Thanks toy Jim and Michael for their response. The suggestion to use regular
if/else was spot on.
> From: tyler_rin...@hotmail.com
> To: r-help@r-project.org
> Date: Sat, 30 Jul 2011 01:48:21 -0400
> Subject: [R] ifelse returns
>
>
> Greetings R Community,
>
> I am working with the ifelse
So the thing is this, ifelse puts together a vector elementwise from two
other vectors (or other things, but vectors is vague enough for now) based
on a series of T/F values: you only put in one logical test, so you only get
a vector of length 1 back, here MODE1[1]=4. It seems like a regular if/els
You might want to breakdown some of the expressions you are using; for example:
> x<-c(2,3,4,4,5,5,6,6,8,10)
>
> df<-as.data.frame(table(x))
> df<-df[order(df$Freq),]
> m<-max(df$Freq)
> (MODE1<-as.vector(as.numeric(as.character(subset(df,Freq==m)[,1]
[1] 4 5 6
> ifelse(sum(df$Freq)/length(df$
Greetings R Community,
I am working with the ifelse function and it is returning something unexpected.
In the code the line with the MODE1 assignment the output is a vector [1] 4 5
6 but when I put the MODE1 object into the ifelse function [R}'s output for
MODE1 is the first number from the
Also, if it works for you and there are no data irregularities, might I
suggest you look at making use of the runSD() and runCov or runCor functions
of the TTR package. If you know the relationship between slope of an
intercept -- beta = cov(x,y) / var(x) -- it should be pretty easy to
implement fo
Beyond that, the files at the URL given are not zip files, so we are
not being told the whole truth.
My guess is that 'Lei Liu' is using Windows without telling us. In
which case (s)he needs to follow the rw-FAQ about installing packages
from the source.
Also note that CRAN packages are arc
Hi Lei,
This is likely a problem related to your OS or your version of R;
however, you have not provided us with either (as the posting guide
linked to below requests), so it is impossible to tell. The easiest
way to let us know is to copy and paste the output from running:
sessionInfo()
at you
Hi there,
I want to install the "adapt" package, which is available at
http://cran.r-project.org/src/contrib/Archive/adapt/. This package
cannot be directly installed by "install packages" menu in R. So I
downloaded the zip file into the hard drive. But I couldn't install
it using "install pa
On 29/07/11 19:57, Martin Maechler wrote:
but *also* see the *workaround* for the bug that has been on
R's ?pdf help page for years []
--> search for "q" (including the quotes).
Why oh why oh why ... are people always thinkg of bugs in R
when they notice something and
why oh w
Hi Billy,
Can you provide your data? You could attach it as a text file or
provide it by pasting the output of:
dput(Q)
into an email. It would help if we could reproduce what you are
doing. You might also consider a list or forum that is more
statistics oriented than Rhelp, as your questions
On Jul 29, 2011, at 8:22 PM, Byerly, Mike M (DFG) wrote:
Is there a function in R that will calculate a running linear slope
similar to the way the function filter() will calculate a moving
average?
If you are happy with the return from filter, then why not take the
first differences as th
Is there a function in R that will calculate a running linear slope
similar to the way the function filter() will calculate a moving
average?
Mike Byerly
Fisheries Biologist
Commercial Fisheries Division
Alaska Dept. of Fish and Game
3298 Douglas Place
Homer, AK 99603
mike.bye...@alask
On Fri, Jul 29, 2011 at 6:00 PM, Sarah Goslee wrote:
> Hi Peter,
>
> I'm not going to look at your large file on what for me is Friday evening, but
> the usual cause of that kind of problem is a single or double quote in the
> text.
bingo! Setting quote = "" solved the problem.
Thanks,
Peter
Hi Peter,
I'm not going to look at your large file on what for me is Friday evening, but
the usual cause of that kind of problem is a single or double quote in the text.
