[correcting a stupid error in my previous post]
testTwoStages <- function(x, y, head.stop = 100){
if(!isTRUE(all(head(x, head.stop) == head(y, head.stop
{
print(paste("quick test returned FALSE"))
return(FALSE)
} else {
full.test = isTRUE(all(tail(x, length(x) - head.stop) == tail(
Steve Jaffe wrote:
The situation is that I know there is a function and know approximately what
the name is, and want to find the exact name. Is there a way of searching
for near-matches (similar to unix apropos). For example, I know there is a
function called something like allequal (or allequal
Hi,
You seem to have a glitch in one file,
testread12=read.table("Test100.txt", head = T)
str(testread12) # Column131 is converted to a factor
# $ Column131: Factor w/ 2920 levels "--","-0.000122393",..
combining the second data set with this one will convert the new data
into factor for this
Hi,
Grid offers several functions to help drawing such graphs,
see Paul Murrell's "Can R Draw Graphs? " (useR! 2006)
I came up with this, as a quick example,
vp <- viewport(
x = unit(0, "npc"),
y = unit(0, "npc"),
just = c("left", "bottom"),
xscale = c(-1, 1) ,
yscale = c(-1, 1))
vp2 <- vi
Wacek helped me out on a similar topic a while back,
ize =
function (d, columns = names(d), izer = as.factor)
{
d[columns] = lapply(d[columns], izer)
d
}
d = data.frame(x=1:10, y=1:10, z =1:10)
str( ize(d, 'y') ) # y is now a factor
str( ize(d, 1:2, `cumsum`) ) # x and y are affected
etc
You're right, I meant to write "lines"not "line". The strange thing is,
although "line" isn't listed in ?unit, it doesn't return an error on my
machine,
> unit(-1, "line")
[1] -1line
> unit(-1, "lines")
[1] -1lines
Reading from
http://svn.r-project.org/R/trunk/src/library/grid/src/unit.c "l
I'm not sure I understood your problem (can you provide an
reproducible example?), but perhaps you can try useOuterStrips() in
the latticeExtra package (the formatting becomes similar to that of
the ggplot2 package, perhaps another option to consider)
Hope this helps,
baptiste
On 23 Mar
On 23 Mar 2009, at 11:52, johnhj wrote:
I have still the same problem... As you said I tried with
par(mfrow=c(2,1))
and par(mfrow=c(1,2)) but without success. Could R compiler be the
problem ?
Why not? But may I suggest you try first the following example I sent
you yesterday,
pn
You need only one loop,
year <- 1951:2000
filelist <- paste("C:\\Documents and Settings\\Data\
\table_",year,".txt", sep="")
filelist
for (ii in seq_along(year)) {
assign(paste("table_", year[ii], sep=""),
read.table(file=ifile[ii], header=TRUE,
sep
On 23 Mar 2009, at 17:39, Jason Rupert wrote:
I would like to replace a few varaibles within a data frame.
For example, in the dataframe below (contrived) I would like to
replace the current housesize value only if the Location is HSV.
However, I would like to leave the other values int
(a <- replicate(5,rnorm(10)))
colMeans(a)
should get you started.
