How to use the predict function with glm in R (With Examples)

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The glm() function in R can be used to fit generalized linear models. This function is particularly useful for fitting , , and other complex models.

Once we’ve fit a model, we can then use the predict() function to predict the response value of a new observation.

This function uses the following syntax:

predict(object, newdata, type=”response”)

where:

• object: The name of the model fit using the glm() function
• newdata: The name of the new data frame to make predictions for
• type: The type of prediction to make.

The following example shows how to fit a generalized linear model in R and how to then use the model to predict the response value of a new observation it hasn’t seen before.

Example: Using the predict function with glm in R

For this example, we’ll use the built-in R dataset called mtcars:

#view first six rows of mtcars data frame
head(mtcars)

                   mpg cyl disp  hp drat    wt  qsec vs am gear carb
Mazda RX4         21.0   6  160 110 3.90 2.620 16.46  0  1    4    4
Mazda RX4 Wag     21.0   6  160 110 3.90 2.875 17.02  0  1    4    4
Datsun 710        22.8   4  108  93 3.85 2.320 18.61  1  1    4    1
Hornet 4 Drive    21.4   6  258 110 3.08 3.215 19.44  1  0    3    1
Hornet Sportabout 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2
Valiant           18.1   6  225 105 2.76 3.460 20.22  1  0    3    1

We’ll fit the following logistic regression model in which we use the variables disp and hp to predict the response variable am (the transmission type of the car: 0 = automatic, 1 = manual).

#fit logistic regression model
model <- glm(am ~ disp + hp, data=mtcars, family=binomial)

#view model summary
summary(model)

Call:
glm(formula = am ~ disp + hp, family = binomial, data = mtcars)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-1.9665  -0.3090  -0.0017   0.3934   1.3682  

Coefficients:
            Estimate Std. Error z value Pr(>|z|)  
(Intercept)  1.40342    1.36757   1.026   0.3048  
disp        -0.09518    0.04800  -1.983   0.0474 *
hp           0.12170    0.06777   1.796   0.0725 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 43.230  on 31  degrees of freedom
Residual deviance: 16.713  on 29  degrees of freedom
AIC: 22.713

Number of Fisher Scoring iterations: 8

We can then use this model to predict the probability that a new car has an automatic transmission (am=0) or a manual transmission (am=1) by using the following code:

#define new observation
newdata = data.frame(disp=200, hp= 100)

#use model to predict value of am
predict(model, newdata, type="response")

         1 
0.00422564

The model predicts the probability of the new car having a manual transmission (am=1) to be 0.004. This means it’s highly like that this new car has an automatic transmission.

Note that we can also make several predictions at once if we have a data frame that has multiple new cars.

For example, the following code shows how to use the fitted model to predict the probability of a manual transmission for three new cars:

#define new data frame of three cars
newdata = data.frame(disp=c(200, 180, 160),
                     hp=c(100, 90, 108))

#view data frame
newdata

  disp  hp
1  200 100
2  180  90
3  160 108

#use model to predict value of am for all three cars
predict(model, newdata, type="response")

          1           2           3 
0.004225640 0.008361069 0.335916069 

Here’s how to interpret the output:

• The probability that car 1 has a manual transmission is .004.
• The probability that car 2 has a manual transmission is .008.
• The probability that car 3 has a manual transmission is .336.

Notes

The names of the columns in the new data frame should exactly match the names of the columns in the data frame that were used to build the model.

Notice that in our previous example, the data frame we used to build the model contained the following column names for our predictor variables:

• disp
• hp

Thus, when we created the new data frame called newdata we made sure to also name the columns:

• disp
• hp

If the names of the columns do not match, you’ll receive the following error message:

Error in eval(predvars, data, env) 

Keep this in mind when using the predict() function.

The following tutorials explain how to perform other common tasks in R: