You could draw a grid with grid, using grid.grill,
library(grid)
pdf("grid.pdf", width=21/2.54,height=29.7/2.54)
grid.grill(h = unit(seq(0, 297, by=1), "mm"),
v = unit(seq(0, 210, by=1), "mm"), gp=gpar(col="grey",lwd=0.1))
grid.grill(h = unit(seq(0, 297, by=5), "mm"),
v = un
One reason might be that you can easily fool the user into running
unexpected/unreadable commands. Guess what this does:
cmd <- paste(c(letters[c(19L, 25L, 19L, 20L, 5L, 13L)], "(' ",
letters[c(19L, 21L, 4L, 15L)], " ", letters[c(4L,
5L, 19L, 20L, 18L, 15L, 25L)], " ", letters[c(1L, 12L, 12L)], "
The scales package?
gradient_n_pal(c("red","white","blue"), c(0,0.5,1)) (seq(0,1,length=100))
HTH,
baptiste
On 26 January 2012 10:36, Jeffrey Joh wrote:
>
> The gray (level) function returns different shades of gray, where level is a
> vector of numbers ranging from 0 to 1. 0 is white and 1
Hi,
which(x < 15)
omits the NA (treated as false).
HTH,
b.
On 29 January 2012 09:36, Federico Calboli wrote:
> Dear All,
>
> just a quick example:
>
>
>> x = 1:25
>> x[12] = NA
>
>> x
> [1] 1 2 3 4 5 6 7 8 9 10 11 NA 13 14 15 16 17 18 19 20 21 22 23 24 25
>
>> y = x[x<10]
>> y
> [1] 1 2
Try this,
plot(1:10)
img <- grid::grid.cap()
# grid.raster(img)
stream <- col2rgb(img)
write.table(stream, file="dam.txt", row.names = FALSE,
col.names = FALSE)
(you'll have to restore the dimensions of the matrix once you've read
the rgb values for each pixel)
HTH,
baptiste
On 30
Hi,
If you don't mind having NAs for missing values, try the following,
mylist = list(1:3, 4:7)
library(plyr)
write.csv(do.call(rbind.fill.matrix, lapply(mylist, matrix, nrow=1)), file="")
HTH,
b.
On 6 February 2012 15:01, Michael wrote:
> Hi all,
>
> I have a list of vector of numbers - the
Hi,
If you read French, you might find the following discussion interesting,
http://www.forum.math.ulg.ac.be/viewthread.html?id=45765
It contains some good suggestions to project an ellipsoid onto a
plane, which as I understand might be related to your question.
HTH,
b.
On 8 February 2012 0
On 9 February 2012 13:02, David Winsemius wrote:
>
> On Feb 8, 2012, at 6:33 PM, Tom Roche wrote:
>
>>
>> Peter Langfelder Thu Feb 9 00:01:31 CET 2012
>>>
>>> I'm exploring using a version control system
>>
>>
>> +1! welcome to the new millenium :-)
>>
>>> to keep better track of changes to the [R
Once upon a time r-forge had the option to sync from an external svn
repository, e.g. hosted on googlecode. I haven't seen it available for
some time, sadly. I'm sure many users would appreciate if this feature
came back with the new interface. Not sure if it could work with git
as well, though.
b
A minimum code to plot a coloured matrix with text labels could be the
following:
library(grid)
library(scales)
library(RColorBrewer)
diverging_palette <- function(d = NULL, centered = FALSE, midpoint = 0,
colors = brewer.pal(7,"PRGn")){
half <- length(colors)/2
The default units of polygonGrob are "npc", I think you want "native" instead.
Try the following,
library(grid)
d = data.frame(x=rnorm(100, 10), y=rnorm(100, -100))
v = dataViewport(xData=d$x, yData=d$y)
grid.points(d$x,d$y, default.units="native", vp=v)
HTH,
b.
On 17 February 2012 02:47,
Hi,
the grImport package provides some tools for this.
HTH,
b.
