing values with zero
    spread(feature_value, indicator, fill=0)

The output:

   id Fifth_Edward Fifth_Frank Fifth_Georgia Fifth_Hank Fifth_Isaac First_2 First_3 First_4 ...
1   1            1           0             0          0           0       0       0       0
2   2            0           1             0          0           0       0       0       0
3   3            0           0             1          0           0       0       0       0
4   4            0           0             0          1           0       0       0       0
5   5            0           0             0          0           1       0       0       0
6   6            1           0             0          0           0       0       0       0
7   7            0           1             0          0           0       0       1       0
8   8            0           0             1          0           0       1       0       0
9   9            0           0             0          1           0       0       0       0
10 10            0           0             0          0           1       0       0       0
11 11            1           0             0          0           0       0       0       0
12 12            0           1             0          0           0       0       0       0
...
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Paul Avatar answered Oct 14 '22 23:10

Paul