R Language Linear Models (Regression) Using the 'predict' function
Example
Once a model is built predict is the main function to test with new data. Our example will use the mtcars built-in dataset to regress miles per gallon against displacement:
my_mdl <- lm(mpg ~ disp, data=mtcars)
my_mdl
Call:
lm(formula = mpg ~ disp, data = mtcars)
Coefficients:
(Intercept) disp
29.59985 -0.04122
If I had a new data source with displacement I could see the estimated miles per gallon.
set.seed(1234)
newdata <- sample(mtcars$disp, 5)
newdata
[1] 258.0 71.1 75.7 145.0 400.0
newdf <- data.frame(disp=newdata)
predict(my_mdl, newdf)
1 2 3 4 5
18.96635 26.66946 26.47987 23.62366 13.11381
The most important part of the process is to create a new data frame with the same column names as the original data. In this case, the original data had a column labeled disp, I was sure to call the new data that same name.
Caution
Let's look at a few common pitfalls:
-
not using a data.frame in the new object:
predict(my_mdl, newdata) Error in eval(predvars, data, env) : numeric 'envir' arg not of length one -
not using same names in new data frame:
newdf2 <- data.frame(newdata) predict(my_mdl, newdf2) Error in eval(expr, envir, enclos) : object 'disp' not found
Accuracy
To check the accuracy of the prediction you will need the actual y values of the new data. In this example, newdf will need a column for 'mpg' and 'disp'.
newdf <- data.frame(mpg=mtcars$mpg[1:10], disp=mtcars$disp[1:10])
# mpg disp
# 1 21.0 160.0
# 2 21.0 160.0
# 3 22.8 108.0
# 4 21.4 258.0
# 5 18.7 360.0
# 6 18.1 225.0
# 7 14.3 360.0
# 8 24.4 146.7
# 9 22.8 140.8
# 10 19.2 167.6
p <- predict(my_mdl, newdf)
#root mean square error
sqrt(mean((p - newdf$mpg)^2, na.rm=TRUE))
[1] 2.325148