On 19 February 2012 16:06, Nick Matzke wrote:
> Hi,
>
> Is there a way to parse a postscript (*.ps) file with R (or perhaps with
> some other command-line utility)?
>
> E.g., I have a map in postscript format with lots of features,
library(plyr)
adply(my.array,1:3)
HTH,
baptiste
On 20 April 2012 08:46, Emmanuel Levy wrote:
> Hi,
>
> I have a three dimensional array, e.g.,
>
> my.array = array(0, dim=c(2,3,4), dimnames=list( d1=c("A1","A2"),
> d2=c("B1","B2","B3"), d3=c("C1","C2","C3","C4")) )
>
> what I would like to get
Hi,
On 24 April 2012 05:00, Duncan Murdoch wrote:
> On 23/04/2012 10:49 AM, Adam Wilson wrote:
>>
>> I routinely write graphics into multi-page PDFs, but some graphics (i.e.
>> plots of large spatial datasets using levelplot()) can result in enormous
>> files. I'm curious if there is a better wa
Hi,
try also grid.table() in gridExtra
b.
On 27 April 2012 01:26, statquant2 wrote:
> Hello,
> I would like to be able to plot an array on a plot, something like:
> |arg1 | arg2 | arg3
> val1| 0.9 | 1.1 | 2.4
> val2| 0.33 | 0.23 | -1.4
> val3| hello| stop | test
> I know Rwave is
For better typography, try tikzDevice, it uses LaTeX to render the text.
b.
On 16 May 2012 07:23, Fisher Dennis wrote:
> David
>
> You missed the point -- the issue was not the spacing between WORDS. It was
> the spacing between LETTERS (as noted in the original email)
> Other suggestions woul
Your log10(HCO3 and sqrt(HCO3 seem to be missing closing brackets.
HTH,
baptiste
On 18 May 2012 11:34, Rich Shepard wrote:
> One of many scripts to produce 4 lattice plots on one page keeps throwing
> an error. I've tried manipulating the file to eliminate the error, but have
> not been able t
Try this,
bquote(.(L1) * .(L2))
HTH,
b.
On 19 May 2012 16:17, Rolf Turner wrote:
>
> In the context in which I am actually working it is necessary for
> me to build parts of a text string to place on a plot in two separate
> stages.
>
> The following toy example illustrates what I am trying to
You can rotate the viewport to flip around the horizontal axis,
library(grid)
grid.text("Chiral")
grid.text("Chiral", vp=viewport(angle=180, y=unit(0.5,"npc")-unit(1,"line")))
HTH,
b.
On 23 May 2012 05:34, Thomas Zumbrunn wrote:
> Maybe my question was not concise enough. I was referring to ob
Oops, sent too early; this obviously just a rotation, not a mirror
image. It illustrates the problem though ;)
b.
On 23 May 2012 07:32, baptiste auguie wrote:
> You can rotate the viewport to flip around the horizontal axis,
>
> library(grid)
> grid.text("Chiral")
&g
you can open a device that has the exact dimensions of the table,
g = tableGrob(head(iris, 4))
png("test.png", width=convertWidth(grobWidth(g), "in", value=TRUE),
height=convertHeight(grobHeight(g), "in",
value=TRUE),units="in", res=150)
grid.draw(g)
dev.off()
Doing this with knitr might
2012 09:43, Alexander Shenkin wrote:
> this works - thanks baptiste! i'm working in Sweave right now - perhaps
> it will be tough in knitr as you mention.