HTH,
baptiste
On 23 Mar 2009, at 18:29, pfc_ivan wrote:
I tried using the for (i..) to make 1000 differents sets of numbers,
but then
I dont know how to get the mean value of all of them... because I
dont even
think 1000
well, the literal answer is that paste("arunoff_",table_year,"_temp")
is a character vector of length 1 so your indexing cannot work. What
you want is to index the data that corresponds to this variable name,
?get
But I should stress that this manipulation with assign and get seems
complete
Dear all,
Trying to extract a few rows for each element of a list of
data.frames, I'm puzzled by the following behaviour,
d <- lapply(1:4, function(i) data.frame(x=rnorm(5), y=rnorm(5)))
str(d)
lapply(d, "[", i= c(1)) # fine, this extracts the first columns
lapply(d, "[", j= c(1, 3)) #
thanks to some off-list replies, I got this to work,
On 25 Mar 2009, at 18:56, baptiste auguie wrote:
Dear all,
Trying to extract a few rows for each element of a list of
data.frames, I'm puzzled by the following behaviour,
d <- lapply(1:4, function(i) data.frame(x=rnorm(5), y
-0.25873225
>
> However, we would still have this, which is expected (same as
> d[1:2] ):
>
>> `[.data.frame`(d, i=1:2)
>x y
> 1 0.45141341 0.03943654
> 2 -0.87954548 1.83690210
> 3 -0.91083710 0.22758584
> 4 0.06924279 1.26799176
> 5 -
Hi,
If your directory contains only files you want to load anyway, then
list.files() is your friend,
list.files(pattern = "comp") # or pattern =".asc" for example
If you do need to create the names manually, then you could create the
combinations with expand.grid, as in,
do.call(paste
Something like this perhaps,
a <- matrix(rnorm(5*49), ncol=49)
pdf(width=15, height=15)
par(mfrow= c(8,6))
apply(a[,-1], 2, plot, x= a[,1])
dev.off()
HTH,
baptiste
On 27 Mar 2009, at 11:05, skrug wrote:
Hi evrybody,
in a matrix consisting of 49 columns, I would like to plot all column
?colorRamp
Hope this helps,
baptiste
On 27 Mar 2009, at 13:16, Paulo E. Cardoso wrote:
I'm trying to create a graph where different cells of a grid (a
shapefile)
will be painted with a color share scale, where the most easy way is
to use
gray().
Can I somehow get a vector (gradient) of c
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0
I need to discriminate shading level accordingly to the abundance
value
(level).
I don't know how to proceed.
Paulo E. Cardoso
-Mensagem original-
De: baptiste auguie [mailto:ba...@exeter.ac.uk]
Enviada: sexta-feira,
wing
columns with vectors.
apply(b[,-1], 2, plot, x= b[,1])
Also all columns have the same length, [R] states that the length are
different.
Can you help me?
baptiste auguie schrieb:
Something like this perhaps,
a <- matrix(rnorm(5*49), ncol=49)
pdf(width=15, height=15)
par(mfrow= c(8,6))
ap
3699 8375 19200 36563 83993 123167
$ X6 : int 248 730 3056 2327 4092 8905 15931 37895 84565
Thanks
baptiste auguie schrieb:
the result of read.table is a data.frame, not a matrix as you first
suggested. Can you copy the result of str(b) so we know what your
data
is made of?
I'
The variable name in your call to assign should vary within the for
loop, otherwise you're always assigning the value to the same variable.
Consider the following example,
listOfNames <- c("a", "b", "c")
listOfVariables <- c("vara", "varb", "varc")
for(index in seq_along(listOfNames)){
yet another attempt,
colours <- as.character(paste(letters,colours(),"stuff",LETTERS))
target <- c("red","blue","green","gray")
matches <- melt(sapply(target, grep, x=colours))
colours[matches$value] <- matches$L1
(probably a worse idea than a straight for loop, though)
baptiste
On 2
may i suggest the following,
a <- do.call(rbind, lapply(cust1_files, read.table))
(i believe expanding objects in a for loop belong to the R Inferno)
baptiste
On 30 Mar 2009, at 12:58, Mike Lawrence wrote:
cust1_files =
list.files(path=path_to_my_files,pattern='cust1',full.names=TRUE)
a
Not exactly the output you asked for, but perhaps you can consider,
library(doBy)
> summaryBy(x3~x2+x1,data=x,FUN=mean)
x2 x1 x3.mean
1 1 A 1.5
2 1 B 2.0
3 1 C 3.5
4 2 A 4.0
5 2 B 5.5
6 2 C 6.0
the plyr package also provides similar functionality, as do t
perhaps,
unlist(d, use.names=F)
baptiste
On 1 Apr 2009, at 22:15, oscar linares wrote:
Dear Rxperts
I have a data.frame as follows
ABCD
14 710
25 811
36 912
I want to convert it to a data frame with a single row (i.e., stack
the
columns w
try this,
d = read.table(textConnection("USER NAME
12 admin
12 admin
10 admin
10 advertising
61 process
17snapshot
61ticket
61ticket
30snapshot
10advertising
10advertising
10advertising
10advertising
"),head=T)
str(d) # note that NAME is a fact
just for curiosity,
`%ni%` <- Negate(`%in%`)
> 1 %ni% c(2,1)
[1] FALSE
d1[id %ni% c(1,4), ]
baptiste
On 2 Apr 2009, at 22:17, gina patel wrote:
I have another question, if I now want to remove multiple id's e.g.
id=1 or 4 is there a simple OR command I can use?
