On 16/01/12 02:08, J Toll wrote:
> Hi,
>
> I'm trying to create a stacked bar plot using ggplot2. Rather than
> plotting the count of each of the 13 "Bar" factors on the Y axis, I
> would like to represent the sum of the Values associated with each of
> the 13 "Bar" factors. Is there a way to do that?  Given the following
> data, that would obviously mean that there would be some negative sums
> represented.  Here's a bit of example data along with the command I've
> been using.
>
>> library(ggplot2)
>> x
>           Value Bar Segment
> 1   1.10020075   1       1
> 2  -1.37734577   2       1
> 3   2.50702876   3       1
> 4   0.58737028   3       2
> 5   0.21106851   3       3
> 6  -2.50119261   4       1
> 7   1.34984831   5       1
> 8  -0.27556149   6       1
> 9  -1.54401647   6       2
> 10 -2.75975562   6       3
> 11 -0.09527123   6       4
> 12  1.36331646   7       1
> 13 -0.36051429   8       1
> 14  1.36790999   9       1
> 15  0.15064633   9       2
> 16  0.34022421   9       3
> 17 -0.64512970  10       1
> 18  0.83268199  11       1
> 19 -1.50117728  12       1
> 20  1.09004959  13       1
>> qplot(factor(Bar), data = x, geom = "bar", fill = factor(Segment))
> Thanks for any suggestions you might have.
>
> James
>
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> and provide commented, minimal, self-contained, reproducible code.
I'm not at my usual computer, but in ggplot2 the geom_bar geom is 
designed for stats, so by default does report count.  You can change the 
stat method (ie to identity to just use the values in the column) but 
the easiest (IIRC) is to give a weight;

#Gives count
||

qplot  <http://had.co.nz/ggplot2/qplot.html>(color,  data=diamonds,  geom="bar")

#Gives sum of carat variable
qplot  <http://had.co.nz/ggplot2/qplot.html>(color,  data=diamonds,  geom="bar",  weight=carat,  ylab="carat")

#just gives raw values from meanprice column
||

qplot  <http://had.co.nz/ggplot2/qplot.html>(cut,  meanprice,  geom="bar",  stat="identity")

Check out the ggplot2 help page (http://had.co.nz/ggplot2/geom_bar.html) for more info.

Regards,

Paul.




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