>
> On 5/25/2012 4:31 PM, baptiste auguie wrote:
>> you can open a device that has the exact dimensions of the table,
&
Dear list,
I have produced a fairly intricate plot arrangement for use in a
publication using layout() and gridBase, and out of curiosity I'd like
to learn whether a more elegant and robust solution could be obtained
with grid to avoid the layout() function.
library(gridBase)
x <- seq(
Hi,
I think you want to Vectorize integrate(), not integrand(). Here's a
way using mapply,
integrand <- function(z)
{
return(z * z)
}
vec1<-1:3
vec2<-2:4
mapply(integrate, lower=vec1, upper=vec2, MoreArgs=list(f=integrand) )
baptiste
On 20 Sep 2008, at 13:08, Andreas Wittmann wrote:
D
Hi,
you can use par.settings to get a consistent color scheme,
pal1 <- rgb(196, 255, 255, max = 255)
pal2 <- rgb( 0, 35, 196, max = 255)
df <- data.frame( Gruppe = c("A", "B", "A", "B"),
Kat = c("x1", "x1", "w1", "w1"),
value= c(1,2, 4, 5))
barchart(value ~ Kat, group=
just a thought: you might want to have a look at the brew package (R-
forge). Instead of source()ing the document in the R console, you'd
use brew(). By default, any text is considered a comment, only lines
contained between <% %> will be run as R code.
baptiste
On 24 Sep 2008, at 21:03, M
Hi,
I'm not sure I understand your example, it'd be easier with some
reproducible code. I think you'd better off using grid, in particular
the figure 5.22 from the book R Graphics provides an example of
embedding a lattice graph in a page. Here's a simplified version,
library(grid)
librar
Hi,
I think you want the following,
df <- data.frame(x=rnorm(100), y=rnorm(100))
plot(df)
legend("topright", title="Land Use Type", cex=0.75, pch=16,
col=c("red", "green"),legend=c("Urban", "green"), ncol=2)
I could not find a way to have a different text size for the title,
Dear list,
I'm using solve(A,b) to solve (!) a linear system, where A is a large
(100 x 100) symmetric, dense, complex matrix. The precise structure of
A is block-Toeplitz, which means that the best algorithm for inversion
would probably be based on fft.
Anyway, I'm looking for an iterativ
you mean like ?source ?
On 26 Sep 2008, at 13:49, Michael Schulte wrote:
Dear R-people,
I want to use an idea from LaTeX in the work flow with R.
It is possible in LaTeX to have a main file from which other files
are called (ie included).
So for example if you have book, the main index file
?attach
or ?with
Hope this helps,
baptiste
On 26 Sep 2008, at 14:57, Stefan Fritsch wrote:
Dear R Users,
another problem for me is the output of a function.
I have several output variables which I give back with the list
command.
test <- function {return(list(a,b,c,d,e,f,g,...))}
Hi,
?unique should work fine for this.
I think the problem in your implementation is that you modify x in the
loop, in particular its length may become shorter than the stopping
condition.
HTH,
baptiste
On 27 Sep 2008, at 14:30, Bastian Offermann wrote:
Hello all,
one brief question
Thanks Hadley, for some reason I didn't see your email until now. It
works fine with the development version,
library(plyr)
df <- data.frame(a=1:10 , b=1:10)
foo1 <- function(a, b, cc=0, d=0){
a + b + cc + d
}
mdply(data. = df, foo1, cc=1, d=2)
I think using . prefixes is a safer optio
Dear list,
I've been trying this for a few hours and I just don't understand how
lattice works with groups and subscripts.
Consider the following example,
xx <- seq(1, 10, length=100)
x <- rep(xx, 4)
y <- c(cos(xx), sin(xx), xx, xx^2/10)
fact <- factor(rep(c("cos", "sin", "id", "square")
Hi, and thanks for your email,
I realise my example was not very good. The actual dataset I'm trying
to plot is rather big and this oversimplified example did not make
much sense.
I actually do need to color different subsets of the data differently
in each panel, that's why I thought of
Many thanks, I think I got the spirit of 'capturing and overriding'
the arguments which was the bit i was missing. It's much clearer now
with a working example.
Thanks again,
baptiste
On 7 Oct 2008, at 21:19, Deepayan Sarkar wrote:
On Tue, Oct 7, 2008 at 8:54 AM, baptiste
It worked for me, do you have the latest version of ggplot2 released a
few days ago (ggplot2_0.7) ?
