# How to compute adjusted R-squared

## Description

If we have fit a multivariate linear model, how can we compute the Adjusted $R^2$ for that model, to measure its goodness of fit?

Related tasks:

## Using statsmodels, in Python

View this solution alone.

We assume you have already fit a multivariate linear model to some data, as in the code below. (If you’re unfamiliar with how to do so, see how to fit a multivariate linear model.) The data shown below is fake, and we assume you will replace it with your own real data if you use this code.

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


You can get a lot of information about your model from its summary.

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model.summary()

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/opt/conda/lib/python3.11/site-packages/scipy/stats/_stats_py.py:1806: UserWarning: kurtosistest only valid for n>=20 ... continuing anyway, n=10
warnings.warn("kurtosistest only valid for n>=20 ... continuing "

Dep. Variable: R-squared: y 0.594 OLS 0.390 Least Squares 2.921 Mon, 24 Jul 2023 0.122 17:47:21 -45.689 10 99.38 6 100.6 3 nonrobust
coef std err t P>|t| [0.025 0.975] 77.2443 27.366 2.823 0.030 10.282 144.206 -2.7009 2.855 -0.946 0.381 -9.686 4.284 7.2989 2.875 2.539 0.044 0.265 14.333 -4.8607 2.187 -2.223 0.068 -10.211 0.490
 Omnibus: Durbin-Watson: 2.691 2.123 0.26 1.251 0.524 0.535 1.62 58.2

Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.

In particular, that printout contains the Adjusted $R^2$ value; it is the second value in the right-hand column, near the top.

You can also obtain it directly, as follows:

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model.rsquared_adj

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0.390392407508503


In this case, the Adjusted $R^2$ is $0.3904$.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

## Solution, in R

View this solution alone.

We assume you have already fit a multivariate linear model to the data, as in the code below. (If you’re unfamiliar with how to do so, see how to fit a multivariate linear model.) The data shown below is fake, and we assume you will replace it with your own real data if you use this code.

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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)


You can get a lot of information about your model from its summary.

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summary(model)

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Call:
lm(formula = y ~ x1 + x2 + x3)

Residuals:
Min      1Q  Median      3Q     Max
-25.031 -20.218  -8.373  22.937  35.640

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)   77.244     27.366   2.823   0.0302 *
x1            -2.701      2.855  -0.946   0.3806
x2             7.299      2.875   2.539   0.0441 *
x3            -4.861      2.187  -2.223   0.0679 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 30.13 on 6 degrees of freedom
Multiple R-squared:  0.5936,	Adjusted R-squared:  0.3904
F-statistic: 2.921 on 3 and 6 DF,  p-value: 0.1222


In particular, that printout contains the Adjusted $R^2$ value; it is the second value in the right-hand column, near the top.

You can also obtain it directly, as follows:

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summary(model)$adj.r.squared  1 [1] 0.3903924  In this case, the Adjusted$R^2$is$0.3904\$.

Content last modified on 24 July 2023.

See a problem? Tell us or edit the source.

## Opportunities

This website does not yet contain a solution for this task in any of the following software packages.

• Excel
• Julia

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