问题
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?
回答1:
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))))
回答2:
(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))
回答3:
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/
回答4:
dummyVars from caret could also be used. http://caret.r-forge.r-project.org/preprocess.html
回答5:
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))
回答6:
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
回答7:
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)
回答8:
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
回答9:
model.matrix(~ First + Second + Third + Fourth + Fifth - 1, data=testFrame)
or
model.matrix(~ First + Second + Third + Fourth + Fifth + 0, data=testFrame)
should be the most straightforward
回答10:
A stats package answer:
new_tr <- model.matrix(~.+0,data = testFrame)
Adding +0 (or -1) to a model formula (e.g., in lm()) in R suppresses the intercept.
Please see
来源：https://stackoverflow.com/questions/4560459/all-levels-of-a-factor-in-a-model-matrix-in-r