Baptiste
On 9 Oct 2008, at 20:55, stephen sefick wrote:
Error in `[.data.frame`(df, , var) : undefined columns selected
I got this error in a fresh R session after rerunning all of the
comm
Hi
it might be as simple as adding type = "b" in your call, however if
you need more help you'll have to provide a reproducible example and
explain what package you used (I think several packages define a
plotCI function).
Hope this helps,
Baptiste
On 10 Oct 2008, at 22:15, Caio Azeved
Hi,
I believe you want to look at ?strip.custom
someStuff <- data.frame(area = rep(c("SOUTH", "NORTH", "EAST",
"WEST"), each
= 25),
group = rep(c("A","B","C","D"), each = 5),
mytime = rep(1:4),
val1 = sample(1:100, size=100, replace=TRUE),
Hi,
I use Textmate, but every now and then I like to try out aquamacs.
I've just downloaded it from < http://aquamacs.org/ >, where ESS is
part of the package. It runs flawlessly for me, out of the box. I just
opened a r file, clicked the big R icon, then simply highlighted part
of the cod
Hi,
I feel that your example isn't exactly minimal so I may be completely
overlooking your question. Would the following do?
library(lattice)
mdf <- data.frame(x <- seq(0, pi, l=100), y=sin(x))
xyplot(y~x, data=mdf, type="b",
par.settings=list(plot.symbol=list(pch=21, col="r
Dear list,
I've generated a list of 3D coordinates representing ellipsoids in
arbitrary orientations. I'm now trying to obtain a 2D projection of
the scene, that is to draw the silhouette of each object on a plane
(x,y). The only way I could think of is to compute the convex hull of
the
I wrote a dirty hack last time I faced this problem, I'll be curious
to see what is the proper way of dealing with the scoping and
evaluation rules.
library(datasets)
myfunction<-function(table, extraction) {
table2<-subset(table,extraction)
return(table2)
}
condition1 <- quote(iris$S
I thought this was a good candidate for the plyr package, but it seems
that l*ply functions are meant to operate only on separate list
elements:
> Lists are the simplest type of input to deal with because they are
> already naturally
> divided into pieces: the elements of the list. For this
at 18:53, hadley wickham wrote:
On Tue, Dec 30, 2008 at 10:21 AM, baptiste auguie
wrote:
I thought this was a good candidate for the plyr package, but it
seems that
l*ply functions are meant to operate only on separate list elements:
Lists are the simplest type of input to deal with
PM, Gabor Grothendieck
wrote:
Try:
do.call(abind, c(foo, along = 3))
On Tue, Dec 30, 2008 at 1:15 PM, baptiste auguie
wrote:
In fact, when writing my post I tried to do exactly what you did
in creating
a 3d array from the list, and I failed miserably! This is (imho)
partly
because the syn
That's really a scilab question, nothing to do with R as far as I can
see. Moreover, you haven't provided any of the information requested
in the posting guide (OS, example, ...).
i'm guessing that something along those lines should work,
system("scilab -nw -f yourscript.sce")
that is, p
Dear list,
I'm having second thoughts after solving a very trivial problem: I
want to extend the relevel() function to reorder an arbitrary number
of levels of a factor in one go. I could not find a trivial way of
using the code obtained by getS3method("relevel","factor"). Instead, I
thou
was missing something obvious, but I was
wrongly assuming that the changes needed were more drastic. I'm now
wondering why this wasn't implemented in relevel() in the first place.
Perhaps such a small modification could be useful, at least in the
documentation?
Thanks,
bapt
nner
The combination of some data and an aching desire for an answer does
not ensure that a reasonable answer can be extracted from a given
body of data.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] Namens baptiste auguie
you can also look at subset,
my.data.frame <- data.frame(a=rnorm(10),
b=factor(sample(letters[1:4], 10, replace=T)))
str(my.data.frame)
my.data.frame[my.data.frame$b == "a", ]
subset(my.data.frame, b == "a")
by the way, it is probably safer not to use "data" as a va
Hi,
I think this is a very common question on this list. I've just created
a page in the R wiki (inspired by https://stat.ethz.ch/pipermail/r-help/2007-May/132466.html)
. With some suggestions and improvements, hopefully we can make a good
reference for others to refer to in the future:
ht
Your dummy data is not a reproducible example. I'm guessing ?unique
could help you.
Hope this helps,
baptiste
On 14 Jan 2009, at 13:19, glenn wrote:
For a list say;
list1<-{1,2,3,4,5,2,1}
How do I remove the duplicates please?
My real list is 20,000 obs long of dates with many duplicates
Dear list,
This is a bit of an off-topic question, but I'm hoping to get some
advice from more experienced people. I've used the website "Web of
Science" to manually collect publication counts responding to several
keywords as a function of date, since the 1960s.
http://apps.isiknowledge.
_ylo=&as_yhi=1960&as_allsubj=all&hl=en&lr=
here it is trivial to create such a string with the desired keyword
and dates, and retrieve the number of results using readLines(url) and
grep.
Thanks to Phil Spector for some pointers.
Best wishes,
baptiste
On 14 Jan 2009, at 13:44,
s not
do what I wanted
# the bit to retrieve is the number below
# 10 of about 21,900 for photonics
z
}
retrieveNumberPublications( makeURLsearch("photonics", c(2008,
NULL)) )
I can isolate the long string containing the result I want, but not
single out t
Have you tried c() from the latticeExtra package?
It worked for me (see below)
library(grid)
library(lattice)
x <- seq(0, 10, length=100)
y <- sin(x)
y2 <- 10*sin(x)
f <- rep(c("1", "2"), each=50)
p1 <- xyplot(y~x,groups=f, ylab="BIG LABEL",
# auto.key=list(space="right"),
par.settings = list
I guess by workspace you mean global environment. I believe this is
generally considered a bad practice, but see ?assign and ?"<<-"
baptiste
On 27 Jan 2009, at 13:54, diego Diego wrote:
Hello experts!
Is there a way to send an internal variable from a function to the
workspace, besides
Try this:
plot(1:20)
axis(3, at=seq(0,20), label=FALSE)
A better description of your plot would be useful if ?axis is not
enough to help you out.
hope this helps,
baptiste
On 27 Jan 2009, at 14:13, mau...@alice.it wrote:
Is there a way to force the number of ticks along an axis ?
I re
Hi,
If all else fails, you could consider using LaTeX itself with psfrag,
or perhaps a similar idea involving eps2pgf.
http://biostat.mc.vanderbilt.edu/twiki/bin/view/Main/PsFrag
Hope this helps,
baptiste
On 29 Jan 2009, at 11:24, Rau, Roland wrote:
Dear all,
I would like to plot the da
Hi,
Perhaps this can help if you don't want to manually specify the
permutation of indices,
A=matrix(10,ncol=2,nrow=2)
B <- 2*A
C <- rbind(A, B)
C[ as.vector(t(matrix(seq(1,nrow(C)),ncol=2))), ] # trick to create
the vector of permutations
[,1] [,2]
[1,] 10 10
[2,] 20 20
[
Another option,
library(ggplot2)
qplot(year, value, data=melt(foo), color= L1)
which can also be achieved "by hand",
test<- do.call(rbind,foo) # combines all data.sets
test$name <- do.call(rep, list(x=names(foo), times =
unlist(lapply(foo,nrow # append the name of the original dataset
I don't know about the maptools package but one general way to do this
would be to compute the convex hull (?chull) of the augmented set of
points and test if the point belongs to it.
Hope this helps,
baptiste
On 5 Feb 2009, at 13:21, Aleksandr Andreev wrote:
In R's maptools package, i
Perhaps this is what was intended?
sims <- list(length=100)
do.call(seq, sims)
seq by itself does not expect a list, but do.call() can create the
appropriate call if a list is what you want to pass to the function.
