I see what you mean now. You are lucky I did this kind of plot in one of my
previous assignment. I have modified the data to suit your needs. The data is
directly quoted from the paper 475.340 Designs of Surveys and Experiments in the
University of Auckland, New Zealand.
# Enter the data, count.
count <- c( 11, 14, 15, 17, 18, 21, 25,
39, 43, 46, 50, 52, 61, 67,
88, 92, 104, 113, 119, 120, 130,
222, 251, 259, 283, 299, 312, 337,
562,,604, 689, 702, 710, 739, 786 )
# Enter the ordered factor, Test.
# Since the data, count, is kind of like a 7 by 5 matrix, you need to
# repeat Test 1 ~ Test 5 by that amount.
# In other words, "Test" will look something like:
# 1 1 1 1 1 1 1
# 2 2 2 2 2 2 2
# 3 3 3 3 3 3 3
# 4 4 4 4 4 4 4
# 5 5 5 5 5 5 5
Test <- ordered( factor( rep( c( 1, 2, 3, 4, 5 ),
rep( 7, 5 )
)
)
)
# To draw a boxplot that has Test 1 ~ Test 5 on its x-axis and
# count on its y-axis.
# There should be 5 box&whiskers within the SAME plot box.
boxplot( split( count, Test ),
xlab = "Test", ylab = "counts",
main = "A Boxplot" )
To make is easier for you, I have even saved it as an R source file and attached
it
below for you. Just saved it to your disk, use "source()" function in
R to source
it in directly. From the R commands, it should be pretty easy to see how you
should change it to get the plot that you want. (The # means R comments, it
will
NOT affect the plots).
Cheers,
Ko-Kang Wang
Christine Serres wrote:
> First of all, I thanks you all for replying me .
> I've not asked my question in good terms, so I've made a little
drawing
> of what I want to do in my boxplot : - )
>
> | - -
> | | |
> | ---- ----
> | | | | |
> | | 2 | | 2 |
> | ---- ----
> | | |
> | - -
> |
> |
> | - -
> | | |
> | ---- ----
> | | | | |
> | | 1 | | 1 |
> | ---- ----
> | | |
> | - -
> |
> ____________________________________________________________________
>
> Test X Test Y
>
> (I don't want my boxes to be shifted for the same Test value because
I
> know that boxplot 1 and 2 don't overlap)
> I'm not sure that the boxplot function can do that.
> But perhaps someone know a trick to help me.
>
> Thanks
> Christine
>
>
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