Dear list,
I cannot find an elegant solution to this problem. I have a factor f
containing several levels (5) and I wish to create a new factor of the
same length with fewer levels (2). This new factor should therefore
group together some levels of the original data. Ideally this grouping
would be
As always the question seems silly after you've learned the answer!
Thanks a lot,
baptiste
On 3 February 2010 15:06, Petr PIKAL wrote:
> # order levels
> f.t<-factor(f, levels=disorder)
> # change levels
> levels(f.t) <- new.lev
> all.equal(f.t,f2)
> [1] TRUE
>
>
> Regards
> Petr
>
>
>>
>> Bes
On 3 February 2010 15:17, Duncan Murdoch wrote:
> On 03/02/2010 8:50 AM, Ken Knoblauch wrote:
>>
>> baptiste auguie googlemail.com> writes:
>>>
>>> Adding two semi-transparent colours results in non-intuitive colour
>>> mixing (a mystery for
ource code later (I haven't found any documentation on
this so far).
Thanks,
baptiste
On 3 February 2010 16:38, Duncan Murdoch wrote:
> On 03/02/2010 9:38 AM, baptiste auguie wrote:
>>
>> That makes perfect sense, thank you, except that I'm not sure where
>> the whit
Thanks for this complementary information. My head itches slightly
when reading about these virtual layers with unidirectional absorption
and reflection properties but I guess that's imputable to my personal
background as a physicist.
I still have a few questions,
- is this behavior documented? (
Dear all,
I use the following to create a list of identical grobs,
require(grid)
rep.grob <- function(g, n){
replicate(n, g, simplify=FALSE)
}
This approach suffers two problems:
1- R CMD check is not happy about the S3-like name. How can / Should I
make this a real S3 method?
2- I don't know
Hi,
Try this,
## create a 2D grid with random point sizes
d = expand.grid(x=1:10,y=1:10)
d$size = runif(nrow(d), 1,5)
## create some random links
links= d[sample(seq(1,nrow(d)),20),]
links$id = sample(2:6, nrow(links),repl=T)
## plot
library(ggplot2)
ggplot(d, mapping=aes(x,y)) +
geom_point(
Try this,
qplot(factor(0), mpg, data=mtcars, geom="boxplot", xlab="")+
coord_flip() + scale_x_discrete(breaks=NA)
HTH,
baptiste
On 18 June 2010 16:47, Jacob Wegelin wrote:
>
> In ggplot2, I would like to make a boxplot that has the following
> properties:
>
> (1) Contrary to default, the meanin
Hi,
Another option would be the tikzDevice package, which lets you process
all the text of your plot with LaTeX. I think the XeTeX variant might
be the most straight-forward to mix different fonts using this
approach.
HTH,
baptiste
On 29 June 2010 16:17, Jinsong Zhao wrote:
> Hi there,
>
> I
Hi,
try adding asp=1 in symbols() to set the aspect ratio of the plotting
region to 1.
HTH,
baptiste
On 17 July 2010 18:21, wrote:
> Hello, I submitted this bug report to r-core and got a rejection saying I
> should post to r-help.
> This is my first time ever submitting a bug report, so for
Hi,
There may be a simpler way but try this,
plot(10^jitter(seq(-2,4,length=10)), 1:10, log="x", xaxt="n")
axis(1, at = axTicks(1),labels = format(axTicks(1),scientific=FALSE))
HTH,
baptiste
On 18 July 2010 10:58, Timothy O'Brien wrote:
> Dear All,
>
> I've done some searching, but to no avai
Dear list,
I read in ?plotmath that I can use bgroup to draw scalable delimiters
such as [ ] and ( ). The same technique fails with < > however, and I
cannot find a workaround,
grid.text(expression(bgroup("<",atop(x,y),">")))
Error in bgroup("<", atop(x, y), ">") : invalid group delimiter
Regar
What do people use to show angle brackets < > in R graphics? Have I
missed something obvious?
