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Here is an absolutely horrible way to confuse yourself and get an inflated reported R-squared on a simple linear regression model in R.

We have written about this before, but we found a new twist on the problem (interactions with categorical variable encoding) which we would like to call out here.

First let’s set up our problem with a data set where the quantity to be predicted (y) has no real relation to the independent variable (x). We will first build our example data:

library("sigr")
library("broom")
set.seed(23255)

d <- data.frame(y= runif(100),
x= ifelse(runif(100)>=0.5, 'a', 'b'),
stringsAsFactors = FALSE)

Now let’s build a model and look at the summary statistics returned as part of the model fitting process:

m1 <- lm(y~x, data=d)
t(broom::glance(m1))
##                        [,1]
## r.squared       0.002177326
## sigma           0.302851476
## statistic       0.213843593
## p.value         0.644796456
## df              2.000000000
## logLik        -21.432440763
## AIC            48.864881526
## BIC            56.680392084
## deviance        8.988463618
## df.residual    98.000000000
d$pred1 <- predict(m1, newdata = d) I strongly prefer to directly calculate the the model performance statistics off the predictions (it lets us easily compare different modeling methods), so let’s also do that also: cat(render(sigr::wrapFTest(d, 'pred1', 'y'), format='markdown')) F Test summary: (R2=0.0022, F(1,98)=0.21, p=n.s.). So far so good. Let’s now remove the "intercept term" by adding the "0+" from the fitting command. m2 <- lm(y~0+x, data=d) t(broom::glance(m2)) ## [,1] ## r.squared 7.524811e-01 ## adj.r.squared 7.474297e-01 ## sigma 3.028515e-01 ## statistic 1.489647e+02 ## p.value 1.935559e-30 ## df 2.000000e+00 ## logLik -2.143244e+01 ## AIC 4.886488e+01 ## BIC 5.668039e+01 ## deviance 8.988464e+00 ## df.residual 9.800000e+01 d$pred2 <- predict(m2, newdata = d)

Uh oh. That appeared to vastly improve the reported R-squared and the significance ("p.value")!

That does not make sense, anything m2 can do m1 can also do. In fact the two models make essentially identical predictions, which we confirm below:

sum((d$pred1 - d$y)^2)
## [1] 8.988464
sum((d$pred2 - d$y)^2)
## [1] 8.988464
max(abs(d$pred1 - d$pred2))
## [1] 4.440892e-16
head(d)
##             y x     pred1     pred2
## 1 0.007509118 b 0.5098853 0.5098853
## 2 0.980353615 a 0.5380361 0.5380361
## 3 0.055880927 b 0.5098853 0.5098853
## 4 0.993814410 a 0.5380361 0.5380361
## 5 0.636617385 b 0.5098853 0.5098853
## 6 0.154032277 a 0.5380361 0.5380361

Let’s double check the fit quality of the predictions.

cat(render(sigr::wrapFTest(d, 'pred2', 'y'),
format='markdown'))

F Test summary: (R2=0.0022, F(1,98)=0.21, p=n.s.).

Ah. The prediction fit quality is identical to the first time (as one would expect). This is yet another reason to directly calculate model fit quality from the predictions: it isolates you from any foibles of how the modeling software does it.

The answer to our puzzles of "what went wrong" is something we have written about before here.

Roughly what is going on is:

If the fit formula sent lm() has no intercept (triggered by the "0+") notation then summary.lm() changes how it computes r.squared as follows (from help(summary.lm)):

r.squared   R^2, the ‘fraction of variance explained by the model’,
R^2 = 1 - Sum(R[i]^2) / Sum((y[i]- y*)^2),
where y* is the mean of y[i] if there is an intercept and zero otherwise.


This is pretty bad.

Then to add insult to injury the "0+" notation also changes how R encodes the categorical variable x.

Compare:

summary(m1)
##
## Call:
## lm(formula = y ~ x, data = d)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -0.53739 -0.23265 -0.02039  0.27247  0.47111
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)  0.53804    0.04515  11.918   <2e-16 ***
## xb          -0.02815    0.06088  -0.462    0.645
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3029 on 98 degrees of freedom
## Multiple R-squared:  0.002177,   Adjusted R-squared:  -0.008005
## F-statistic: 0.2138 on 1 and 98 DF,  p-value: 0.6448
summary(m2)
##
## Call:
## lm(formula = y ~ 0 + x, data = d)
##
## Residuals:
##      Min       1Q   Median       3Q      Max
## -0.53739 -0.23265 -0.02039  0.27247  0.47111
##
## Coefficients:
##    Estimate Std. Error t value Pr(>|t|)
## xa  0.53804    0.04515   11.92   <2e-16 ***
## xb  0.50989    0.04084   12.49   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.3029 on 98 degrees of freedom
## Multiple R-squared:  0.7525, Adjusted R-squared:  0.7474
## F-statistic:   149 on 2 and 98 DF,  p-value: < 2.2e-16

Notice the second model directly encodes both levels of x. This means that if m1 has pred1 = a + b * (x=='b') we can reproduce this model (intercept and all) as m2: pred2 = a * (x=='a') + (a+b) * (x=='b'). I.e., the invariant (x=='a') + (x=='b') == 1 means m2 can imitate the model with the intercept term.

The presumed (and I think weak) justification of summary.lm() switching the model quality assessment method is something along the lines that mean(y) may not be in the model’s concept space and this might lead to reporting negative R-squared. I don’t have any problem with negative R-squared, it can be taken to mean you did worse than the unconditional average. However, even if you accept the (no-warning) scoring method switch: that argument doesn’t apply here. m2 can imitate having an intercept, so it isn’t unfair to check if it is better than using only the intercept.

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