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problem with qq( ) (PR#1729)

2 messages · Jarno.Tuimala@Helsinki.Fi, Brian Ripley

Original
Jarno.Tuimala@Helsinki.Fi
# 
Full_Name: Jarno Tuimala
Version: 1.5.0
OS: Windows Nt
Submission from: (NULL) (193.166.1.21)


Running the following analysis gives as a result kukot vs. kisut.

aikaero <- -27.5+5*(0:10)
frekvenssi.m <- c(0,1,15,37,53,45,23,18,4,1,1)
frekvenssi.n <- c(2,8,12,19,33,47,42,15,2,1,0)
win.graph()
aikaero <- rep(c(aikaero,aikaero), c(frekvenssi.m,frekvenssi.n))
 a.ero <- data.frame(aikaero,
            sukupuoli=rep(c("kukot","kisut"),
                          c(sum(frekvenssi.m),sum(frekvenssi.n))))
    qq(sukupuoli ~ aikaero, data=a.ero)

However, running the following analysis, gives miehet vs. miehet.

 aikaero <- -27.5+5*(0:10)
 frekvenssi.m <- c(0,1,15,37,53,45,23,18,4,1,1)
 frekvenssi.n <- c(2,8,12,19,33,47,42,15,2,1,0)
 win.graph()
 aikaero <- rep(c(aikaero,aikaero), c(frekvenssi.m,frekvenssi.n))
 a.ero <- data.frame(aikaero,
        sukupuoli=rep(c("miehet","naiset"),
                      c(sum(frekvenssi.m),sum(frekvenssi.n))))
     qq(sukupuoli ~ aikaero, data=a.ero)

This seems to happen everytime "sukupuoli" contains one variable name, which
starts with a letter "n". If both names start with "n" or other letters the
results is miehet vs naiset as expected.

So this works fine:

a.ero <- data.frame(aikaero,
        sukupuoli=rep(c("nisset","naiset"),
                      c(sum(frekvenssi.m),sum(frekvenssi.n))))
     qq(sukupuoli ~ aikaero, data=a.ero)

And this also:

a.ero <- data.frame(aikaero,
        sukupuoli=rep(c("miehet","maiset"),
                      c(sum(frekvenssi.m),sum(frekvenssi.n))))
     qq(sukupuoli ~ aikaero, data=a.ero)

Where's the problem?
 

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Brian Ripley
# 
1) This is a report on the lattice package, and we both need you to
mention that and to give the version number of lattice that you used
to be sure of beeing able to reproduce this.

2) It seems that the plots are actually correct, just the labels are
wrong. You can correct that by specifying xlab and ylab in the call.

3) The error appears to be a typo:

    if (missing(ylab))
        ylab <- if (is.f.y)
            unique(levels(y))[y]
        else paste("y:", as.character(unique(levels(y)[[2]])))

should be [2] for consistency.
On Tue, 2 Jul 2002 Jarno.Tuimala@Helsinki.Fi wrote: