How to force R to use a specified factor level as reference in a regression?

Question

How can I tell R to use a certain level as reference if I use binary explanatory variables in a regression?

It's just using some level by default.

lm(x ~ y + as.factor(b)) 

with b {0, 1, 2, 3, 4}. Let's say I want to use 3 instead of the zero that is used by R.

Maishamaisie

You should do the data processing step outside of the model formula/fitting. When creating the factor from b you can specify the ordering of the levels using factor(b, levels = c(3,1,2,4,5)). Do this in a data processing step outside the lm() call though. My answer below uses the relevel() function so you can create a factor and then shift the reference level around to suit as you need to. – Pandarus

I reworded your question. You're actually after changing the reference level, not leaving one out. – Inharmonious

thx for rewording my question. Indeed, relevel() was what I was looking for. Thx for the detailed answer and the example though. I am not sure if the linear-regression tag is a bit misleading because this applies to all kinds of regression using dummy explanatories... – Maishamaisie

Answer 1

See the relevel() function. Here is an example:

set.seed(123)
x <- rnorm(100)
DF <- data.frame(x = x,
                 y = 4 + (1.5*x) + rnorm(100, sd = 2),
                 b = gl(5, 20))
head(DF)
str(DF)

m1 <- lm(y ~ x + b, data = DF)
summary(m1)

Now alter the factor b in DF by use of the relevel() function:

DF <- within(DF, b <- relevel(b, ref = 3))
m2 <- lm(y ~ x + b, data = DF)
summary(m2)

The models have estimated different reference levels.

> coef(m1)
(Intercept)           x          b2          b3          b4          b5 
  3.2903239   1.4358520   0.6296896   0.3698343   1.0357633   0.4666219 
> coef(m2)
(Intercept)           x          b1          b2          b4          b5 
 3.66015826  1.43585196 -0.36983433  0.25985529  0.66592898  0.09678759

Pandarus

To preseve the original variable, just don't use the within, but df$bR = relevel(df$b, ref=3). – Bilek

You can use relevel() inside your formula, wouldn't affect the original dataset... – Wigan

Can one use this approach to plot all factor levels together in a coefficient plot? – Tomcat

Answer 2

I know this is an old question, but I had a similar issue and found that:

lm(x ~ y + relevel(b, ref = "3")) 

does exactly what you asked.

Polygon

This was a big help! Only solution that included a way to do it within the lm() command which was exactly what I needed. Thanks! – Kitchenware

This is a very flexible way of working with factors. I like the fact that I can combine it with as.factor() if needed, for instance by using ...+relevel(as.factor(mycol), ref = "myref")+... – Elemental

This is by far the best solution here! I love it. – Canice

Answer 3

Others have mentioned the relevel command which is the best solution if you want to change the base level for all analyses on your data (or are willing to live with changing the data).

If you don't want to change the data (this is a one time change, but in the future you want the default behavior again), then you can use a combination of the C (note uppercase) function to set contrasts and the contr.treatments function with the base argument for choosing which level you want to be the baseline.

For example:

lm( Sepal.Width ~ C(Species,contr.treatment(3, base=2)), data=iris )

Remanence

Answer 4

The relevel() command is a shorthand method to your question. What it does is reorder the factor so that whatever is the ref level is first. Therefore, reordering your factor levels will also have the same effect but gives you more control. Perhaps you wanted to have levels 3,4,0,1,2. In that case...

bFactor <- factor(b, levels = c(3,4,0,1,2))

I prefer this method because it's easier for me to see in my code not only what the reference was but the position of the other values as well (rather than having to look at the results for that).

NOTE: DO NOT make it an ordered factor. A factor with a specified order and an ordered factor are not the same thing. lm() may start to think you want polynomial contrasts if you do that.

Brahe

Polynomial contrasts, not a polynomial regression. – Stadium

Is there a way to set the reference level at the same time that you define the factor, rather than in a subsequent call to relevel? – Flitter

For some reason R is treating my variable as being ordered (it is just a bunch of strings), any idea of how to fix this? – Toffic

It's best to ask new questions separately. You can link to this one and say you are expending it but it's best not to hide new stuff in the comments. – Brahe

Answer 5

You can also manually tag the column with a contrasts attribute, which seems to be respected by the regression functions:

contrasts(df$factorcol) <- contr.treatment(levels(df$factorcol),
   base=which(levels(df$factorcol) == 'RefLevel'))

Melvinmelvina

Answer 6

For those looking for a dplyr/tidyverse version. Building on Gavin Simpson solution:

# Create DF
set.seed(123)
x <- rnorm(100)
DF <- data.frame(x = x,
                 y = 4 + (1.5*x) + rnorm(100, sd = 2),
                 b = gl(5, 20))

# Change reference level
DF = DF %>% mutate(b = relevel(b, 3))

m2 <- lm(y ~ x + b, data = DF)
summary(m2)

Derman

I'm confused why you put "If the variable is a factor" where you did... this is necessary whether you use relevel() or forcats::fct_relevel() – Hamlani

You are correct, thanks! I added "you can also use", because, afaik, fct_relevel only works with factors. – Derman

relevel only works with factors. fct_relevel only works with factors. There isn't any difference between the functions except the name, AFAIK. Saying "If the variable is a factor you can also use fct_relevel" implies that if the variable is not a factor you could use relevel, but that is not true. – Hamlani