Displaying 3 results from an estimated 3 matches for "plotdata2".
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plotdata
2008 Mar 25
2
ggplot2 - facetting
...g, but here also I can not find how
to do this. I do not want to use qplot, but rather the more flexible
options. However, it seems I am doing still something pretty stupid,
because I always get an error, even if it seems I am doing everything
like the examples.
My code is below.
set.seed(123)
plotdata2<-data.frame(x=rep((1:4),4), y=rep(0.1*(1:4),4),
group=sample(factor(rep(1:4,rep(4,4)), labels=c('Class1', 'Class2',
'Class3', 'Class4'))))
plotdata2 <- plotdata2[order(plotdata2[,'group']),]
plot0<-ggplot()
layer1<-layer(data=plotdata2,
mapping...
2008 Mar 23
2
ggplot2 - legend for fill coulours
...ll colours. When the
sequence is different, the legend gets "scrambled", in that the order
of the colours does not match the labels.
My code is below. Can anyone tell me how to get around this? (R 2.2.6
for Windows, ggplot2 version 2_0.5.7)
Thanks,
Pedro
library(ggplot2)
Data:
> plotdata2
x y group
1 1 0.1 grey30
2 2 0.2 grey30
3 3 0.3 grey10
4 4 0.4 grey90
5 1 0.1 grey30
6 2 0.2 grey60
7 3 0.3 grey60
8 4 0.4 grey90
9 1 0.1 grey60
10 2 0.2 grey10
11 3 0.3 grey90
12 4 0.4 grey30
13 1 0.1 grey90
14 2 0.2 grey60
15 3 0.3 grey10
16 4 0.4 grey10
> levels(plotdata2$gro...
2000 Jul 07
2
Question of programming style
...:
mmtop94.2[lower.tri(mmtop94.2)] <- NA
# Here i is the row (y variable), j is the column (x var)
plotdata <- matrix(0, ncol=3, nrow=121)
for (i in 1:11)
{
for (j in 1:11)
{
k <- (i-1) * 11 + j
plotdata[k,] <- c(j, abs(i - 12), mmtop94.2[i, j])
}
}
plotdata2 <- na.omit(plotdata)
scatterplot3d(plotdata2, type='h')
It seems to me that I should not have to initialize the plotdata matrix. And
are all those for loops really necessary?
Thanks in advance for the help.
______________________________________________________________________
Stuart...