Hope this helps,
baptiste
On 5 Feb 2009, at 19:46, Uwe Ligges wrote:
Dear list,
This is quite a specific question requiring the package orthopolynom.
This package provides a nice implementation of the Legendre
polynomials, however I need the associated Legendre polynomial which
can be readily expressed in terms of the mth order derivative of the
correspon
A powerful scheme for harmonic inversion of time signals known as
"filter diagonalization method" is available from MIT: http://ab-initio.mit.edu/wiki/index.php/Harminv
I don't know of any R interface, but it might be a good option for
your problem.
Cheers,
baptiste
On 10 Feb 2009, at 13
In a different perspective the sage project might also be an option,
it seems to interface to Maxima and R among other things. I haven't
tested it myself though.
http://www.sagemath.org/index.html
Best wishes,
baptiste
PS: sagemath.org is a well-thought website, perhaps a good
inspirat
A somewhat twisted approach that has not been mentioned is to consider
everything a comment unless it is enclosed in special tags, as done in
the brew package,
for example,
brew(textConnection(
"You won't see this R output, but it will run. <% foo <- 'bar' %>
Now foo is <%=foo%> and toda
On 11 Feb 2009, at 13:41, Gustaf Rydevik wrote:
On Wed, Feb 11, 2009 at 2:15 PM, baptiste auguie
wrote:
A somewhat twisted approach that has not been mentioned is to
consider
everything a comment unless it is enclosed in special tags, as done
in the
brew package,
for example,
brew
lattice and ggplot2 also offer a general way of doing this,
# first create a data.frame in the long format containing the two
data sets
x1 <- seq(-10, 10)
x2 <- seq(-8, 12)
y1 <- sin(x1/3)
y2 <- cos(x2/2)
d1 <- data.frame(x=x1, y=y1, var="1")
d2 <- data.frame(x=x2, y=y2, var="2")
library(res
Hi,
I think Reduce could help you.
DF1 <- data.frame(var1 = letters[1:5], a = rnorm(5))
DF2 <- data.frame(var1 = letters[3:7], b = rnorm(5))
DF3 <- data.frame(var1 = letters[6:10], c = rnorm(5))
DF4 <- data.frame(var1 = letters[8:12], d = rnorm(5))
g <- merge(DF1, DF2, by.x="var1", by.y="var1"
Another option using Recall,
merge.rec <- function(.list, ...){
if(length(.list)==1) return(.list[[1]])
Recall(c(list(merge(.list[[1]], .list[[2]], ...)), .list[-(1:2)]), ...)
}
my.list <- list(DF1, DF2, DF3, DF4)
test2 <- merge.rec(my.list, by.x="var1", by.y="var1", all=T)
all
s names
Reduce(function(x, y) merge(x, y, all=T,by.x="var1", by.y="var1"),
list(DF1, DF2, DF3, DF4), accumulate=F)
Error in match.names(clabs, names(xi)) :
names do not match previous names
- Lauri
2009/2/19 baptiste auguie :
Hi,
I think Reduce could help you.
DF1 &l
Perhaps you can try this,
d <- c(0.00377467, 0.00377467, 0.00377467, 0.00380083,
0.00380083, 0.00380083,
0.00380959, 0.00380959, 0.00380959, 0.00380083, 0.00380083,
0.00380083)
c( t( cbind(matrix(d, ncol=3, byrow=T), 0)))
I don't know how to avoid the transpose operation that m
thanks all for the correction, funny how it's often the complicated
solution that comes to mind first.
baptiste
On 19 Feb 2009, at 13:41, Eik Vettorazzi wrote:
actually
c(rbind(0,matrix(d, nrow=3)))
which has the bonus of giving the desired result ;)
baptiste auguie schrieb:
Pe
Another approach using latticeExtra, more ggplot2-like:
p <- xyplot(matter~year|plot,type="l")
p +
latticeExtra::layer(panel.abline(v=1995))
On 20 Feb 2009, at 09:34, Chris Bennett wrote:
Hi,
I want to add a dashed vertical line to a number of xyplots.
