Message-ID: <376F5BBB-F2FD-4297-A626-B16F8F73841D@comcast.net>
Date: 2011-01-13T04:32:17Z
From: David Winsemius
Subject: easy loop question
In-Reply-To: <4D2E776D.8050209@gmail.com>
On Jan 12, 2011, at 10:54 PM, Sebasti?n Daza wrote:
> Hi everyone,
>
> I am new in R and programming. I have tried to remove the values out
> of range in some variables using a loop:
>
> 1)
>
> var <- names(est8vo[, 77:83]) # I got the variable names
>
> > var
> [1] "p16.1" "p16.2" "p16.3" "p16.4" "p16.5" "p16.6" "p16.7"
>
> for (i in 1:7) {
> var.i <- var[i]
> est8vo$var.i[ est8vo$var.i==3] <- 99
You CANNOT use names like that. (It makes no sense to supply a vector
argument to "$".) If you want to change every instance of 3 within a
group of columns. See if this works
est8vo[, 77:83] <- sapply( est8vo[, 77:83], function(x) ifelse(x==3,
99, x) )
> }
>
> I got this error:
>
> Error in `$<-.data.frame`(`*tmp*`, "var.i", value = numeric(0)) :
> replacement has 0 rows, data has 215700
>
>
> 2) The second step would be to define the factors, but I got the
> same error:
est8vo[, 77:83] <- sapply(est8vo[, 77:83] , factor, labels=c("vac?o",
"s?", "no", "doble marca"))
>
> for (i in 1:7) {
> var.i <- var[i]
> est8vo$var.i <- factor(est8vo$var.i,
Wrong. Wrong. Wrong. And please forget about using "$" inside loops in
the left-hand side. It is not designed for that.
> levels=c(0, 1, 2, 99),
> labels=c("vac?o", "s?", "no", "doble marca")
> )
> }
>
>
> I don't know how to do it.
> Thank you in advance!
> Sebastian
>
> ______________________________________________
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> and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT