Once we’ve fit a in R, we can use the predict() function to predict the response value of a new observation that the model has never seen before.

This function uses the following syntax:

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

where:

• object: The name of the logistic regression model
• 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 use this function in practice.

Example: Using predict() with a Logistic Regression Model in R

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

#view first six rows of mtcars dataset
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 create a new data frame that contains information about eight cars the model has never seen before and use the predict() function to predict the probability that a new car has an automatic transmission (am=0) or a manual transmission (am=1):

#define new data frame
newdata = data.frame(disp=c(200, 180, 160, 140, 120, 120, 100, 160),
                     hp=c(100, 90, 108, 90, 80, 90, 80, 90),
                     am=c(0, 0, 0, 1, 0, 1, 1, 1))

#view data frame
newdata

#use model to predict value of am for all new cars
newdata$am_prob <- predict(model, newdata, type="response")

#view updated data frame
newdata

  disp  hp am      am_prob
1  200 100  0  0.004225640
2  180  90  0  0.008361069
3  160 108  0  0.335916069
4  140  90  1  0.275162866
5  120  80  0  0.429961894
6  120  90  1  0.718090728
7  100  80  1  0.835013994
8  160  90  1  0.053546152

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.

And so on.

We can also use the table() function to create a confusion matrix that displays the actual am values vs. the predicted values by the model:

#create vector that contains 0 or 1 depending on predicted value of am
am_pred = rep(0, dim(newdata)[1])
am_pred[newdata$am_prob > .5] = 1

#create confusion matrix
table(am_pred, newdata$am)

am_pred 0 1
      0 4 2
      1 0 2

Lastly, we can use the mean() function to calculate the percentage of observations in the new data frame that the model correctly predicted the value of am for:

#calculate percentage of observations the model correctly predicted response value for
mean(am_pred == newdata$am)

[1] 0.75

We can see that the model correctly predicted the am value for 75% of the cars in the new data frame.