One way to diagnose the problem is to look at the rows in the text file itself
right around 25952 - there's always something th
Hi all,
I encountered a problem when trying to read in an Illumina chip
annotation file. The offending file is large, so I zipped it up and
posted it at
http://www.genetics.ucla.edu/labs/horvath/CoexpressionNetwork/tmp/ProbeInfo_Expression.txt.bz2
Executing this:
annot = read.table(bzfile("Prob
Hi:
This is simple to do in any of the supported graphics systems in R,
and there are a number of kind people who are willing to and capable
of providing help. Unfortunately, since you didn't supply any data or
code, you've made it difficult for them to help you. Please read the
Posting Guide that
Hi:
The reason why you can fit N linear models to the same predictor
quickly is because the model matrix X is fixed and you can store the
multiple Y's as a matrix. Consequently, you can run the N linear
regressions in one shot using least squares, and lm() implements this
efficiently by allowing t
On Jul 29, 2011, at 6:38 PM, Sumukh Sathnur wrote:
Hi all,
I used image.plot() to create a heat map of a matrix:
as.matrix(read.table("Matrix.txt", sep="\t"))->x
HeatBrk<-seq(5,25,2.5)
MyCol= gray((7:0)/7)
library(fields)
image.plot(x, col=MyCol, breaks=HeatBrk, legend.shrink=0.3)
dev.copy(d
On Fri, 29 Jul 2011, Michel Lutz wrote:
Achim,
Thank you so much for this prompt answer. Really appreciated !
Anyway, I am still a bit lost... don't you mind if I ask you somme
additional questions?
* *one standard approach is to employ a HACcovariance matrix* I did many
researches but I nev
Fixed (both). Wait for the autoupdate or check
https://svn.r-project.org/R-project-web/trunk/contributors.html
-pd
On Jul 29, 2011, at 23:23 , Hadley Wickham wrote:
> And I think Uwe is missing from that list!
> Hadley
>
> On Fri, Jul 29, 2011 at 3:34 PM, Paul Menzel
> wrote:
>> Dear R webma
Perfect! Thanks!
By the way, I see that, unlike base rbind, it does not work for vectors and
lists:
rbind(c(a=1),c(b=2)) => matrix(1:2,2,1,dimnames=list(NULL,"a"))
== as.matrix(data.frame(a=1:2))
but
rbind.fill(c(a=1),c(b=2)) => NULL
Shouldn't it give something like
matrix(c(1,
Well there is the "esp" package, though it is still very pre-alpha. Let's see
what it has to suggest for this question:
> source('g:/R/esp/esp.R')
> esp()
[1] "tabulation overabundant positron carfare contaminant sprawl station swat
aversion"
>
Wow, that almost looks like it should make some s
And I think Uwe is missing from that list!
Hadley
On Fri, Jul 29, 2011 at 3:34 PM, Paul Menzel
wrote:
> Dear R webmasters,
>
>
> my browser defaults to the charset UTF-8 and since [1] seems to be
> encoded in ISO-8859-1 the umlauts are not displayed correctly. It would
> be great if the encoding
Dear R webmasters,
my browser defaults to the charset UTF-8 and since [1] seems to be
encoded in ISO-8859-1 the umlauts are not displayed correctly. It would
be great if the encoding could be added to the header [2]. (Or the page
converted to UTF-8 and the problem would be solved for me. ;-) )
Jack,
There is no pch= argument for plot3d() or points3d(). You can change the
color and the size of the symbols. Or you can use characters as symbols
using text3d(). I found that suggestion by searching the rhelp archives.
It is helpful to share code that will work.
x <- c(-4, -2, 0, 2, 4)
On Jul 29, 2011, at 4:02 PM, jack hietpas wrote:
Hello I was wondering if anyone has been able to change the pch
value for
points in 3dplot()? I am able to change point colors just not
symbol types.
I have included a portion of my code below. I realize this won't work
outside of my script
Sarah Goslee wrote:
I get the same error on a clean R session with rioja 0.5-6.
