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How to predict the response variable in a linear model

Description

If we have a linear model and a value for each explanatory variable, how do we predict the corresponding value of the response variable?

Related tasks:

Using statsmodels, in Python

View this solution alone.

Let’s assume that you’ve already built a linear model. We do an example below with fake data, but you can use your own actual data. For more information on the following code, see how to fit a multivariate linear model.

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import pandas as pd
df = pd.DataFrame( {
    'x1' : [ 2,  7,  4,  3, 11, 18,   6, 15,   9,  12],
    'x2' : [ 4,  6, 10,  1, 18, 11,   8, 20,  16,  13],
    'x3' : [11, 16, 20,  6, 14,  8,   5, 23,  13,  10],
    'y'  : [24, 60, 32, 29, 90, 45, 130, 76, 100, 120]
} )

import statsmodels.api as sm
model = sm.OLS( df['y'], sm.add_constant( df[['x1','x2','x3']] ) ).fit()

Let’s say we want to estimate $y$ given that $x_1 = 5$, $x_2 = 12$, and $x_3=50$. We can use the model’s predict() function as shown below, but we must add an entry for the constant term in the model—we can use any value, but we choose 1.

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model.predict( [ 1, 5, 12, 50 ] )

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array([-91.71014402])

For the given values of the explanatory variables, our predicted response variable is $-91.71014402$.

Note that if you want to compute the predicted values for all the data on which the model was trained, simply call model.predict() with no arguments, and it defaults to using the training data.

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model.predict()

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array([ 47.5701159 ,  24.35988296,  42.21531274,  47.27613825,
       110.86526185,  70.03097584,  95.12689978,  70.91290879,
       106.52986696,  91.11263692])

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

Solution, in R

View this solution alone.

Let’s assume that you’ve already built a linear model. We do an example below with fake data, but you can use your own actual data. For more information on the following code, see how to fit a multivariate linear model.

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x1 <- c( 2,  7,  4,  3, 11, 18,   6, 15,   9,  12)
x2 <- c( 4,  6, 10,  1, 18, 11,   8, 20,  16,  13)
x3 <- c(11, 16, 20,  6, 14,  8,   5, 23,  13,  10)
y  <- c(24, 60, 32, 29, 90, 45, 130, 76, 100, 120)
model <- lm(y ~ x1 + x2 + x3)

Let’s say we want to estimate $y$ given that $x_1 = 5$, $x_2 = 12$, and $x_3=50$. We can use R’s predict() function as shown below.

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predict(model, newdata = data.frame(x1 = 5, x2 = 12, x3 = 50))

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-91.71014

For the given values of the explanatory variables, our predicted response variable is $-91.71014$.

Note that if you want to compute the predicted values for all the data on which the model was trained, simply call predict(model) with no new data, and it defaults to using the training data.

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predict(model)

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        1         2         3         4         5         6         7         8 
 47.57012  24.35988  42.21531  47.27614 110.86526  70.03098  95.12690  70.91291 
        9        10 
106.52987  91.11264

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

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