Thanks,
baptiste
On 9 September 2010 17:57, baptiste auguie
wrote:
> Dear list,
>
> I read in ?plotmath that I can use bgroup to draw scalable delimiters
> such as [ ] and ( ).
7;x, y'~symbol("\076")))
>>>
>>> HTH,
>>> Dennis
>>
>> It's a matter of taste, but I would use "\341" and "\361".
>> However, these are still not scalable, AFAICS.
>
> Not exactly scalable angles, but you ca
p 12, 2010, at 6:15 AM, baptiste auguie wrote:
>
>> Thanks everyone. I've also had a look at plotmath.c where bgroup is
>> defined for "[", "{", "(", "." but not "<". It seems quite trivial to
>> add it, at first sight, however
I see, thanks. Looking at this table I guess the short answer is no,
these cannot be made to scale and the only ones that could have
already been implemented in bgroup().
Thanks,
baptiste
On 12 September 2010 22:11, Paul Murrell wrote:
> Hi
>
> On 13/09/2010 7:57 a.m., baptiste aug
Dear list,
I'm seeking some advice regarding a particular numerical integration I
wish to perform.
The integrand f takes two real arguments x and y and returns a vector
of constant length N. The range of integration is [0, infty) for x and
[a,b] (finite) for y. Since the integrand has values in R
Dear list,
I'm calculating the integral of a Gaussian function from 0 to
infinity. I understand from ?integrate that it's usually better to
specify Inf explicitly as a limit rather than an arbitrary large
number, as in this case integrate() performs a trick to do the
integration better.
However,
,0)
> [1] NaN NaN NaN
> Warning message:
> In sqrt(x) : NaNs produced
>> f(-3.9,0.1)
> [1] NaN NaN NaN
> Warning message:
> In sqrt(x) : NaNs produced
> --
> David
> On Sep 21, 2010, at 4:11 AM, baptiste auguie wrote:
>
>> Dear list,
>>
>>
Recipes.
Thanks,
baptiste
On 21 September 2010 14:26, Hans W Borchers wrote:
> baptiste auguie googlemail.com> writes:
>
>>
>> Dear list,
>>
>> I'm seeking some advice regarding a particular numerical integration I
>> wish to perform.
>>
>&
I see, thank you.
I'm still worried by the very dramatic error I obtained just from
shifting so slightly the support of the integrand, it took me a while
to figure what happened even with this basic example (I knew the
integral couldn't be so small!).
For a general integration in [0, infty), ther
adaptIntegrate: Result f(x) is not numeric or has wrong dimension
Best,
baptiste
On 21 September 2010 17:11, Hans W Borchers wrote:
> baptiste auguie googlemail.com> writes:
>
>>
>> Thanks, adaptIntegrate() seems perfectly suited, I'll just need to
>> figure a t
Got it, thanks!
baptiste
On 21 September 2010 22:38, Hans W Borchers wrote:
> baptiste auguie googlemail.com> writes:
>
>>
>> Thanks. I am having trouble getting adaptIntegrate to work with a
>> multivalued integrand though, and cannot find a working example.
>&g
Dear list,
I'm using lattice::levelplot to plot a coloured image of 3D data. The
range of the z values goes from negative to positive, but is not
exactly centred around 0. I would however like to map a diverging
colour scale with white falling exactly at 0, and both extremes being
symmetrical in t
Hi,
I remember a discussion we had on this list a few months ago for a
better way to decide if a point is inside a convex hull. It eventually
lead to a R function in this post,
http://tolstoy.newcastle.edu.au/R/e8/help/09/12/8784.html
I don't know if it was included in the geometry package in th
Try this,
qplot(outcome, counts, data=d.AD, facets=.~ treatment)
HTH,
baptiste
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide c
Hi,
Check the grImport package (I think it has a vignette, perhaps on Paul
Murrell's homepage.)