I tried d2<-(d1[id != 1 |
Dear all,
I'm puzzled by the following example inspired by a recent question on
R-help,
cc <- textConnection("user_id website time
20google0930
21yahoo0935
20facebook1000
25facebook1015
61google
Hi,
Is this what you want?
d <- data.frame(density.AL = seq(1, 10),
density.AK = seq(1, 10), # many others...
Date=letters[1:10]) # dummy example
library(reshape)
melt(subset(d, Date == "b"), id="Date")
BTW, I spotted a few awkward things in your code,
st <- c("AL",
Dear list,
I often need to convert several variables from numeric or integer into
factors (before plotting, for instance), as in the following example,
d <- data.frame(
x = seq(1, 10),
y = seq(1, 10),
z = rnorm(10),
a = letters[1:10])
d2 <-
within(d,
Excellent!
I felt it was fairly trivial but i can be quite dense on Friday
mornings.
I really like the generalisation.
Many thanks,
baptiste
On 3 Apr 2009, at 12:11, Wacek Kusnierczyk wrote:
baptiste auguie wrote:
Dear list,
I often need to convert several variables from numeric or
hoo 935 1
4 25 facebook 1015 1
5 61 google 940 1
Have I missed a built-in function to obtain this result?
Thanks,
baptiste
On 3 Apr 2009, at 14:16, hadley wickham wrote:
On Fri, Apr 3, 2009 at 4:43 AM, baptiste auguie
wrote:
Dear all,
I'm puzzled by the f
of course!
Thanks,
baptiste
On 3 Apr 2009, at 14:48, hadley wickham wrote:
On Fri, Apr 3, 2009 at 8:43 AM, baptiste auguie
wrote:
That makes sense, so I can do something like,
count <- function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d, .(user_id), transf
I had a similar need for conversions in optics. I put together several
functions and data on r-forge, where it does not clutter CRAN but can
still be shared conveniently with others.
baptiste
On 3 Apr 2009, at 19:27, Duncan Murdoch wrote:
stephen sefick wrote:
I am starting to use R fo
Hi,
It seems to me that you could write a generic function myplot() and
have different methods for each class of object (like plot does).
Either S3 or S4 classes would do I think. Then it is only a matter of
making each method work separately. In particular, the method for a
formula coul
Hi,
Have you looked at the compare package? It might do what you want (I
just remember seeing its description on R News recently but I've never
used it),
d <- data.frame(x=1:10,y=sin(1:10),z=factor(letters[1:10]))
d1 <- d
d1$x[2:3] <- jitter(d$x[2:3] )
d2 <- subset(d1, !(z %in% c("a","g")
Here's one attempt with plyr, hopefully Hadley will give you a better
solution ( I could not get cast() to do it either)
test <-
data
.frame
(a=c("A","A","A","A","B","B","B"),b=c(1,1,2,2,1,1,1),c=sample(1:7))
ddply(test,.(a,b),.fun=function(.) paste(.)[3])
a b V1
1 A 1c(2, 4
Hi,
I think it's a FAQ (== vs all.equal to test for equality), and not
related to expand.grid. See ?all.equal and ?"=="
You don't need which, gridd[gridd[,2]=="0.6" , ] would work fine, or
more elegantly (imho),
gridd <- expand.grid(x=x,y=y )
subset(gridd, factor(x) == "0.6")
Hope
On 8 Apr 2009, at 11:44, Duncan Murdoch wrote:
Mark Heckmann wrote:
Dear Duncan,
Thanks for the reply. This works, but unfortunately I need a
different
solution.