Here is a simple script of the typ
Hi, try this:
p <- xyplot(matter~year|plot,type="l")
update(p, panel=function(...){
panel.xyplot(...)
panel.abline(v=1995)
} )
On 20 Feb 2009, at 09:34, Chris Bennett wrote:
Hi,
I want to add a dashed vertical line to a number of xyplots.
Here is a simple script of the t
Hi,
something like this perhaps,
create_string <- function(.s){
result <- read.table(textConnection(.s))
sapply(result, as.character)
}
(test <- create_string("ab cd ef"))
hope this helps
baptiste
On 20 Feb 2009, at 16:38, Sean Zhang wrote:
My dear R-helpers:
I am a novic
Paul Murrell's book provides such an example using Grid (figure 5.22).
A short example is available on his website:
http://www.stat.auckland.ac.nz/~paul/grid/doc/moveline.pdf
It may be possible to use this in conjunction with gridBase.
baptiste
On 22 Feb 2009, at 20:43, Eik Vettorazzi wrot
Hi,
I got this problem once, and Prof. Ripley kindly added an example in
the help page of ?cut,
>
> aaa <- c(1,2,3,4,5,2,3,4,5,6,7)
>
> ## one way to extract the breakpoints
> labs <- levels(cut(aaa, 3))
> cbind(lower = as.numeric( sub("\\((.+),.*", "\\1", labs) ),
> upper = as.numeric(
Hi,
Perhaps Hadley's plyr package can help,
library(plyr)
temp <- list(x=2,y=3,x=4)
llply(temp, function(x) x^2 )
$x
[1] 4
$y
[1] 9
$x
[1] 16
baptiste
On 27 Feb 2009, at 03:07, Alexy Khrabrov wrote:
Sometimes I'm iterating over a list where names are keys into another
data structure, e
Hi,
you probably want to use ?all.equal instead of "=="
I couldn't run your example, though
Hope this helps,
baptiste
On 27 Feb 2009, at 10:32, Peterko wrote:
hi i am creating some variables from same data, but somewhere is
different
rouding.
look:
P = abs(fft(d.zlato)/480)^2
hladane= s
Hi,
you could do one of the following,
1) combine a, b, c, d, e in a list and use ?lapply
my.list <- list(a,b,c,d,e)
lapply(my.list, foo)
where foo() is a function to be applied to each individual element
2) alternatively, see ?get to retrieve the value of a variable from
its name. Your
If you have many such repetitions it can be annoying to type the
rbind(cbind ... sequence. Perhaps this would help,
m <- read.table(textConnection("
IndividualsValue
A3
B4
C5
D2"), head=T)
data.frame(groups= rep(c("group1", "group2", "group3"),
each=nrow(m)), Ind=m[, 1]
What's wrong with geom_text?
my.value = 0.65
qplot(1,1)+geom_hline(v=0)+
geom_text(mapping=aes(x=1,y=0),label=paste(my.value),vjust=-1)
baptiste
On 3 Mar 2009, at 18:10, Dave Murray-Rust wrote:
Hello,
I'm using geom_hline to add a minimum line to my plot (representing
the best s
"test")
I'm sure Hadley will come up with a better explanation.
Best,
baptiste
On 3 Mar 2009, at 19:51, Dave Murray-Rust wrote:
On 3 Mar 2009, at 18:41, baptiste auguie wrote:
What's wrong with geom_text?
my.value = 0.65
qplot(1,1)+geom_hline(v=0)+
geom_text(mapping
Dear list,
I'd like to ask for some advice in creating a wrapper function for
this C code,
http://www.fourmilab.ch/documents/specrend/
http://www.fourmilab.ch/documents/specrend/specrend.c
I could probably port it in R, but I've been hoping to use compiled
code for a while and this looks
Hi,
Try:
polygon(c(x1, rev(x2)), c(y1, rev(y2)), col="grey")
In general you might need to make sure the data is well ordered.