HTH,
baptiste
On 3 October 2010 14:52, Tal Galili wrote:
> Hello Dieter,
>
> Looking at this thread (from 2005)
> http://tolstoy.newcastle.edu.au/R/help/05/10/14320.html
> It seems you can't read
Hi,
grid.table in gridExtra might give you some inspiration.
HTH,
baptiste
On 13 October 2010 10:14, Joel wrote:
>
> It should look something like this (not at all relevant except the look)
>
> http://r.789695.n4.nabble.com/file/n2993297/tableR19.jpg
> --
> View this message in context:
> htt
Hi,
As an alternative, maybe you could use lattice::panel.levelplot.raster
which I think doesn't have this problem in pdf viewers.
HTH,
baptiste
On 26 October 2010 02:30, David Winsemius wrote:
>
> On Oct 25, 2010, at 6:50 AM, Mario Valle wrote:
>
>> Dear all,
>> I'm using R 2.12.0 on Windows
Hi,
Have a look at the directlabels package; it does just that for lattice
and ggplot2.
HTH,
baptiste
On 26 October 2010 08:02, Jeffrey Spies wrote:
> Hi, all,
>
> Let's say I have some time series data--10 subjects measured 20
> times--that I plot as follows:
>
> library(ggplot2)
> dat <- dat
Hi,
I think you want ?unlist
d = data.frame(x=1, y=2, z=3)
v = unlist(d)
is(v)
[1] "numeric" "vector"
HTH,
baptiste
On 31 October 2010 16:54, James Hirschorn wrote:
> Suppose df is a dataframe with one named row of numeric observations. I want
> to coerce df into a named vector.
>
>
>
> as.ve
Hi,
Regarding your '10 commandments' in Q3, you may find useful tips in
"the R inferno" by Pat Burns.
HTH,
baptiste
On 2 November 2010 05:04, Santosh Srinivas wrote:
> Hello Group,
>
> This is an open-ended question.
>
> Quite fascinated by the things I can do and the control I have on my
> ac
Hi,
try this,
xyplot(Time~Chromosome|factor(Elements),
data = mtx[order(mtx$Chromosome), ], ... [snipped])
HTH,
baptiste
On 7 November 2010 13:17, Alex Reynolds wrote:
> I have the following xyplot figure:
>
> http://img577.imageshack.us/img577/686/filesizeresults1200
Hi,
The easiest way might be the directlabels package from R-forge.
Otherwise, you could write your own panel function.
HTH,
baptiste
On 10 November 2010 23:01, Joon Yeong Kim wrote:
> Hi,
>
> I've been trying to find a way to label the the peak or mean of a
> densityplot for a while but have
Hi,
For curiosity's sake, below is another version with ggplot2 and Grid graphics,
library(pixmap)
logo <- read.pnm(system.file("pictures/logo.ppm", package="pixmap")[1])
library(ggExtra) # r-forge, requires gridExtra
qplot(rnorm(100),rnorm(100)) +
annotate("pixmap", x=-Inf, y=-Inf, picture=lo
Hi,
I'm joining in with a question -- is it possible to vary the color of
the lines along z? The 'colors' argument doesn't seem to allow a
vector in this situation.
Thanks,
baptiste
On 21 November 2010 21:02, Carl Witthoft wrote:
> Thanks, Dennis. Here's an enhanced version:
>
> z <- seq(
Apparently He who starts from 0 needn't be called unfortunate,
fortune('indexed')
baptiste
On 22 November 2010 20:59, Ben Bolker wrote:
> Bert Gunter gene.com> writes:
>
>>
>> Eh??? Why would you want to do that?? (R isn't C).
>>
>> So the simple answer is: you can't.
>>
>> The other answer is
fixing the various typos in your code, this works,
lat_plot() +
scale_fill_manual(value=mycolours)
HTH,
baptiste
On 23 November 2010 19:47, John Kane wrote:
> Someone was asking how to do a 16 category piechart in OpenOffice Calc and it
> appears that it can not be done (which we, probably, s
e any idea why
> scale_colour_manual(value=mycolours)
> does not work. More typos on my part or am I misunderstanding what
> scale_colour_manual(value=mycolours) is supposed to do?
