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The `predict()` function in R is used to generate predictions from a logistic regression model. It takes as input the fitted model and a set of new values for the predictor variables, and returns a vector of predicted probabilities for each observation. The predicted probabilities can then be used to make predictions about class labels.
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.
The following tutorials explain how to perform other common tasks in R: