try this,
library(plyr)
ddply(d, .(block, trial), function(.d) .d[1:2, ])
block trial x y
1 1 1 605 150
2 1 1 603 148
3 1 2 607 148
4 1 2 605 152
HTH,
baptiste
On 11 May 2009, at 13:49, Jens Bölte wrote:
Hello,
I have been struggling for quite
good point, i forgot about head (!),
library(plyr)
ddply(d, .(block, trial), head, 2)
block trial x y
1 1 1 605 150
2 1 1 603 148
3 1 2 607 148
4 1 2 605 152
On 11 May 2009, at 14:04, Gabor Grothendieck wrote:
Try this:
do.call(rbind, by(DF, DF[1:2],
Hi,
I once made this function (essentially the same as Romain),
assignFiles -
function (pattern = csv, strip = (_|.csv|-| ), ...) # strip is
any pattern you want to remove from the filenames
{
listFiles - list.files(pattern = pattern, all.files = FALSE,
full.names = FALSE,
Depending on the nature of your pdf file, it may be possible to use
the grImport package. I've never used it before, but a quick test
seems promising,
# create a test picture
colorStrip -
function (colors, draw = T)
{
x - seq(0, 1 - 1/ncol(colors), length = ncol(colors))
y -
I'd first try plyr and see if it's efficient enough,
library(plyr)
listOfFiles - list.files(pattern= .txt)
d - ldply(listOfFiles, read.table)
str(d)
alternatively,
d - do.call(rbind, lapply(listOfFiles, read.table))
HTH,
baptiste
On 13 May 2009, at 12:45, SYKES, Jennifer wrote:
Alternatively,
signif(c(pi,exp(1)), 3)
?signif # and others in that page
HTH,
baptiste
On 14 May 2009, at 13:47, jim holtman wrote:
Depending on what you want to do, use 'sprintf':
x - 1.23456789
x
[1] 1.234568
as.character(x)
[1] 1.23456789
sprintf(%.1f %.3f %.5f, x,x,x)
[1] 1.2
On 15 May 2009, at 10:01, Prof Brian Ripley wrote:
On Fri, 15 May 2009, Dieter Menne wrote:
Stats Wolf stats.wolf at gmail.com writes:
Postscript, however, does not have to be what I need for two
reasons.
First, it does not accept some special characters from foreign
languages (exactly
If you're desperate for a workaround, you might want to try this
example using pgfSweave,
http://ggplot2.wik.is/Mathematical_annotations
On a similar vein, you could try psfrag replacements with a postscript
device (there is some code for this on the list archives).
Feel free to comment /
I'd suggest you first combine the 12 data.frames into one, using
melt() from the reshape package.
makeDummy - function(.){ # since you don't provide a reproducible
example
data.frame(x=letters[1:10], y=rnorm(10))
}
listOf12DataFrames - lapply(1:12, makeDummy)
Dear list,
This might be a topic for r-devel but i may be missing something
obvious.
I don't understand the rationale in the absolute sizes of the point
symbols, and I couldn't find it documented. The example below uses
Grid to check the size of the symbols against a square of 10mm x
Hi,
Perhaps you can try this,
seq.weave - function(froms, by, length, ... ){
c(
matrix(c(sapply(froms, seq, by=by, length = length/2, ...)),
nrow=length(froms), byrow=T)
)
}
seq.weave(c(2, 3), by=3, length=8)
seq.weave(c(2, 3, 4),
Try this and see if it helps (if not, please help improving it),
http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-misc:legendoutside
HTH,
baptiste
On 27 May 2009, at 20:31, Wade Wall wrote:
Hi all,
I have been trying to figure out how to place a legend beside a plot,
rather than
recycling rule: repeat the shorter element as many times as necessary,
all.equal(1:2 + 1:10 , rep(1:2, length=10) + 1:10)
# TRUE
HTH,
baptiste
On 28 May 2009, at 22:00, bogaso.christofer wrote:
I have following addition :
1:2 + 1:10
[1] 2 4 4 6 6 8 8 10 10 12
I could not
Dear all,
I'm trying to access and modify grobs in a ggplot2 plot. The basic
idea for raw Grid objects I understand from Paul Murrell's R graphics
book, or this page of examples,
http://www.stat.auckland.ac.nz/~paul/grid/copygrob/copygrobs.R
However I can't figure out how to apply this
I'm not sure it's currently possible with ggplot2 (lattice and
latticeExtra offer some workarounds for this).