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-redhat-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETAR
Hello I was wondering if anyone has been able to change the pch value for
points in 3dplot()? I am able to change point colors just not symbol types.
I have included a portion of my code below. I realize this won't work
outside of my script, but thought, maybe there is something obvious that I
a
I get the same error on a clean R session with rioja 0.5-6.
> sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-redhat-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSA
Dear Simon,
Am Freitag, den 29.07.2011, 12:32 -0700 schrieb Crock:
> i`m trying to use the plot function to show more variable-funktions in one
> graphic. so for example i would like to show Chinas Income in different
> sectors in one graphic. obviousely typing plot(ChinaIncome) the graphics are
Can you be a little more specific as to the structure of your data - time
series, single point, levels, relative values, etc? I think that will let us
help you along the way to a specific set of plot tools...For now check out the
R plots/graphics gallery (easy google). Also see ?plot
Michael We
Hi all
I write below code for simulation and forecasting a stochastic mortality
model with parameters kappa2(t) and gamma3(t-x) where t is the time and x
the age but I got this message:"Error in G3.sim[, k] = G3.aug :
number of items to replace is not a multiple of replacement length"
Any sugge
hi,
i`m trying to use the plot function to show more variable-funktions in one
graphic. so for example i would like to show Chinas Income in different
sectors in one graphic. obviousely typing plot(ChinaIncome) the graphics are
split and there is one graphic for agriculture, one for industry i
I'm using rioja 0.5-6 with R 2.12.1 / x86_64-pc-linux-gnu on
Kubuntu 11.04.
When I run this example from the rioja reference manual:
library(rioja)
# compare diatom data from core from Round Loch of Glenhead
# with SWAP surface sample dataset
data(RLGH)
data(SWAP)
result <- compare.data
Hi all,
I am attempting to run PCA on a matrix (nrow=66, ncol=84) using 'prcomp'
(stats package). My data (referred to as 'Q' in the code below) are
separate river streamflow gaging stations (columns) and peak instantaneous
discharge (rows). I am attempting to use PCA to identify regions of that
Thank you so much for the help!
I got it down to 1sec per run.
-Chris
On Fri, Jul 29, 2011 at 1:12 PM, "Dénes TÓTH" wrote:
>
> Hi,
>
> you can solve the task by simple matrix algebra.
> Note that I find it really inconvenient to have a matrix with variables in
> rows and cases in columns, so I t
Thanks again,
I like the idea of adding transparency. I will try it. For now, I resigned
myself to add points to the original barplot using points as such:
plotV<-barplot(Value[,1],space=0)
dev.off()
par(mar=c(4,5,3,1))
barplot(Value[,1],space=0, col="grey30",axis.lty=2)
points(plotV, sampledept
On Jul 29, 2011, at 1:23 PM, marcel wrote:
This solution worked well,
There is no "this solution" visible to most of us. Please learn to
post edited context.
and the key section under ?xyplot was very helpful
in adjusting lots of details in the key, including position and even
the
dist
This solution worked well, and the key section under ?xyplot was very helpful
in adjusting lots of details in the key, including position and even the
distance between text and key symbols, using "between". One thing I didn't
see described very well was how to toggle from a transparent key symbol t
> Personally, given that R can't do much anything without all its base
> packages (ie the single /usr/local/bin/R.bin executable can't do
> anything on its own) I can't see the point in a single static binary.
> If the client is worried about library compatibility then build R with
> dynamic linki
Barry Rowlingson wrote:
>
> Personally, given that R can't do much anything without all its base
> packages (ie the single /usr/local/bin/R.bin executable can't do
> anything on its own) I can't see the point in a single static binary.
> If the client is worried about library compatibility then b
Hi,
you can solve the task by simple matrix algebra.
Note that I find it really inconvenient to have a matrix with variables in
rows and cases in columns, so I transposed your predictors matrix.