>
> Thanks
>
>
> --- On Tue, 11/23/10, baptiste auguie wrote:
>
>> From: baptiste
Hi,
Try this,
do.call(`+`, x) / length(x)
HTH,
baptiste
On 24 November 2010 20:37, Tim Howard wrote:
> R users,
> This probably involves a simple incantation of one of the flavors of apply...
> that I can't yet figure out. Consider a list of data frames. I'd like to
> apply a function (mean
Hi,
Your function fails for a number of reasons. One of them is your
comparison (use browser() to see what is the value taken by f in your
function). Also, n, mean, min and max could not be extracted from ...
with your construction.
Here's my suggestion,
randomIra = function(f="runif", ...){
swi
Hi,
Embarrassingly enough, it was quite straight-forward in the first
versions of grid.table(). You might want to try with version r11 for
example,
source("http://gridextra.googlecode.com/svn-history/r11/trunk/R/tableGrob.r";)
library(grid)
tc = textConnection("
carat VeryLongWordIndee
Hi,
The easiest way to get the wide curly braces in your plot might be the
tikzDevice package. In its vignette you'll find an example of placing
an arbitrary tikz element in a plot, featuring a curly bracket. Sadly,
the internal coordinate system used by ggplot2 might make the
positioning a little
Hi,
Try the package orthopolynom on CRAN.
HTH,
baptiste
On 8 December 2010 12:51, Alaios wrote:
> Hello everyone,
> I would like to find out if there are already implemented function for
> legendre
> polynomials. I tried google but returns nothing. How do you suggest me to
> search
> for th
Hi,
Try the following,
plot(1:10,rnorm(10),t="o")
## fill the points in white
plot(1:10,rnorm(10),t="o",pch=21,bg="white")
You could also try this with Grid graphics,
library(gridExtra)
# like type="o"
grid.barbed(space=0)
# like type="b"
grid.barbed(space=1)
# like the example above, but with
Hi,
The fastest way seems to be,
all(x[1] == x)
HTH,
baptiste
On 16 December 2010 15:17, Jannis wrote:
> Dear list,
>
>
> this might be an easy one, but I could figure out a solution (or how to
> google the right term).
>
> Is there any way to test whether all elements of a vector are identi
Hi,
Also, see the gridbase package that gives you the flexibility of Grid viewports.
HTH,
baptiste
On 17 December 2010 09:33, Jim Lemon wrote:
> On 12/17/2010 02:00 PM, Dario Strbenac wrote:
>>
>> Hello,
>>
>> Is it possible to call a graphing function that uses layout() multiple
>> times and
Hi,
this seems to work,
plot.new()
legend("topleft", legend=as.expression(c(bquote(.(txt) ==
.(obv)*degree), "Von Mises distribution")))
HTH,
baptiste
On 28 December 2010 20:17, Tyler Dean Rudolph
wrote:
> legend("topleft", legend=c(bquote(.(txt) == .(obv)*degree), "Von Mises
> distribution"
See aes_string(), perhaps.