Perhaps you can try this,
http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-ggplot2:sharelegend
(neater here: )
you can use title() with the sub argument,
title(sub=x label, outer=T) # you might want to play around with
line argument
baptiste
On 1 Jun 2009, at 22:03, Andre Nathan wrote:
Hello
On Wed, 2009-05-27 at 13:38 -0600, Greg Snow wrote:
Create an outer margin (see ?par), then use mtext to
Dear all,
I feel like I've been reinventing the wheel with this code (implementing
type = 'b' for Grid graphics),
http://econum.umh.ac.be/rwiki/doku.php?id=tips:graphics-grid:linesandpointsgrob
Has anyone here attempted this with success before? I found suggestions
of overlapping large
Jonathan Williams wrote:
Dear R-helpers,
I have been trying to figure out how to plot a graph with an axis label
consisting of a mixture of Latin, Greek and subscript characters.
Specifically, I need to write A[beta]{1-42}, where A is Latin script A,
[beta] is Greek lower case beta and {1-42}
Ben Bolker wrote:
amor Gandhi wrote:
Hi,
I have gote the following data
x1 - c(rep(1,6),rep(4,7),rep(6,10))
x2 - rnorm(length(x1),6,1)
data - data.frame(x1,x2)
and I would like to compute the mean of the x2 for each individual of x1,
i. e. x1=1,4 and 6?
You'll probably get seven
Marie Sivertsen wrote:
Dear list,
I have a vector of elements which I want to combined each with each, but
none with itself. For example,
v - c(a, b, c)
and I need a function 'combine' such that
combine(v)
[[1]]
[1] a b
[[2]]
[1] a b
[[3]]
[1] b c
I am not very
Dear list,
I'm quite surprised by this,
unit(1:5,char)[-c(1:2)]
#4char 3char # what's going on??
while I expected something like,
c(1:5)[-c(1:2)]
# 3 4 5
Note that,
unit(1:5,char)[c(1:2)]
# 1char 2char # fine
?unit warns about unit.c for concatenating, but also says,
It is possible to
jim holtman wrote:
try this:
Oh well, i spent the time writing this so i might as well post my
(almost identical) solution,
x-c(1:3, 6: 7, 10:13)
breaks = c(TRUE, diff(x) != 1)
data.frame(start = x[breaks], length = tabulate(cumsum(breaks)))
Hoping this works,
baptiste
x
Paul Murrell wrote:
Hi
The bug is now fixed in the development version
(thanks to Duncan for the diagnosis and suggested fix).
Paul
Thank you both for your help and dedication!
Best regards,
baptiste
--
_
Baptiste Auguié
School of Physics
University of
Kenny Larsen wrote:
Hi,
A fairly basic problem I think, here although searching the inetrnet doesn't
seem to reveal a solution. I have a dataset with two columns of real
numbers. It is read in via read.table. I simply need to square the data in
the second column and then plot it. I have tried
I knew I had seen this in action! But as you mention, most pages only
display ~~RDOC~~ at the moment.
I second the idea of using the wiki for such collaborative work. If the
current (r-devel) version of all help pages could be automatically
copied to the wiki, users would have a convenient way
Kenny Larsen wrote:
Hi All,
I have hunted high and low and tried dozens of things but have yet to
achieve the result I require. Below is my code (taken mostly from another
thread on here) thus far:
files-list.files()
files-files[grep('.wm4', files)]
labels-gsub('.wm4', '',files)
for(i in
Hi,
the grImport package provides some functionality for svg and postscript
graphics,
http://cran.r-project.org/web/packages/grImport/index.html
Best,
baptiste
Robbie Morrison wrote:
Dear R-help
I want to display an image file in a new plot frame.
SVG is my preferred format, but I
Hi,
I tend to use a slightly modified version of stats::relevel, (from an
old thread on this list),
relevel =
function (x, ref, ...)
{
lev - levels(x)
if (is.character(ref))
ref - match(ref, lev)
if (any(is.na(ref)))
stop('ref' must be an existing level)
nlev - length(lev)
Commenting on this, is there a strong argument against modifying
relevel() to reorder more than one level at a time?