# your data
regress.y = rnorm(150)
predictors = matrix(rnorm(6000*150), ncol=150, nrow=6000)
tpreds <
Dear Simon and David,
The Rcmdr does indeed use the recode() function in the car package for recodes.
When recode() is called directly, it works to escape the apostrophe, but this
doesn't work in the Rcmdr dialog, which then sees the apostrophe and complains.
I can suggest two solutions, of whi
Why dont you plot a stacked barplot with the two values for each depth
(or whatever is on the x axis) besides each other? Another option would
be to add transparency to the fill color of the plot (col=rgb(1,1,1,0.5)).
On 07/29/2011 05:21 PM, Colin Bergeron wrote:
Thanks Jannis,
I am including
I keep the following function handy so that I can change the numeric
value from POSIXct back to POSIXct; I tend to keep the numeric in
matrices since it allows faster operations in some cases. This just
puts the appropriate classes on the object; quicker than using
as.POSIXct(z, origin="1970-01-01
Thanks a lot, Jim!
Dimitri
On Fri, Jul 29, 2011 at 12:48 PM, jim holtman wrote:
> But your requirements said "last underscore"; so we now have a change
> that really says the first underscore. So we change the regex to:
>
>> strsplit(sub("[^_]+_(.+)_(.+).o$", "\\1 \\2", x), ' ')
> [[1]]
> [1] "a
But your requirements said "last underscore"; so we now have a change
that really says the first underscore. So we change the regex to:
> strsplit(sub("[^_]+_(.+)_(.+).o$", "\\1 \\2", x), ' ')
[[1]]
[1] "a1" "2.5"
[[2]]
[1] "a2" "2.53"
[[3]]
[1] "a3_bla" "1"
>
On Fri, Jul 29, 2011 at 12:4
Thank you, Jim.
It's close but not all the way. The desired result is:
> [[1]]
> [1] "a1" "2.5"
>
> [[2]]
> [1] "a2" "2.53"
>
> [[3]]
> [1] "a3_bla" "1"
Dimitri
On Fri, Jul 29, 2011 at 12:35 PM, jim holtman wrote:
> try this:
>
>> x<-c("name_a1_2.5.o","name_a2_2.53.o","name_a3_bla_1.o")
>> st
Thanks David and Janis,
now the result is logic!
have a nice WE,
Cristabel.
On 07/29/2011 06:32 PM, David Winsemius wrote:
On Jul 29, 2011, at 12:21 PM, cristabel.duran wrote:
Hi Janis,
thank you for you answer.
Actually already I did exactly what you said. However, when I convert
the num
Hi,
If you are doing repeated fits in this specialized case, you can
benefit by skipping some of the fancy overhead and going straight to
lm.fit.
## what you had
data.residuals <- t(apply(predictors, 1, function(x)( lm(regress.y ~
-1 + as.vector(x))$residuals)))
## passing the matrices directly
try this:
> x<-c("name_a1_2.5.o","name_a2_2.53.o","name_a3_bla_1.o")
> strsplit(sub(".*?([^_]+)_([^-]+).o$", "\\1 \\2", x), ' ')
[[1]]
[1] "a1" "2.5"
[[2]]
[1] "a2" "2.53"
[[3]]
[1] "bla" "1"
>
On Fri, Jul 29, 2011 at 12:08 PM, Dimitri Liakhovitski
wrote:
> Hello!
> Hope you could help me
On Jul 29, 2011, at 12:21 PM, cristabel.duran wrote:
Hi Janis,
thank you for you answer.
Actually already I did exactly what you said. However, when I
convert the number of "seconds" from the first measurement (some day
in July 2010) then a I get a date in the year 2050 which is
impossib
Hi Janis,
thank you for you answer.
Actually already I did exactly what you said. However, when I convert
the number of "seconds" from the first measurement (some day in July
2010) then a I get a date in the year 2050 which is impossible. The
latest date possible is 29March2011.
my value in
Oh darn, I missed the recursive-ness entirely. You condition on the filtered
series, not the signal itself.