My script is supposed to run completely automated and the graphics
I produce
vary in size each time I run the script. But I wan
clr <-
function ()
rm(list = ls(pos = .GlobalEnv), pos = .GlobalEnv)
this works for me
HTH,
baptiste
On 8 Apr 2009, at 10:50, Taraxacum88 wrote:
why
rm(list=ls(all=TRUE),envir=globalenv())
is ok but
ca<-function() rm(list=ls(all=TRUE),envir=globalenv())
ca()
does not work?
Thank yo
with base graphics,
par(bg=NA)
see ?par
Hope this helps,
baptiste
On 8 Apr 2009, at 12:54, Gundala Viswanath wrote:
Is there a way to do it in R?
Especially generating plot in EPS/PDF format.
By transparent I mean clear (not white) background.
I want to attached it to dark PPT slides.
-
with ggplot2,
d <- melt(df2,id="year")
qplot(year,value,data=d,colour=variable,geom=c("line","point")) +
geom_text(data= subset(d, variable == "cars"), aes(label=value))
with lattice, my best guess would be to use grid.text in a custom
panel function.
Hope this helps,
baptiste
On 8 Apr 2
panel.xyplot(x,y,...)
grid.text(unit(x,"native"),unit(y,"native"),label=y, just="top")}
)
baptiste auguie-2 wrote:
with ggplot2,
d <- melt(df2,id="year")
qplot(year,value,data=d,colour=variable,geom=c("line","point"))
try this,
do.call(rbind, resample2)
#or simply,
replicate(1000, Cusum(sample(lambs,replace=F)))
you could also look at the plyr package.
Hope this helps,
baptiste
On 10 Apr 2009, at 09:03, Melissa2k9 wrote:
Hi,
I have used this command :
resamples<-lapply(1:1000,function(i) sample(lam
Hi,
Two thoughts I'd like to share on this subject:
1) Something really cool for conversions between units is the Google
search bar: type in " 3 inches in cm" and you get,
3 inches = 7.62 centimeters
or, " 3 £ in dollar",
3 UK£ = 4.4007 U.S. dollars
or "12 cubic meters to pi
Try this,
numbers <- c("one","two","three","four")
values <- c(10,20,30,40)
v <- list(sample(numbers,3),sample(numbers,2))
v
sapply(v,function(.l) values[match(.l, numbers)] )
HTH,
baptiste
On 14 Apr 2009, at 13:01, Manoel Silva wrote:
Dear All,
Here's my problem. I have two lists:
v
[[
Is this what you want?
plotNames <- c("plot1", "plot2", "plot3") # plot is probably best
left as the name of the base function
full.data[full.data$PLOTID %in% plotNames, ] # note the comma
HTH,
baptiste
On 14 Apr 2009, at 15:20, zack holden wrote:
Dear List,
I'm stuck on what seems
glad it was helpful.
%in% is a logical operator, so you can use "!" to negate the result
(with parentheses),
! ( 4 %in% 1:3)
Alternatively, define a new operator,
`%ni%` <- Negate(`%in%`)
1 %ni% c(2,1)
Next time you ask a follow-up question please send it to the r-help
list so others ca
?assign
(but are you sure you really want to name all these objects
separately? Usually in R you would put them together in a list or a
data.frame, it is much more convenient for later manipulations)
On 14 Apr 2009, at 18:32, Zachary Patterson wrote:
I am new to R. I would like to automat
I think you want to have a look at the plyr or doBy packages.
It would be easier to give a precise answer with a minimal example.
HTH,
baptiste
On 15 Apr 2009, at 18:03, Lane, Jim wrote:
Hi, All
Forgive me if this is a stupid newbie question. I'm having no luck
googling an answer to this, p
Do you want abind?
http://cran.r-project.org/web/packages/abind/index.html
baptiste
On 15 Apr 2009, at 19:33, Cable, Samuel B Civ USAF AFMC AFRL/RVBXI
wrote:
I have a multidimensional array "a", for example,
a
, , 1
[,1] [,2]
[1,]13
[2,]24
, , 2
[,1] [,2]
[1,]
Hi,
You should give a minimal reproducible example so that we know more
precisely what you want to do (what's a "multigraph"?).
