1
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> dt = data.table(x = rnorm(100))
> dt[, y := 1+0.2*x + rnorm(100)]
> 
> fit = lm(y~x, data = dt)
> summary(fit)

Call:
lm(formula = y ~ x, data = dt)

Residuals:
     Min       1Q   Median       3Q      Max 
-2.19493 -0.75218 -0.03459  0.64181  2.38214 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   0.8561     0.1036   8.266 6.86e-13 ***
x             0.3025     0.1056   2.865   0.0051 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.022 on 98 degrees of freedom
Multiple R-squared:  0.07731,   Adjusted R-squared:  0.06789 
F-statistic: 8.211 on 1 and 98 DF,  p-value: 0.005095

> dt[, 1 - sum(fit$residuals^2) / sum(y^2)]
[1] 0.419261

Shouldn't R-squared be 0.419? Or at least close to it, not 0.07?

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0
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Your formula

1 - sum(fit$residuals^2) / sum(y^2)

is wrong. It should be

$$ R^2 = 1 - \frac{\sum_i (y_i-\hat{y_i})^2}{\sum_i (y_i-\bar{y})^2} $$

where $\hat y_i$ is prediction and $\bar y$ is mean of $y$. Check Wikipedia entry on $R^2$ to learn more.

Notice that this could be easily generalized since $\bar y$ is basically the same as prediction from intercept-only regression, so the dividend is residuals from your model and divisor is residuals from the null model (the most basic one that we can imagine).

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4
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You're missing an important part of the $R^2$ computation

> library(data.table)
> set.seed(154)
> dt = data.table(x = rnorm(100))
> dt[, y := 1+0.2*x + rnorm(100)]
> fit = lm(y~x, data = dt)
> summary(fit)

Call:
lm(formula = y ~ x, data = dt)

Residuals:
    Min      1Q  Median      3Q     Max 
-2.6419 -0.6176 -0.0164  0.6459  2.6444 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  1.09677    0.10214  10.738   <2e-16 ***
x            0.21187    0.09974   2.124   0.0362 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.021 on 98 degrees of freedom
Multiple R-squared:  0.04402,   Adjusted R-squared:  0.03427 
F-statistic: 4.513 on 1 and 98 DF,  p-value: 0.03616

> dt[, 1 - sum(fit$residuals^2) / sum((y - mean(y))^2)]
[1] 0.04402207

The difference is in the very last line.

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