baptiste
On 1 January 2011 18:56, Mark Sharp wrote:
> I am wanting to change arguments to a function dynamically. For example, in
> making a call to qplot, I want to dynamically define all of the arguments so
> that I can create the plot dependent on user input. I ha
Hi,
embed() seemed well-suited, but I couldn't figure out an elegant way to use it
embed(c(A,A), 4)[1:4, 4:1]
HTH,
baptiste
On 6 January 2011 22:34, ADias wrote:
>
> Hi
>
> Suppose we have an object with strings:
>
> A<-c("a","b","c","d")
>
> Now I do:
>
> B<-matrix(A,4,4, byrow=F)
>
> and I
Hi,
Try this,
mode(m) <- "integer"
HTH,
baptiste
On 10 January 2011 10:17, emj83 wrote:
>
> Hi,
>
> I would like to turn my TRUE/FALSE matrix into a 1/0 matrix (i.e. True=1 and
> False=0)
>
> [,1] [,2] [,3]
> [1,] TRUE FALSE FALSE
> [2,] TRUE TRUE FALSE
> [3,] TRUE TRUE TRUE
>
>
Hi,
Try this,
m = melt(dat, id="Date")
head(m)
qplot(Date, value, data=m, colour=variable, geom="line")
ggplot(m) + facet_grid(variable~., scales="free_y") +
geom_path(aes(Date, value))
HTH,
baptiste
On 10 January 2011 14:12, Santosh Srinivas wrote:
> Hello R-Group,
>
> I am trying plott
yers,
qplot(Date, Close, data=dat,geom="line") +
geom_line(aes(Date, vol), colour="red")
HTH,
baptiste
>
>
> -Original Message-
> From: baptiste auguie [mailto:baptiste.aug...@googlemail.com]
> Sent: 10 January 2011 18:59
> To: Santosh Srinivas
>
Hi,
do.call(sum, mylist)
?do.call
baptiste
On 27 July 2010 14:36, Nicola Sturaro Sommacal
wrote:
> Hi!
>
> I have a list of 24 elements, all of the same type (dataframe, for example).
>
> I am looking for an alternative to mylist[[1]] + mylist[[2]] + ... +
> mylist[[24]] to obtain the sum.
>
>
Hi,
library(gridExtra)
example(patternGrob)
provides some patterns to fill a rectangular area using Grid graphics.
It could in theory be used in lattice. I wouldn't use it either, but I
can imagine how it might be useful on very special occasions.
Best,
baptiste
On 28 July 2010 06:11, HC
Hi,
To add tables, the gplots package has a textplot() function, and for
Grid graphics there is a grid.table() function in gridExtra.
HTH,
baptiste
On 5 August 2010 00:02, Ralf B wrote:
> Hi R Users,
>
> I need to produce a simple report consisting of some graphs and a
> statistic. Here simpli
Hi,
One way you could do it is to create a separate graph for each
category. The y axis labels would replace the strip labels. You could
then stack the graphs on the page, and add a common legend. The tricky
part would be to make sure the different panels have the same width
and height.
Another o
## hack: tweak ggplot2's axis.title.y option to use our gTree
foo <- function()
function(label, x, y)
ylab
p + opts(strip.text.y =theme_blank(),
strip.background=theme_blank()) +
opts( axis.title.y = foo())
HTH,
baptiste
On 12 August 2010 07:44, baptiste auguie wrot
Hi,
I see no need to construct the vector, try this instead,
belong = function(x=4, y=c(1,10)) x <= y[2] && x >= y[1]
see also ?findInterval
HTH,
baptiste
On 13 August 2010 01:10, fishkbob wrote:
>
> So basically I want to do this -
>
> 4 %in% 1:10
> should return true
>
> Would there be an
Dear list,
I wish to use a specific driver to process an sweave document in the
inst/doc directory of a package. Specifically, I would like to use
either cacheSweave or pgfSweave to speed up the creation of the
vignette which requires lengthy computations. The same request would
also apply to the
light and, say, pgfSweave in the same
document.
Sincerely,
baptiste
On 13 August 2010 11:10, Romain Francois wrote:
>
> Hi,
>
> I've been meaning to ask the same question before.
>
> Le 13/08/10 11:01, baptiste auguie a écrit :
>>
>> Dear list,
>>
>&
LEPHONE=C
> [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
>
> attached base packages:
> [1] grid stats graphics grDevices utils datasets methods
> [8] base
>
> other attached packages:
> [1] gridExtra_0.7
>
> loaded via a namespace (and not attach
Try this,
b = 20
plot(1, ylab= bquote(italic(P) * .(b)) )
HTH,
baptiste
On 19 August 2010 20:02, array chip wrote:
> Hi all, let me give a simple example:
>
> b<-20
> I would like to print ylab as "P20" where "P" is printed in Italic font. When
> I
> do the following:
>
> plot(1, ylab=expre
Dear list,
I'm using the brew package to generate a report containing various
plots. I wrote a function that creates a plot in png and pdf formats,
and outputs a suitable text string to insert the file in the final
document using the asciidoc syntax,
<%
tmp <- 1
makePlot = function(p, name=paste(
wrote:
> Not sure what you want. Plot does that automatically. It seems to use
> path.expand() to make the %03d expansion. Not that path.expand() is
> documented to do this, but it seem to work.