Perhaps you can get inspiration from Paul Murrell's R graphics book,
in particular Figure 5.22,
http://www.stat.auckland.ac.nz/~paul/RGraphics/customgrid-lattice
at 12:11, rajesh j wrote:
> multigraph has several graphs in 1 layout.I'm drawing a large graph
> and plotting small portions of it below.I'm drawing lines from the
> large graph to the smaller ones to show which part of the large
> graph it is.
>
> On Thu, Apr 16, 20
Perhaps,
apply(combn(letters[1:4],2), 2, paste,collapse="")
Hope this helps,
baptiste
On 16 Apr 2009, at 17:33, Juergen Rose wrote:
Am Donnerstag, den 16.04.2009, 10:59 -0400 schrieb David Winsemius:
Thanks David,
is there also a shorter way to get the columns names of the new data
frames?
The grImport package seems to provide such possibility for vector
graphics,
http://www.stat.auckland.ac.nz/~paul/Talks/gddg.pdf
imageJ is another open-source option.
baptiste
On 16 Apr 2009, at 16:44, Shubha Vishwanath Karanth wrote:
Hi R,
Wanted to check if there are any packages avail
Perhaps try the pgfSweave package on r-forge?
http://r-forge.r-project.org/projects/pgfsweave/
HTH,
baptiste
On 19 Apr 2009, at 22:15, Jonas Stein wrote:
Hi,
i use Sweave to put plots in my .tex Documents. (pdflatex)
Is there a nice solution for this:
a) get a real LaTeX formula in the plo
I vaguely recall thinking with such convoluted constructs when
switching from Matlab to R. The lack of generic data structures such
as lists makes you define variable names that you can identify and
manipulate. There are structures in Matlab, but I think they are much
less used than lists i
if it hasn't been suggested already:
ggplot2 also has a book and a website: http://had.co.nz/ggplot2/
I would recommend any newcomer to have a look as it provides a clear,
consistent and elegant syntax to produce very nice plots.
baptiste
On 22 Apr 2009, at 03:14, Erik Iverson wrote:
In add
If most of the functions are quite stable (you don't change them too
often), you could also consider creating a R package with
package.skeleton.
baptiste
On 23 Apr 2009, at 10:39, jgar...@ija.csic.es wrote:
source() and the use of functions
...
Javier
---
I am working on a program to
It is an R command (package utils), see ?package.skeleton
baptiste
On 23 Apr 2009, at 10:51, mau...@alice.it wrote:
>
> Is that an R command ?
> I browswd for the on-line hlp about such a command but could not
> find it.
> Thank you.
> maura
>
>
> -Messaggio or
>
>
> -Messaggio originale-
> Da: Duncan Murdoch [mailto:murd...@stats.uwo.ca]
> Inviato: gio 23/04/2009 14.21
> A: mau...@alice.it
> Cc: baptiste auguie; r-help Help
> Oggetto: Re: [R] R: R: how to split and handle a big R program into
> multiple files
>
on suitable devices, you could consider transparency,
plot(f,col=alpha("grey",0.8),pch=19)
baptiste
On 24 Apr 2009, at 14:09, Knut Krueger wrote:
f<- data.frame("x"=c(1,3,5,6,1),"y"=c(1,2,3,4,1))
plot(f)
__
R-help@r-project.org mailing list
http
Hi,
Have you considered using high-level plotting functions provided by
the ggplot2 or lattice package? Here's a dummy example,
x <- seq(0, 10, length=100)
y1 <- sin(x)
y2 <- cos(x)
y3 <- x^2/100
y4 <- 1/x
d <- data.frame(x, y1, y2, y3, y4)
library(reshape)
dm <- melt(d, id="x")
dm$type1
Hi,
You could do this very easily using ggplot2,
#install.packages("ggplot2", dep=TRUE)
library(ggplot2)
c <- ggplot(mtcars, aes(y=wt, x=mpg)) + facet_grid(. ~ cyl)
c + stat_smooth(method=lm) + geom_point()
See more examples on Hadley's website: http://had.co.nz/ggplot2/
Hope this helps
I'm not sure I'm following you but have you tried,