In that case, I have a solution which is pretty fast, but not particularly
R-esque.
In effect your filter just says, take x but if you see a 1, sit out for the
next wait periods. This seems
I wasn't at my normal computer yesterday, so I didn't run the example. I
thought "rgl" was one of those color palette generator packages. So, my
suggestion of using animation was completely off base (oops).
But I did notice two other things:
1) your color column changes from a factor to a charac
On Jul 29, 2011, at 11:32 AM, Simon Kiss wrote:
Dear colleagues,
I'm using R64 (2.13) on Mac OS 10.6.8 and I've encountered a problem
with the recode function in Rcommander.
It's probably in a package named 'car'.
library(car)
?recode
The application cannot deal with apostrpohes ( ' ) d
Hi, everyone.
I need to run lm with the same response vector but with varying predictor
vectors. (i.e. 1 response vector on each individual 6,000 predictor vectors)
After looking through the R archive, I found roughly 3 methods that has been
suggested.
Unfortunately, I need to run this task multi
Hello!
Hope you could help me split the strings.
I have a set of strings:
x<-c("name_a1_2.5.o","name_a2_2.53.o","name_a3_bla_1.o")
I need to extract from each string:
1. Its unique part that comes before the last "_", i.e.: "a1","a2","a3_bla".
2. The part that comes after the last "_" and befor
Thank you for all your help. The unexpected solution to my predict()
problem was to use the Misc menu and remove all objects before
proceeding further. The final digits of the trees continue to be
clipped but I can live with that.
Mark
__
R-help@r-pr
I'm not sure I understand what your filter intends to do, but could this not
be done more efficiently with logicals and the which? You also might need
the cumsum() function and diff() with the optional lag argument if I've
misunderstood your filter.
Specifically try this:
res = c(x,rep(NA,wait))
On Fri, Jul 29, 2011 at 4:32 PM, Roy Mendelssohn
wrote:
> Actually this file, as it contents state, follow the CF convention, which is
> widely used in climate science, oceanography and atmospheric science (see
> http://cf-pcmdi.llnl.gov/documents/cf-conventions). The particular
> projection
On Fri, Jul 29, 2011 at 11:26 AM, wrote:
> data=read.table("http://statcourse.com/research/boston.csv";, ,
> sep=",", header = TRUE)
> library(rpart)
> fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+
> PT+B+LSTAT)
> predict(fit,data[4,])
>
> plot only reve
Hi! Thank you all. I finally understood the way to solve the problem
from Barry's mail.
>From the file created with coordinates(Lat,Long) using fan, everything
looks ok (wgs84). I will now try Bryan's approach in R.
Ana
On Fri, Jul 29, 2011 at 4:22 PM, Barry Rowlingson
wrote:
>> float l
>
>
> The rlat and rlon variables are the grid coordinate system, which
> appears to be some non-standard conical projection. Climate
> scientists!
>
Actually this file, as it contents state, follow the CF convention, which is
widely used in climate science, oceanography and atmospheric scienc
Dear colleagues,
I'm using R64 (2.13) on Mac OS 10.6.8 and I've encountered a problem with the
recode function in Rcommander. The application cannot deal with apostrpohes (
' ) do not. I've got a factor from the 2008 Canada Election study (highest
level of schooling) and some of the values i
Your plot only shows part of the tree provided by print(fit).
I would like to plot the entire tree.
Original Message
Subject: Re: [R] help with plot.rpart
From: Joshua Wiley <[1]jwiley.ps...@gmail.com>
Date: Fri, July 29, 2011 8:25 am
To: [2]m...@statcourse
I understand, that eta^2 presents a useful measure of effect sizes.
