What does the capital letter "I" in R linear regression formula mean?

Nancy picture Nancy · Jun 12, 2014 · Viewed 17.2k times · Source

I haven't been able to find an answer to this question, largely because googling anything with a standalone letter (like "I") causes issues.

What does the "I" do in a model like this?

data(rock)
lm(area~I(peri - mean(peri)), data = rock)

Considering that the following does NOT work:

lm(area ~ (peri - mean(peri)), data = rock)

and that this does work:

rock$peri - mean(rock$peri)

Any key words on how to research this myself would also be very helpful.

Answer

Gavin Simpson picture Gavin Simpson · Jun 12, 2014

I isolates or insulates the contents of I( ... ) from the gaze of R's formula parsing code. It allows the standard R operators to work as they would if you used them outside of a formula, rather than being treated as special formula operators.

For example:

y ~ x + x^2

would, to R, mean "give me:

  1. x = the main effect of x, and
  2. x^2 = the main effect and the second order interaction of x",

not the intended x plus x-squared:

> model.frame( y ~ x + x^2, data = data.frame(x = rnorm(5), y = rnorm(5)))
           y           x
1 -1.4355144 -1.85374045
2  0.3620872 -0.07794607
3 -1.7590868  0.96856634
4 -0.3245440  0.18492596
5 -0.6515630 -1.37994358

This is because ^ is a special operator in a formula, as described in ?formula. You end up only including x in the model frame because the main effect of x is already included from the x term in the formula, and there is nothing to cross x with to get the second-order interactions in the x^2 term.

To get the usual operator, you need to use I() to isolate the call from the formula code:

> model.frame( y ~ x + I(x^2), data = data.frame(x = rnorm(5), y = rnorm(5)))
            y          x       I(x^2)
1 -0.02881534  1.0865514 1.180593....
2  0.23252515 -0.7625449 0.581474....
3 -0.30120868 -0.8286625 0.686681....
4 -0.67761458  0.8344739 0.696346....
5  0.65522764 -0.9676520 0.936350....

(that last column is correct, it just looks odd because it is of class AsIs.)

In your example, - when used in a formula would indicate removal of a term from the model, where you wanted - to have it's usual binary operator meaning of subtraction:

> model.frame( y ~ x - mean(x), data = data.frame(x = rnorm(5), y = rnorm(5)))
Error in model.frame.default(y ~ x - mean(x), data = data.frame(x = rnorm(5),  : 
  variable lengths differ (found for 'mean(x)')

This fails for reason that mean(x) is a length 1 vector and model.frame() quite rightly tells you this doesn't match the length of the other variables. A way round this is I():

> model.frame( y ~ I(x - mean(x)), data = data.frame(x = rnorm(5), y = rnorm(5)))
           y I(x - mean(x))
1  1.1727063   1.142200....
2 -1.4798270   -0.66914....
3 -0.4303878   -0.28716....
4 -1.0516386   0.542774....
5  1.5225863   -0.72865....

Hence, where you want to use an operator that has special meaning in a formula, but you need its non-formula meaning, you need to wrap the elements of the operation in I( ).

Read ?formula for more on the special operators, and ?I for more details on the function itself and its other main use-case within data frames (which is where the AsIs bit originates from, if you are interested).