identical(matrix(c(1,1,1,1),ncol=2), matrix(c(1,1,1,1),ncol=2))
?all.equal
?isTRUE
?identical
and possibly the compare package,
compare(matrix(c(1,1,1,1),ncol=2),matrix(c(1,1,1,1),ncol=2))
HTH,
baptiste
On 26 Apr 2009, at 18:02, Esmail wro
ER TABLE JOIN IN R
(Gabor Grothendieck)
12. Re: THE EQUIVALENT OF SQL INNER TABLE JOIN IN R (Peter Dalgaard)
13. Re: 3 questions regarding matrix copy/shuffle/compares (Esmail)
14. Re: Conditional plot labels (baptiste auguie)
15. Re: Scatterplot of two groups side-by-side? (baptiste auguie)
Try this,
a = matrix(rnorm(10*10),ncol=10)
a[, seq(1,ncol(a),by=2)]
baptiste
On 29 Apr 2009, at 09:28, Umesh Srinivasan wrote:
Hi all,
If I have a huge matrix/ dataframe and I want to create a new matrix/
dataframe with every second (or third, or fourth etc.) row of the
original
mat
Hi,
Here's one approach that I find perhaps more elegant than sweeping
through the columns by their index,
library(ggplot2)
data(economics)
str(economics)
# library(reshape)
d <- melt(economics, id="date")
foo <- function(var="pop", d, smooth=FALSE, ... ) {
p <- qplot( data=subs
Try this,
items <- c(list(col=2),list(pch=2))
par(mfrow=c(2, 1))
for (ii in seq(2)) {
do.call(function(x, y, ... ) plot(1:10, ...), items[ii])
}
?do.call
HTH,
baptiste
On 29 Apr 2009, at 19:30, Sebastien Bihorel wrote:
Dear R-users,
I would like to know if is it possible to set a fun
Try this,
# d <- read.table(pipe("pbpaste"), head=T) # read your data
table(d)
# library(reshape)
cast(as.data.frame(table(d)), .~Firm, fun=sum)
HTH,
baptiste
On 4 May 2009, at 13:19, Cecilia Carmo wrote:
Hi everyone:
I need to count the number of banks of each firm in my
data. The firm
your code is missing a closing bracket for the text labels,
legend(1,4, c("Simulation", "Observation"), lty=1:2, col=2:3)
baptiste
On 4 May 2009, at 20:46, Steve Murray wrote:
Dear all,
I'm attempting to insert a legend into a line graph. I've sorted out
the positioning, but I'm unable t
On 5 May 2009, at 19:28, Duncan Murdoch wrote:
On 5/5/2009 1:05 PM, Markus Loecher wrote:
Dear R users,
while I enjoy the built-in log argument to the plot() function, I
wished it
would be as easy to create more general custom transformed axes
such as
sqrt(), logit, etc...
for example, i
I think the pgfSweave project on R-forge is working on this (as far as
i know it currently relies on eps2pgf)
http://r-forge.r-project.org/R/?group_id=331
HTH,
Baptiste
On 6 May 2009, at 15:37, Lasse Bombien wrote:
Hi all,
I saw a thread from 2007 about the possible implementation of a PG
Hi,
Try this,
x <- matrix(1:9, ncol=3, byrow=T)
sca <- c(2.5, 1.7, 3.6)
x %*% diag(1/sca)
HTH,
baptiste
2009/12/27 Muhammad Rahiz :
> Hi useRs,
>
> I ran into an inconsistent output problem again. Here is the simplify
> illustration
>
> I've got a matrix as follows
>
>> x
> V1 V2
Hi,
Try print(p) instead of plot(p)
HTH,
baptiste
2009/12/29 Bryan Hanson :
> I¹m trying to build a simple formula interface to work with a function using
> ggplot2. The following scheme ³works² up until the plot(p) request, at which
> point there are complaints about xlim¹s and a blank graphic
Hi,
I think you can also use plyr for this,
dft <- read.table(textConnection("P1idVeg1Veg2AreaPoly2 P2ID
1 p p 1 1
1 p p 1.5 2
2 p p 2 3
2 p h 3.5 4")
Hi,
Here is some artificial data followed by minimal ggplot2 and lattice examples,
makeUpData <- function(){
data.frame(x=sample(letters[1:4], 100, repl=TRUE), y=rnorm(100))
}
datasets <- replicate(15, makeUpData(), simplify=FALSE)
names(datasets) <- paste("dataset", seq_along(datasets), sep
as well as a book.
