[R] A problem of splitting the right screen in 3 or more independent vertical boxes:
Aldi Kraja
aldi at wustl.edu
Sat May 11 16:01:48 CEST 2013
For those that may have this question in the future, here are two solutions:
As suggested from David and Sarah,
One has to remove par function from defining screen splits, instead use
layout function.
For example:
layout(matrix(c(1,1,2,3),2,2,byrow=T))
which says, split the screen in 4 blocks, of them use block space 1 and
2 oriented by row for picture 1, and the two remaining blocks for
picture 2 and picture 3. Similarly, one can split the screen into 6, 8
blocks and so on by changing also how many blocks of space you want to
assign to a specific picture for example.
layout(matrix(c(1,1,1,2,3,4,5),2,3,byrow=T)) ## six splits, of those
first 3 belong to picture 1
layout(matrix(c(1,1,2,3,4,5,6,7),2,4,byrow=T)) ## 8 splits of those
first 2 belong to picture 1 and so on.
Dennis Murphey provided also another beautiful solution via ggplot2. See
following.
Thank you,
Aldi
On 5/3/2013 7:57 PM, Dennis Murphy wrote:
> Hi:
>
> Here are a couple of ways you could go using the ggplot2 and gridExtra
> packages, but it requires that you rearrange your data.
>
> ggplot2 prefers data frame input, so we convert the histogram data to
> a simple data frame and apply a couple of tricks to get an appropriate
> data frame for the box plot data.
>
> # Cast x into a data frame
> DF0 <- data.frame(x = rnorm(1000,mean=0,sd=1))
>
>
> wheat1 = rnorm(100,mean=0,sd=1)
> wheat2 = rnorm(150,mean=0,sd=2)
> tomatos3 = rnorm(200,mean=0,sd=3)
> tomatos4 = rnorm(250,mean=0,sd=4)
> cucumbers5 <- rnorm(300,mean=0,sd=5)
> cucumbers6 <- rnorm(400,mean=0,sd=6)
>
> # commodity is defined so that products are paired
> DF1 <- data.frame(
> commodity = factor(rep(c("wheat", "tomatos", "cucumbers"), c(250, 450, 700))),
> product = factor(rep(c("wheat1", "wheat2", "tomatos3", "tomatos4",
> "cucumbers5", "cucumbers6"),
> times = c(100, 150, 200, 250, 300, 400))),
> value = c(wheat1, wheat2, tomatos3, tomatos4, cucumbers5, cucumbers6) )
>
> library(ggplot2)
>
> # Simple histogram with ggplot2
> p0 <- ggplot(DF0, aes(x = x)) + geom_histogram()
>
> # Paired box plots by commodity
> p1 <- ggplot(DF1, aes(x = product, y = value)) +
> geom_boxplot() +
> facet_wrap( ~ commodity, scales = "free_x", ncol = 3)
>
> # gridExtra lets you combine multiple grid graphics plots together
> # (ggplot2 and lattice are both written in grid)
> library(gridExtra)
> grid.arrange(p0, p1, nrow = 2) # stack histogram on top of boxplots
> grid.arrange(p0, p1, nrow = 1) # what you asked for
>
> I prefer the former, but it's easy enough to fix the problems with the latter.
>
> Dennis
On 5/3/2013 6:07 PM, David Winsemius wrote:
> On May 3, 2013, at 3:21 PM, Sarah Goslee wrote:
>
>> Hi Aldi,
>>
>> You might want
>> ?layout
>> instead.
>>
> Indeed. In particular a matrix argument might be:
>
> matrix(c(1,2,3, 4,4,4)
>
>
>> Sarah
>>
>> On Fri, May 3, 2013 at 5:54 PM, Aldi Kraja <aldi at wustl.edu> wrote:
>>> Hmm,
>>> I had a typo paste by mistake in my x vector
>>> It has to be:
>>>
>>> x<-rnorm(1000,mean=0,sd=1)
>>> wheat1<-rnorm(100,mean=0,sd=1)
>>> wheat2<-rnorm(150,mean=0,sd=2)
>>> tomatos3<-rnorm(200,mean=0,sd=3)
>>> tomatos4<-rnorm(250,mean=0,sd=4)
>>> cucumbers5<-rnorm(300,mean=0,sd=5)
>>> cucumbers6<-rnorm(400,mean=0,sd=6)
>>> par(mfrow=c(1,2))
>>>
>>> hist(x, main="Left screen OK")
>>>
>>> boxplot(wheat1,wheat2,tomatos3,tomatos4,cucumbers5,cucumbers6)
> I think you will need a separate call to boxplot for each grouping. The `boxplot` function will nto be able to access the device specifications.
>
>
--
--------------------------------------------------------------------------------
Aldi T. Kraja, DSc, PhD aldi at dsgmail.wustl.edu
Research Associate Professor of Genetics
Division of Statistical Genomics http://dsgweb.wustl.edu/aldi
Washington University School of Medicine, Phone:(314)362-2498 Fax:(314)362-4227
4444 Forest Park Blvd., Campus Box 8506 Saint Louis, MO, 63110
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