However, the power-calculation-tool G*Power I intend to use, requires an
"Effect size f" which is calculated like this: "Variance explained by
special effect" / "Variance explained by special effect" + "Variance within
groups"
I
Hi all,
table() did the trick, and very efficiently, too! Thanks for the
advice,
Dave
On Thu, Jul 28, 2011 at 5:39 PM, David Winsemius wrote:
>
> On Jul 28, 2011, at 4:24 PM, David Warren wrote:
>
> Hi all,
>>
>>I'm working with a sizable dataset that I'd like to summarize, but I
>>
Hi,
I have a question about a special recursive filter problem.
What I have:
- given variables:
x: time series with rather randomly occuring '0' and '1'
wait: non negative integer
- a working but ineffectiv implementation (see below)
How the implementation works (what I want):
The filter s
data=read.table("http://statcourse.com/research/boston.csv";, ,
sep=",", header = TRUE)
library(rpart)
fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+
PT+B+LSTAT)
predict(fit,data[4,])
plot only reveals part of the tree in contrast to the results on obtai
Mark,
The below is not directly reproducible. There is no "MV" or "PT"
variable in the dataset you reference. I am assuming you meant:
dat <- read.table("http://statcourse.com/research/boston.csv";, ,
sep=",", header = TRUE)
fit <- rpart(MEDV ~ CRIM + ZN + INDUS + CHAS + NOX + RM + AGE + DIS
Ah, now we're getting somewhere. Think how easy this would be if
you'd provided this information initially.
On Fri, Jul 29, 2011 at 11:11 AM, wrote:
> Ø data=read.table("http://statcourse.com/research/boston.csv";, , sep=",",
> header = TRUE)
> Ø library(rpart)
> Ø fit=rpart (MV~ CRIM+ZN+INDU
> float lat(rlat, rlon) ;
> lat:standard_name = "latitude" ;
> lat:long_name = "latitude" ;
> lat:units = "degrees_north" ;
> float lon(rlat, rlon) ;
> lon:standard_name = "longitude" ;
> lon:long_name = "longi
Thanks Jannis,
I am including an example code of what I am trying to do:
par(mar=c(4,5,3,1))
barplot(Value[,1],space=0, col="grey30",axis.lty=2)
barplot(sampledepth[,1]/2, space=0, col="grey30", inside=FALSE,
add=TRUE)
sampledepth is divided by two to have the same axis scale as Valu
Well, the CF ncdf convention (for example) suggests to use variables
with identical names as the dimensions as coordinate variables. In your
case the dimensions are called rlat/rlong so the software uses these
rotated axes. In your file you however also have variables lat and long
and they seem
à data=read.table("http://statcourse.com/research/boston.csv";, , sep=",",
header = TRUE)
à library(rpart)
à fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT)
Please: Show me the tree.
Mark
Original Message
Subject: Re: [R] help
Hi Ana:
On Jul 29, 2011, at 8:03 AM, Ana wrote:
> netcdf precip.DMI.HS1 {
> dimensions:
> rlat = 84 ;
> rlon = 90 ;
> time = 10800 ;
> time_bnds = 2 ;
> variables:
> float lat(rlat, rlon) ;
> lat:standard_name = "latitude" ;
> lat:long_name
Hi Roy and David,
My main problem is how to interpret the coordinate system.
If I try to open it directly in a GIS software that supports necdf
format, the location is far away from where it should be (plots
rlong,rlat).
All I want to do is to get a data.frame with a column for long,
another for l
Dear Collin,
as always, a reproducible example code would help us to understand what
you want to do. This way I can only guess
And my guess would be that it is much easier to use:
par(new=TRUE)
and to superimpose the barplot with whatever curve you want to include.
You may need to set
On Fri, Jul 29, 2011 at 7:44 AM, Ana wrote:
> Can someone help me out with a "small" problem?
>
> I've started using netcdf files recently, and I want to extract the grid id
> and also the coordinates from a HIRHAM netcdf file.
> I know how to extract a slice of dataset both in space and in time
Thanks to several who responded--I should have looked (doh!) in the
documentation. play3d() and movie3d() were just the ticket!