Lattice also has a dedicated book, and a companion website with the figures,
http://r-forge.r-project.org/projects/lmdvr/
HTH,
baptiste
2009/12/29 baptiste auguie :
> Hi,
>
> Here is some artificial data followed by minimal ggplot2 and lattice examples,
>
> make
Hi,
You can set up a Grid layout with one viewport at the bottom and
another on the left and use grid.text to add your labels. An example
is given below using grid.pack.
The gridExtra package provides a convenient wrapper for these regular
arrangements of plots,
##library(gridExtra) #http://grid
Hi,
Using backticks might work to some extent,
library(lattice)
`my variable` = 1:10
y=rnorm(10)
xyplot(`my variable` ~ y)
but if your data is in a data.frame the names should have been converted,
make.names('my variable')
[1] "my.variable"
HTH,
baptiste
2010/1/3 Jay :
> Hello!
>
> one more
Hi,
Thank you for this fun package.
I recently posted a related question on R-help that seemed to pass
unnoticed. Basically, I suggested that the functionality of fortunes
could be extended to R FAQ entries, also allowing contributed packages
to provide their own (fortune or) faq data file. My ini
Hi,
Something like this maybe,
plot.new()
lab = expression(bar(T)*"("*-x*" ; "*alpha*")"-G*"("*x*" ; "*alpha*" , "*J*")")
text(0.5,0.5,lab)
?plotmath
HTH,
baptiste
2010/1/8 bernardo lagos alvarez :
> Dear useRs,
>
> How can I, writting the correct greek letter using postscrip or pdf functio
...@r-project.org [mailto:r-help-boun...@r-
>> project.org] On Behalf Of Brian Diggs
>> Sent: Thursday, December 31, 2009 3:08 PM
>> To: baptiste auguie; David Winsemius
>> Cc: r-help
>> Subject: Re: [R] expand.grid game
>>
>> baptiste auguie wrote:
>> &
It did take me a good night's sleep to understand it. I was stuck with
the exact same question but I see now how the remaining balls are
shared among all 8 urns (therefore cases with 11, 12, 13, ... 17 balls
are also dealt with).
Thanks again,
baptiste
2010/1/12 Rolf Turner :
>
> On 13/01/2010,
Hi,
You could play with the splitTextGrob() function from the RGraphics package,
string <- "Lorem ipsum dolor sit amet, consectetur adipiscing elit.
Quisque leo ipsum, ultricies scelerisque volutpat non, volutpat et
nulla. Curabitur consequat ullamcorper tellus id imperdiet. Duis
semper malesuada
Hi,
?eval seems like a good candidate
HTH,
baptiste
2010/1/15 Jiiindo :
>
> Hello all,
> I want to call R from java. And I have a expression in Java as a String,
> example : (variable 1 + variable 2)* variable 3 and i want R calculate this
> expression. How can I do?
> ex:
> Java
> -int x1,x2;
Hi,
I think you could use iapply (search the archives) or the plyr package
to save you from transposing the result.
HTH,
baptiste
2010/1/19 marco salvini :
> Can you please help on the issue?
> I using the apply command on a matrix below the example:
>
> Create a vector
> x =c(5, 3, 2:4, NA, 7,
Hi,
try c.trellis() from the latticeExtra package.
