CodeGuy - 4 months ago 15

R Question

I learned to get a linear fit with some points using lm in my R script. So, I did that (which worked nice), and printed out the fit:

`lm(formula = y2 ~ x2)`

Residuals:

1 2 3 4

5.000e+00 -1.000e+01 5.000e+00 7.327e-15

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 70.000 17.958 3.898 0.05996 .

x2 85.000 3.873 21.947 0.00207 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 8.66 on 2 degrees of freedom

Multiple R-squared: 0.9959, Adjusted R-squared: 0.9938

F-statistic: 481.7 on 1 and 2 DF, p-value: 0.00207

I'm trying to determine the best way to judge how great this fit is. I need to compare this fit with a few others (which are also linear using

`lm()`

Answer

If you want to access the pieces produced by `summary`

directly, you can just call `summary`

and store the result in a variable and then inspect the resulting object:

```
rs <- summary(lm1)
names(rs)
```

Perhaps `rs$sigma`

is what you're looking for?

**EDIT**

Before someone chides me, I should point out that for some of this information, this is not the recommended way to access it. Rather you should use the designated extractors like `residuals()`

or `coef`

.