>
> Kees
>
> On Sat, 21 Aug 2010 13:04:54 +0200, baptiste auguie
> wrote:
>
&
27;') # function needed here
> png(real.name.png)
> print(p)
> dev.off()
> pdf(real.name.pdf)
> print(p)
> dev.off()
> cat(noquote(paste('image:',real.name.png,',width=',width,',link=real.name.pdf)
> }
>
> On Sat, 21 Aug 2010 14:02:04 +0200
(I really should start using these reading glasses).
>>
>> My apologies
>>
>> On Sat, 21 Aug 2010 14:20:17 +0200, baptiste auguie
>> wrote:
>>
>>> I dunno, it doesn't seem to do it for me,
>>>
>>> name = "Rplot%03d.png&q
Hi,
I think you could do it quite easily with lattice,
library(lattice)
latticeGrob <- function(p, ...){
grob(p=p, ..., cl="lattice")
}
drawDetails.lattice <- function(x, recording=FALSE){
lattice:::plot.trellis(x$p, newpage=FALSE)
}
plots <- replicate(4, xyplot(rnorm(10)~rnorm(10),xlab=
hi,
also, make sure you have set the aspect ratio to 1:1 when plotting (asp=1).
HTH,
baptiste
On 25 August 2010 10:20, Benno Pütz wrote:
> Maybe
>
> perp.slope = -1/slope
> abline(cy - cx*perp.slope, perp.slope)
>
> where cx, cy are x- and y-coordinate of C, resp., and slope the slope you
> c
Dear list,
I wish to visualise some 4D data as a kind of colour / translucent
cloud in 3D. I haven't seen such plots in R (but perhaps I missed a
feature of rgl). The easiest option I found would be to export the
data in povray's df3 (density file) format and visualise it with
povray.
The format
hi,
try this
lab =bquote(paste("Estimated ", t[50]," from ",.(what)))
HTH,
baptiste
On 27 August 2010 20:19, Dieter Menne wrote:
>
> plot.new()
> lab =expression(paste("Estimated ", t[50]," from tgv"))
> text(0.5,0.5,lab)
> # Should look the same as above. I could not get the substitute righ
Hi,
It's easy with ggplot2,
library(ggplot2)
## create an empty plot
p <- ggplot(map=aes(x,y))
## create a dummy list of data.frames with different ranges
d <- replicate(4, data.frame(x=sample(1:10,1)+rnorm(10),
y=sample(1:10,1)+rnorm(10)),
simplify=
Another way that I like is reshape::melt.list() because it keeps track
of the name of the original data.frames,
l = replicate(1e4, data.frame(x=rnorm(100),y=rnorm(100)), simplify=FALSE)
system.time(a <- rbind.fill(l))
# user system elapsed
# 2.482 0.111 2.597
system.time(b <- melt(l,id=1:2)
On 7 September 2010 17:19, Erik Iverson wrote:
> See ?grid.layout or perhaps ?arrange from the gridExtra package.
>
gridExtra::grid.arrange(), rather.
baptiste
> Abhijit Dasgupta wrote:
>>
>> Hi,
>>
>> Is there a function similar to the layout function in base graphics in
>> either lattice or gg
arrange() was renamed grid.arrange() when plyr started using this name
for a different function. I think it happened in version 0.6.5 of
gridExtra. The current version on CRAN is 0.7.
baptiste
On 7 September 2010 17:46, Erik Iverson wrote:
>
>
> baptiste auguie wrote:
>>
>>
Hi,
You can have each cell of a matrix contain a matrix, but for a reason
that is just not clear to me the matrices are wrapped in a list,
m = matrix(replicate(4,matrix(1:9,3,3),simplify=FALSE), 2,2)
m[1,2][[1]]
str(m)
and even more surprising to me, m itself has become a list for some reason.