HTH,
baptiste
2010/1/20 George Chen :
> Hello,
>
> I would like to juxtapose two lattice graphs with common X axes such that the
> X axes line up. I am using plot right now but the edges are not neat and it
> would be nice if I could just d
Hi,
One approach could be a while loop as follows. Note that your circle
should have radius 0.5 if I understood the problem correctly.
N <- 5
npoints <- 0
ntests <- 0
points.in.circle <- matrix(NA, ncol=2, nrow=N)
while (npoints < N) {
test.point <- runif(2, -0.5, 0.5) # generate new point
Hi,
Try this,
a = replicate(3, data.frame(x=1:10, y=rnorm(10)), simplify=FALSE)
lapply(a, "[", "y")
HTH,
baptiste
2010/1/22 Ivan Calandra :
> Hi everybody!
>
> I have a (stupid) question but I cannot find a way to do it!
>
> I have a list like:
>> SPECSHOR_tx_Asfc
> $cotau
> SPECSHOR Asfc.m
lected ("y").
Best,
baptiste
2010/1/22 Ivan Calandra :
> Thanks Baptiste, it does help.
>
> However, I don't really understand what "[" means. Could you please tell me
> more about it? I didn't find anything helpful on that in the help.
>
> Thanks in
Hi,
Try this,
x <- seq(0, 10, len = 100)
y <- jitter(sin(x), 1000)
old.par <- par()
par(bg=grey(0.5))
plot(x, y, new = TRUE, t = "n")
lims <- par("usr")
plot(x, y, col = 1,
panel.first = {
rect(lims[1], lims[3], lims[2], lims[4],
col = "lightblue") })
HTH,
baptiste
2010/1/26 :
Hi,
I remember asking a similar question some time ago, I don't know if
the matter has evolved since then,
http://groups.google.com/group/ggplot2/browse_frm/thread/a3df8a0d1ee335fb/e3bedd50fb9bd567?lnk=gst&q=theme#e3bedd50fb9bd567
There's also set_default_scale, somewhat related to your question
Hi,
It's easy with ggplot2,
cols = matrix(c("#F7FBFF", "#DEEBF7", "#C6DBEF", "#9ECAE1", "#6BAED6",
"#4292C6", "#2171B5", "#08519C" ,"#08306B"), ncol=3)
library(ggplot2)
m = melt(cols)
qplot(factor(X1),factor(X2),data=m, fill=value, geom="tile") +
scale_fill_identity()
HTH,
baptiste
2010/1/2
Hi,
I think it's a bug in quartz(). The following example uses png() with
cairo or quartz backends, and only cairo respects the size and
resolution (as verified in Adobe Photoshop).
png(file="foo-300.png", type="quartz", units="in",width=5, height=3, res=300)
plot(1,1)
dev.off()
png(file="fo
Hi,
Would this do as an alternative syntax?
g1 <- quote(1/Tm)
mat <- list(0, bquote(f1*s1*.(g1)))
vals <- data.frame(f1=1, s1=.5, Tm=2)
sapply(mat, eval, vals)
HTH,
baptiste
On 29 January 2010 17:51, Jennifer Young
wrote:
> Hallo
>
> I'm having trouble figuring out how to evaluate an expres
Hi,
Hadley recently proposed a strategy using plyr for a very similar problem,
listOfFiles <- list.files()
names(listOfFiles) <- basename(listOfFiles)
library(plyr)
d <- ldply(listOfFiles, scan)
Even if you don't want to use plyr, it's always better to group things
in a list rather than clutter
d(comb,combadd)
>
> }
> boxplot(dats ~ index, data = comb)
>
>
> works just great. There is no additional files in the folder. But look, how
> much code for such a simple task. I'd definitely prefer the plyr solution.
>
> Maxim
>
>
> 2010/1/30 baptiste auguie
&g
Hi,
Adding two semi-transparent colours results in non-intuitive colour
mixing (a mystery for me anyway). Is it additive (light), substractive
(paint), or something else? Consider the following example, depending
on the order of the two "layers" the overlap region is either purple
or dark red. I h
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