Hi:
It wasn't clear to me precisely what you wanted, but here are a couple of
ideas in the hope that it will help.
I used ggplot2 for the graphics, so it requires some manipulation of your
dataset from 'wide' format to 'long'.
I also add an indicator for side of the ear (odd is side one (L?), even is
side 2) and a variable I call 'loc' to
indicate the value associated with the splxx variable.
I read the data into a data frame called spikelets. The first step is to
remove the rows of missing responses:
naind <- apply(spikelets[, -1], 1, function(x) all(is.na(x)))
spikelets2 <- spikelets[!naind, ]
Next, I use the plyr package and its melt() function to convert the data
frame from 'wide' to 'long' form:
library(ggplot2) # attaches the plyr package in the loading process
spikes.long <- melt(spikelets2, id = 'cn')
The variable 'variable' contains the variable names as a vector (spl01,
spl02, ..., spl14)
Next, I create a variable called loc, which represents the numeric part of
the spl variables, and then
create a variable side to distinguish one side of the awn from the other.
'variable' is then removed...
spikes.long$loc <- as.numeric(substring(spikes.long$variable, 4))
spikes.long$side <- factor(2 - spikes.long$loc %% 2)
spikes.long$variable <- NULL
Now we're in a position to plot. The first is a scatterplot of the response
by location, stratified by cultivar;
it contains color to distinguish sides.
# With color:
p <- qplot(loc, value, data = spikes.long, group = cn,
colour = side)
p + facet_grid(cn ~ .)
The color is not terribly informative, so to get rid of it, remove the
colour = side argument. One could
also merge the plots together and fit smooths to the different cultivars.
ggplot(spikes.long, aes(loc, value, colour = cn)) +
geom_point() + geom_smooth(se = FALSE)
I also came up with boxplot pairs by side for each cultivar, which is shown
below:
q <- ggplot(spikes.long, aes(side, value))
q + geom_boxplot() + facet_grid(~ cultivar)
For some reason, I kept getting these messages from every ggplot2 call:
Error in recordGraphics(drawGTree(x), list(x = x), getNamespace("grid")) :
invalid graphics state
but all of the plots rendered as expected.
HTH,
Dennis
2010/1/10 Carl-Göran CG. Pettersson <cg.petters...@vpe.slu.se>
> Dear all
>
> R2.10 WinXP
>
> I have a dataset dealing with the way different wheat cultivars build their
> yield.
> Wheat ears are organised in spikelets where the spikelets can be numbered
> from the bottom, with even numbers on one side and odd on the other.
> I know how many kernels there were in each spikelet after some months spent
> counting them...
>
> Now I want to illustrate the differences between the cultivars in how the
> kernels are distributed in the ears.
> In the best of all possible worlds it would be possible to place histograms
> or boxplots on adjecent sides of vertical lines representing different
> cultivars.
> I have done some experimenting using boxplot() but I am stuck and out of
> ideas right now.
>
> All ideas are welcome!
> /CG
>
>
> Here is a sample dataset with the countings of kernels for the first 14
> spikelets:
>
> cn spl01 spl02 spl03 spl04 spl05 spl06 spl07 spl08
> spl09 spl10 spl11 spl12 spl13 spl14
> Lans 1.8 3.1 3.5 3.8 3.8 4.1 4.2 4.3 4.4
> 4.5 4.2 4.1 3.9 3.8
> Kranich 0.6 2.4 3.4 4.2 4.5 4.7 4.9 4.9 4.8
> 4.7 4.4 4.1 4.1 3.9
> Loyal 1.1 2.7 3.6 3.7 4.1 4.4 4.4 4.6 4.3
> 4.5 4.3 4.1 3.8 3.7
> Boomer NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Oakley NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Hereford 0.6 2.3 3.3 3.6 3.9 4 4.2 4.1
> 4.1 3.9 3.9 3.6 3.4 3.2
> Kranich 0.3 2.5 3.6 4 4.4 4.5 4.3 4.8 4.7
> 4.6 4.4 4.3 4.1 4
> Oakley 0.5 2.1 3.2 3.4 3.8 4.4 4.3 4.3 4.3
> 4.2 4.2 3.9 3.8 3.6
> Loyal 1.6 3.3 3.9 4.2 4.3 4.4 4.4 4.6 4.6
> 4.5 4.3 4.3 4.2 4
> Hereford NA NA NA NA NA NA NA NA
> NA NA NA NA NA NA
> Oakley 0.5 2.1 3.2 3.6 4 4 4.1 4.4 4.4
> 4.2 4.1 3.8 3.8 4
> Kranich NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Lans 1.4 3 3.3 3.8 3.9 4.3 4 4.3 4.3
> 4.3 4 4.1 4 4
> Hereford 1.2 2.7 3.6 3.8 4 4 4.1 4.2
> 4.1 4.1 3.9 3.6 3.8 3.3
> Boomer 0.3 2.5 3.1 3.8 3.9 4.4 4.1 4.2 4.3
> 4 4.2 4 3.8 3.7
> Lans NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Boomer 0.2 1.9 3 3.4 3.7 3.9 3.9 4 4
> 4 3.8 3.8 3.6 3.4
> Loyal NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Boomer NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Kranich NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Kranich 0.3 1.1 2.9 3.5 3.9 4.3 4.4 4.4 4
> 4.2 4.2 4 3.9 3.8
> Hereford 0.5 2.1 3.1 3.6 3.7 3.9 4 3.8
> 4 3.8 3.6 3.6 3.1 3
> Loyal NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Boomer 0.3 0.8 2.8 3 3.6 3.7 3.8 4 3.8
> 3.5 3.3 3.2 3.2 2.9
> Oakley 0.5 2.7 3.4 3.8 4 3.9 4.2 4.5 4.3
> 4.4 4 4 3.9 3.9
> Loyal 0.9 2.6 3.6 3.8 3.8 4.4 4.2 4.4 4.2
> 3.9 3.8 4 3.4 3.7
> Oakley NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Hereford 0.7 2.9 3.6 4 4 3.9 4 4
> 4 3.9 3.8 3.7 3 3
> Hereford NA NA NA NA NA NA NA NA
> NA NA NA NA NA NA
> Loyal 0.7 2.3 3.5 3.7 3.9 3.8 4.2 4.1 4.1
> 4.1 4 4 3.4 3.6
> Boomer 0.7 2 3.3 3.5 3.9 3.7 4 3.9 3.8
> 4 3.7 3.8 3.5 3.4
> Lans NA NA NA NA NA NA NA NA NA
> NA NA NA NA NA
> Lans 1.9 3 3.7 3.8 3.9 4 3.9 4.3 4.1
> 4.1 4.1 3.8 3.8 3.9
> Lans 1.1 2.6 3.3 3.7 4.1 4 4.2 4.2 4.2
> 4 4.1 4.1 3.8 3.6
> Kranich 0.5 1.3 2.9 3.8 3.8 4.3 4.3 4.4 4.4
> 4 4.3 3.9 3.6 3.4
> Oakley 0.1 2 3.1 3.5 4.1 3.9 4.1 4.2 4.2
> 4.2 4.1 4 3.9 3.8
> ______________________________________________
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