Time Series Graphics - "cannot plot more than 10 series"

Try

library(zoo)
plot(zoo(t.mbaye3), ylim = c(0, 20))

or to put them all on the same plot:

plot(zoo(t.mbaye3), screen = 1, col = 1:10)

On Sat, Feb 7, 2009 at 2:07 PM, Dominik Hattrup
<dominik.hattrup at googlemail.com> wrote:

It's working now! Thank You Guys!!

       t.mbaye3 <- t(mbaye3)
       plot(ts(t.mbaye3))

Now I am searching for an alternative to ylim, which is not working. I
am not the first one :):

https://stat.ethz.ch/pipermail/r-help/2004-October/059376.html

https://stat.ethz.ch/pipermail/r-help/2006-July/109085.html

Dominik

David Winsemius schrieb:

Thanks. It did give me the opportunity to discover one reason why
dput(<object>) may not be a totally general solution to the problem of
minimizing R-help responder time. That structure depends on an object
that is not included, namely the mbaye dataset that you constructed
mbaye3 from. I was able to trim out the call and remove the xtabs class
from that structure and would have been able to test my theory (and that
Gabor already confirmed) that the problem was that the plot.ts required
a column oriented series and that the transpose operation was needed.

This was my trimmed version of mbaye3 which R accepted as valid in the
absence of "mbaye":

mbaye3 <- structure(c(0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0,
0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 1, 0, 0,
0, 1, 3, 1, 0, 0, 4, 0, 0, 0, 3, 0, 0, 0, 0, 1, 6, 0, 0, 0, 2,
0, 0, 0, 0, 1, 8, 0, 0, 0, 0, 0, 0, 1, 0, 2, 13, 3, 0, 0, 2,
1, 2, 3, 0, 1, 7, 2, 0, 0, 3, 2, 5, 1, 0, 1, 13, 0, 0, 0, 1,
1, 5, 6, 3, 1, 6, 1, 0, 0, 2, 1, 1, 6, 3, 1, 20, 4, 0, 0, 4,
1, 2, 3, 4, 0, 15, 0, 0, 0, 1, 0, 2, 2, 9, 1, 9, 2, 0, 0, 0,
2, 3, 5, 6, 3, 5, 1, 0, 0, 0, 2, 6, 3, 6, 2, 16, 4, 0, 4, 4,
0, 3, 3, 5, 0, 5, 1, 1, 3, 2, 1, 3, 1, 5, 3, 8, 3, 0, 2, 0, 0,
1, 4, 1, 2, 5, 1, 1, 5, 3, 0, 2, 2, 1, 0, 5, 3, 0, 3, 0), .Dim = c(10L,
22L), .Dimnames = structure(list(name = c("Denmark", "France",
"Germany", "Greece", "Ireland", "Italy", "Netherlands", "Portugal",
"Spain", "UK"), year = c("1972", "1973", "1974", "1975", "1976",
"1977", "1978", "1979", "1980", "1981", "1982", "1983", "1984",
"1985", "1986", "1987", "1988", "1989", "1990", "1991", "1992",
"1993")), .Names = c("name", "year")), class = c( "table"
))

--
Dominik Hattrup