Hi,
I get the same crash with x11() with sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-apple-darwin9.8.0
locale:
[1] en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.UTF-8/en_GB.UTF-8
attached base packages:
[1] grid stats graphics grDevices utils datasets methods
[8] base
However it works
Unfortunately, it seems that vcd doesn't return grobs but draws
directly to the device, which prevents a concise solution. You could
try the following,
library(gridExtra)
library(vcd)
data("Titanic")
p = grid.grabExpr(mosaic(Titanic))
grid.arrange(p, p, p, ncol=2)
Or, more versatile but also mor
Hi,
The package maintainer is aware of this feature request. In the
meantime, I've used Currying,
require(cubature)
f <- function(x, a) cos(2*pi*x*a) # a simple test function
adaptIntegrate(roxygen::Curry(f, a=0.2), lower=0, upper=2)
HTH,
baptiste
On 4 May 2011 05:57, Ravi Varadhan wrote
Hi,
Try this,
cat(format("The TITLE", width=80, justify="centre"))
HTH,
baptiste
On 5 May 2011 19:28, Dan Abner wrote:
> Hello everyone,
>
> I have a few questions about the print() fn:
>
> 1) I have the following code that does not center the character string:
>
> print("The TITLE",quote=FAL
Hi,
On 8 May 2011 21:18, Berwin A Turlach wrote:
> G'day Dan,
>
> On Sun, 8 May 2011 05:06:27 -0400
> Dan Abner wrote:
>
>> Hello everyone,
>>
>> I am attempting to use the %in% operator with the ! to produce a NOT
>> IN type of operation. Why does this not work? Suggestions?
Alternatively,
ex
I imagine mind_read() easy to implement with Robin Hankin's emulator
package -- under some weak assumptions about the user; mind_write(),
however, seems more involved and might require investing in new
hardware.
Best,
baptiste
On 21 May 2011 12:04, Rolf Turner wrote:
>
> On reflection, it seems
Hi,
There are probably much better ways, but try this
transform(dat, group = as.numeric(factor(paste(A,B,C, sep=""
HTH,
baptiste
On 31 May 2011 09:47, Mendolia, Franco wrote:
> Hello,
>
> I would like to create a group variable that is based on the values of three
> variables:
>
> For ex
I propose a Pi Haiku (PIQ),
Pi is of certain value,
In statistics, invaluable, yet
Transcending numerics.
Best,
baptiste
On 1 June 2011 11:55, Ravi Varadhan wrote:
> Nice to know that the `pi' can be sliced in so many different ways!
>
> There are exactly 3.154 x e08 seconds in a (non-leap) ye
A, B, C should have the same number of rows.
mlist = replicate(3, matrix(rnorm(6), 2), simplify=FALSE)
names(mlist) = LETTERS[seq_along(mlist)]
with(mlist, cbind(A,B,C))
or,
do.call(cbind, mlist)
HTH,
baptiste
On 5 June 2011 11:14, Jim Silverton wrote:
> How can I cbind three or more matrice
Hi,
Try this
ggplot(df, aes(x,y)) + geom_tile(aes(fill=height), colour="white") +
scale_fill_gradientn(colours = c("red", "gold", "green")) +
geom_text(aes(lab=height))
HTH,
baptiste
On 14 June 2011 07:12, idris wrote:
> I have a dataframe df with columns x, y, and height. I want to create a
Hi,
Lattice and ggplot2 are both ideally suited for this task. Consider
this example,
library(ggplot2)
d = data.frame(x=1:10, a1=rnorm(10), b1=rnorm(10))
m = melt(d, id ="x") # reshape into long format
qplot(x, value, data=m, geom="path", colour=variable)
library(lattice)
xyplot(value~x, data=m
Hi,
On 16 May 2010 03:31, michael westphal wrote:
[ snipped ]
> Any suggestions?
