On 16-Mar-07 12:41:50, Alberto Monteiro wrote: > Ted Harding wrote: >> >>> alpha <- 0.3 >>> beta <- 0.4 >>> sigma <- 0.5 >>> err <- rnorm(100) >>> err[15] <- 5; err[25] <- -4; err[50] <- 10 >>> x <- 1:100 >>> y <- alpha + beta * x + sigma * err >>> ll <- lm(y ~ x) >>> plot(ll) >> >> ll is the output of a linear model fiited by lm(), and so has >> several components (see ?lm in the section "Value"), one of >> which is "residuals" (which can be abbreviated to "res"). >> >> So, in the case of your example, >> >> which(abs(ll$res)>2) >> 15 25 50 >> >> extracts the information you want (and the ">2" was inspired by >> looking at the "residuals" plot from your "plot(ll)"). >> > Ok, but how can I grab those points _in general_? What is the > criterium that plot used to mark those points as bad points?
Ahh ... ! I see what you're after. OK, look at the plot method for lm(): ?plot.lm ## S3 method for class 'lm': plot(x, which = 1:4, caption = c("Residuals vs Fitted", "Normal Q-Q plot", "Scale-Location plot", "Cook's distance plot"), panel = points, sub.caption = deparse(x$call), main = "", ask = prod(par("mfcol")) < length(which) && dev.interactive(), ..., id.n = 3, labels.id = names(residuals(x)), cex.id = 0.75) where (see further down): id.n: number of points to be labelled in each plot, starting with the most extreme. and note, in the default parameter-values listing above: id.n = 3 Hence, the 3 most extreme points (according to the criterion being plotted in each plot) are marked in each plot. So, for instance3, try plot(ll,id.n=5) and you will get points 10,15,25,28,50. And so on. But that pre-supposes that you know how many points are "exceptional". What is meant by "extreme"is not stated in the help page ?plot.lm, but can be identified by inspecting the code for plot.lm(), which you can see by entering plot.lm In your example, if you omit the line which assigns anomalous values to err[15[, err[25] and err[50], then you are likely to observe that different points get identified on different plots. For instance, I just got the following results for the default id.n=3: [1] Residuals vs Fitted: 41,53,59 [2] Standardised Residuals: 41,53,59 [3] sqrt(Stand Res) vs Fitted: 41,53,59 [4] Cook's Distance: 59,96,97 There are several approaches (with somewhat different outcomes) to identifying "outliers". If you apply one of these, you will probably get the identities of the points anyway. Again in the context of your example (where in fact you deliberately set 3 points to have exceptional errors, thus coincidentally the same as the default value 3 of id.n), you could try different values for id.n and inspect the graphs to see whether a given value of id.n marks some points that do not look exceptional relative to the mass of the other points. So, the above plot(ll,id.n=5) gave me one point, "10" on the residuals plot, which apparently belonged to the general distribution of residuals. Hoping this helps, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 16-Mar-07 Time: 13:43:54 ------------------------------ XFMail ------------------------------ ______________________________________________ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.