I started a topic a while back (recursive relevel,
https://stat.ethz.ch/pipermail/r-help/2009-January/184397.html) and I've
happily used the proposed change since then by
For the sake of brevity, I like to use this trick,
plot(0, 0)
mtext(~Monthly Precipitation (mm x *10^2*/month))
HTH,
baptiste
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
utkarshsinghal wrote:
Hi Jim,
What you are saying is correct. Although, my computer might not have
same speed and I am getting the following for 10M entries:
user system elapsed
0.559 0.038 0.607
Moreover, in the case of character vectors, it gets more than double.
In my
[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(y,
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
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
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 use
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
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
, 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 some advice on how to wrap a R function
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
very
comfortable with
factors. Could you please give me some advice?
Thank you very much.
baptiste auguie-2 wrote:
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
.
-thomas
On Sun, 15 Mar 2009, baptiste auguie wrote:
I've put together a rough R port of that C code [*] in a package on
r-forge:
http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/pkg/spectral/?root=photonics
http://r-forge.r-project.org/R/?group_id=160 ( package spectral
On 19 Mar 2009, at 07:22, Dieter Menne wrote:
Gundala Viswanath gundalav at gmail.com writes:
I have the following code that try to plot
simple sinus curve into 2x2 grid in 1 page.
But this code of mine create 4 plots in 1 page
each. What's wrong with my approach?
...
library(lattice)
Hi,
you could have a look at the doBy package which makes these operations
easier. Hadley's plyr package is also another option.
baptiste
On 20 Mar 2009, at 01:01, Altaweel, Mark R. wrote:
Hi,
I am trying to perform various functions on a list with a number of
elements. For example. I
Hi,
it would help if you provided a minimal example.
Here is one approach I often use with the plotting package ggplot2,
parameters - expand.grid(m=c(0, 1), s=seq(0.1, 1,length=10))
x - seq(-5, 5, length=300)
foo - function(m, s){
data.frame(x=x, y=dnorm(x, m, s), m=factor(m),
is not
as large as I would expect.
Thanks again in advance.
Mark
-Original Message-
From: baptiste auguie [mailto:ba...@exeter.ac.uk]
Sent: Fri 3/20/2009 4:32 AM
To: Altaweel, Mark R.
Cc: r-help@r-project.org
Subject: Re: [R] functions within a list
Hi,
you could have a look at the doBy
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,
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
(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
completely
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=rnorm(5
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 -0.20477052 -0.25873225
Romain
baptiste auguie wrote:
Dear all,
Trying to extract a few
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
?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
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, 27 de Março de 2009
[,-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))
apply(a[,-1], 2, plot, x= a[,1
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'm guessing the most elegant solution will be to use the reshape
package, followed
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 28 Mar 2009,
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)
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
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
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
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, AK)
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
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 ba...@exeter.ac.uk
wrote:
Dear all,
I'm puzzled
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 ba...@exeter.ac.uk
wrote:
That makes sense, so I can do something like,
count - function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d
I had a similar need for conversions in optics. I put together several
functions and data on r-forge, where it does not clutter CRAN but can
still be shared conveniently with others.
baptiste
On 3 Apr 2009, at 19:27, Duncan Murdoch wrote:
stephen sefick wrote:
I am starting to use R
Hi,
It seems to me that you could write a generic function myplot() and
have different methods for each class of object (like plot does).
Either S3 or S4 classes would do I think. Then it is only a matter of
making each method work separately. In particular, the method for a
formula
Hi,
Have you looked at the compare package? It might do what you want (I
just remember seeing its description on R News recently but I've never
used it),
d - data.frame(x=1:10,y=sin(1:10),z=factor(letters[1:10]))
d1 - d
d1$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)
2 B 1 c(7,
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 this
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
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
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 2009, at
, just=top)}
)
baptiste auguie-2 wrote:
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
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)
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 pints,
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
[[1]]
[1]
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 like
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 can
?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
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,
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,
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, 2009 at 4:34 PM, baptiste auguie
ba...@exeter.ac.uk wrote:
Hi,
You should give a minimal reproducible example
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
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
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
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
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
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
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