>
i'd suggest you
- read the posting guide
- upgrade your R to the latest version
- don't post to two mailing lists
- make your example minimal, self-contained, reproducible
- show the result of sessionInfo()
HTH,
Hi,
Try this,
saveMyWork <- .Last.value
HTH,
baptiste
On 17 May 2010 15:07, math_daddy wrote:
>
> Hello.
>
> I ran a simulation that took a few days to complete, and want to analyze the
> results, but have just realized that I (idiotically) did not assign the
> output to a variable when I int
Hi,
try this,
m = matrix(runif(2000*2400), nrow=2000)
library(grid)
grid.raster(m)
HTH,
baptiste
On 17 May 2010 20:35, tetonedge wrote:
>
> I have a matrix that is 2400x2000 and I would like to display it as an image,
> I have tried image(), but due to the size of the matrix the drawing of t
No, that's only true for lattice and ggplot2 graphics. The problem
here is with this line,
windows(width=5, height=5)
which shouldn't be there.
HTH,
baptiste
On 17 May 2010 22:23, Jun Shen wrote:
> If you do plotting in a loop, then you need to print it to the device.
>
> print(plot(xj,y))
>
Dear all,
I got a couple of warnings using panel.levelplot.raster,
In panel.levelplot.raster(..., interpolate = TRUE) :
'y' values are not equispaced; output will be wrong
although I was quite sure my data were equally spaced (indeed, I
created them with seq()). A closer look at the source cod
On 18 May 2010 15:30, Deepayan Sarkar wrote:
> Maybe a better test would be
>
> isTRUE(all.equal(diff(range(diff(ux))), 0))
>
> I'll try that out for the next release.
>
Sounds good (and works for me), thanks.
baptiste
__
R-help@r-project.org mailin
Dear list,
I am puzzled by this,
substitute(expression(x), list(x = factor(letters[1:2])))
# expression(1:2)
Why do I get back the factor levels inside the expression and not the
labels? The following work as I expected,
substitute(expression(x), list(x = letters[1:2]))
# expression(c("a", "b")
Thank you for the explanation, and the fortune-ish quote,
“As the documentation for substitute() says, there is no guarantee
that the result makes sense.”
Best,
baptiste
On 19 May 2010 02:59, Duncan Murdoch wrote:
> On 18/05/2010 4:36 PM, baptiste auguie wrote:
>>
>> Dear l
Hi,
See also ?lattice::xyplot and ?ggplot2::geom_point , either one can do
it automatically.
HTH,
baptiste
On 19 May 2010 12:24, Jannis wrote:
> Dears,
>
> before I start programming my own function I would like to ask you whether
> there is any function already available that lets me plot a x
Hi,
ggplot2 or lattice could help you in creating the plots. Adding a
summary will however require some play with Grid graphics; either
using gridBase to mix lattice / ggplot2 output with base R graphics
(e.g. textplot() from some package I forget), or you'll need to
produce the textual summary i
On 1 June 2010 11:34, Peter Ehlers wrote:
> Or, for a very slight further reduction in time in
> the case of larger matrices/vectors:
>
> as.vector(tcrossprod(V, xyzs))
>
> I mention this merely to remind new users of the
> excellent speed of [t]crossprod().
>
> -Peter Ehlers
Thanks, I've been
Hi,
You could use melt from the reshape package to create a long format
data.frame. This is more easy to plot with lattice or ggplot2, and you
can then use facetting to arrange several plots on the same page. The
dummy example below produces 10 pages of output with 10 graphs per
page.
library(ggp
Hi,
It's not clear what you mean by summary text without a minimal
reproducible example. If your text is ordered as a matrix or a
data.frame, you might want to try this grid function,
gridExtra::grid.table(as.matrix(summary(iris)), theme=theme.white())
If your text has the form of a paragraph, t
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