Hope this helps,
baptiste
On 8 Mar 2009, at 16:52, Martin Batholdy wrote:
hi,
the code below produces me two curved lines.
Now I want to fill the space betwe
Hi,
You could use the reshape package:
d$e <- e
recast(d, variable~e, fun=sum)
The doBy package is another option.
baptiste
On 8 Mar 2009, at 17:14, soeren.vo...@eawag.ch wrote:
A dataframe holds 3 vars, each checked true or false (1, 0). Another
var holds the grouping, r and s:
### s
Hi,
Try this:
DF1 <- data.frame(var1 = letters[1:5], x = rnorm(5), y =2)
DF2 <- data.frame(var1 = letters[3:7], x = rnorm(5), y=3)
DF3 <- data.frame(var1 = letters[6:10], x = rnorm(5), y=0)
# ... DF10 if you wish
( result <- merge_all(list(DF1, DF2, DF3) ))
save( result, file ="merged.rda")
-1.387224 2 0.61761 31
0.9774042 -1.387224
d 0.6459462 0.46152 31
1.3345912 0.645946
e 0.0587832 -0.2531231
0.631676 2 0.058783
baptiste auguie-2 wrote:
Hi,
Hi,
On 9 Mar 2009, at 15:32, Sean Zhang wrote:
> Dear R-helpers:
> I am an R newbie and have a question related to writing functions that
> accept unlimited number of input arguments.
it's usually through the ... argument, e.g in paste(...).
> (I tried to peek into functions such as paste and
) Reduce("append", x)
> add(list(vec1, vec2))
> #add(vec1,vec2) does not work at the moment
>
> -sean
>
>
>
> On Mon, Mar 9, 2009 at 11:50 AM, baptiste auguie
> wrote:
> Hi,
>
>
> On 9 Mar 2009, at 15:32, Sean Zhang wrote:
>
>> Dear R-help
try
?paste
baptiste
On 10 Mar 2009, at 20:01, ig2ar-s...@yahoo.co.uk wrote:
Hi again R-ists,
How do you construct a string that you can pass to system()?
For instance. Say I do
system("echo Hello!")
Hello!
That works. Now the alternative: I need to construct the string like
this
On 12 Mar 2009, at 13:22, richard.cot...@hsl.gov.uk wrote:
I'm trying to write a loop to sum my data in the following way:
(the second - the first) + (the third - the second) + (the fourth -
the third) + ...
for each column.
This is just sum(diff(x)), or even x[length(x)] - x[1].
I think r
Hi,
your example is quite messy (neither reproducible or minimal). I think
you could try the following,
mdf <- data.frame(1:3)
names(mdf) <- "147"
i <- 147
mdf[ as.character(i) ]
Hope this helps,
baptiste
On 11 Mar 2009, at 22:34, Mohan Singh wrote:
Hi everyone
I am trying to
Hi,
For a good discussion of the link between colour and spectra I would
suggest,
http://www.fourmilab.ch/documents/specrend/
which provides an open-source C code to perform the conversion you ask
for. I asked for some advice on how to wrap a R function around this
code last week but sad
On 14 Mar 2009, at 13:08, Jinsong Zhao wrote:
Hi,
For a good discussion of the link between colour and spectra I would
suggest,
http://www.fourmilab.ch/documents/specrend/
which provides an open-source C code to perform the conversion you
ask
for. I asked for some advice on how to wrap a
h/documents/specrend/
On 14 Mar 2009, at 12:33, baptiste auguie wrote:
Hi,
For a good discussion of the link between colour and spectra I would
suggest,
http://www.fourmilab.ch/documents/specrend/
which provides an open-source C code to perform the conversion you
ask for. I asked for s
Hi,
I think you could get a cleaner solution using ?cut to split your data
in given ranges (the break argument), and then using this factor to
give the appropriate percentage.
Hope this helps,
baptiste
On 15 Mar 2009, at 20:12, diegol wrote:
Using R 2.7.0 under WinXP.
I need to write
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