Dear Sebastian,
Consider matplot() for this. Here is an example (taken from Baptiste
Auguie's post):
date <- factor(letters[1:9])
d <- data.frame(x1=seq(1, 9), x2=seq(2, 10), date=date)
matplot(d[,-3],pch=16,xaxt='n',las=1,ylab='Some label here',xlab='Date')
axis(1,d[,3],d[,3])
legend('toplef
Here's my suggestion using the ggplot2 package (but you may prefer to
stick with base functions),
date = factor(letters[1:9])
d <- data.frame(x1=seq(1, 9), x2=seq(2, 10), date=date)
head(d) # dummy data that resembles yours
str(d)
library(reshape)
md <- melt(d, id="date") # creates a data.fr
Sorry for the mistake. As you probably already guesed, I am just
starting using R. I could not name the difference between a matrix and a
data.frame.
> str(b)
'data.frame': 9 obs. of 7 variables:
$ Datum: Factor w/ 9 levels "06.03.","07.03.",..: 1 2 3 4 5 6 7 8 9
$ X1 : int 408 335 2123 4
the result of read.table is a data.frame, not a matrix as you first
suggested. Can you copy the result of str(b) so we know what your data
is made of?
I'm guessing the most elegant solution will be to use the reshape
package, followed by ggplot2 or lattice.
baptiste
On 27 Mar 2009, at 14
Unfortunately, I could not solve the problem of plotting all columns of
a matrix against the first column
I used:
b=read.table("d:\\programme\\R\\übungen\\Block 1b.txt", header=T)
"b" is a table with the first column using Dates and the following
columns with vectors.
apply(b[,-1], 2, plot
Something like this perhaps,
a <- matrix(rnorm(5*49), ncol=49)
pdf(width=15, height=15)
par(mfrow= c(8,6))
apply(a[,-1], 2, plot, x= a[,1])
dev.off()
HTH,
baptiste
On 27 Mar 2009, at 11:05, skrug wrote:
Hi evrybody,
in a matrix consisting of 49 columns, I would like to plot all column
Hi evrybody,
in a matrix consisting of 49 columns, I would like to plot all columns
against the first in 48 different graphs.
Can you help me?
Thank you in advance
Sebastian
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