On 7/28/2011 4:48 PM, Jean V Adams wrote:
>
> Dale,
>
> I do not have the same color problem when I run your code on my PC.
> The colors in both devices look the sam
This problem may be more related to netcdf files than to RAnd the
answer to your question very much depends on which netcdf convention
the AWI folks who created your data followed. Additionally it depends on
the Software you use to read these ncdf files.
Regarding the grid locations there
I am not familiar with the HIRHAM netcdf files. Do you have the netcdf library
installed and can you print the output of the "ncdump -h" command on one of the
files?
-Roy
On Jul 29, 2011, at 7:44 AM, Ana wrote:
> Can someone help me out with a "small" problem?
>
> I've started using netcdf
On Jul 29, 2011, at 10:44 AM, Ana wrote:
Can someone help me out with a "small" problem?
I've started using netcdf files recently, and I want to extract the
grid id
and also the coordinates from a HIRHAM netcdf file.
I know how to extract a slice of dataset both in space and in time
and I
Dear list,
I want to plot a sample depth curve over a barplot. It would be perfect if
the argument "inside" in the barplot function would be functional, cause I
could just add this curve to the actual barplot, but it seems like it is not
(not yet implemented). Argument "inside" would allow not to
Mark,
The R-help list is not telepathic. I asked entirely reasonable follow-up
questions intended to elicit from you a more extensive explanation of
your problem.
On Fri, Jul 29, 2011 at 10:26 AM, wrote:
> 1. I did not receive your reply, which, not incidentally, does not address
> either of my
Can someone help me out with a "small" problem?
I've started using netcdf files recently, and I want to extract the grid id
and also the coordinates from a HIRHAM netcdf file.
I know how to extract a slice of dataset both in space and in time and I
also know the area that this file should cover, h
Well,if you would have had a look at ?POSIXct the documentation would
have given
you quite a comprehensive explanation of the meaning of the values of
your time vector as they are the values these POSIXct values are
actually stored in (they are seconds since 1970-01-01). Additionally you
would
On Fri, Jul 29, 2011 at 3:49 PM, gaiarrido wrote:
> Thanks very much,
> and...Which do you recommend?
>
I am using emacs + ESS - but as you have mentioned "options", I assume you
would like to e.g. manage package installation from the GUI - which you can
do with both mentioned GUIs.
Which one I
Dear Mario,
Have a look at Eclipse with the StatET plugin. One of the benefits is that is
available for both Linux and Windows which simplifies things when you need to
switch often between Windows and Linux (e.g. Windows at work and Linux at home).
Best regards,
Thierry
--
Hi Martin,
As Sarah said, I do not know of any way to "change the default", but
you can certainly do it manually each time. Here is an example:
## generate some data
set.seed(10)
x <- rnorm(101)
## store boxplot results
s <- boxplot(x, plot = FALSE)
## look at them
s
## replace the stats with lo
On Jul 29, 2011, at 10:10 AM, Eik Vettorazzi wrote:
Hi Philip,
have a look at ?plotmath and try
mtext(expression(paste("Total Landings of"~~italic("Pecten
maximus"),"(tonnes)")), side=2,line=2)
"~~" is used for extra space between plain and italic font, but that
might be a matter of taste.
Thanks very much,
and...Which do you recommend?
-
Mario Garrido Escudero
PhD student
Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola
Universidad de Salamanca
--
View this message in context:
http://r.789695.n4.nabble.com/R-in-Linux-Ubuntu-tp3703329p3704107.html
I'm using rioja 0.5-6 with R 2.12.1 / x86_64-pc-linux-gnu on
Kubuntu 11.04.
When I run this example from the rioja reference manual:
library(rioja)
# compare diatom data from core from Round Loch of Glenhead
# with SWAP surface sample dataset
data(RLGH)
data(SWAP)
result <- compare.data
Hi Philip,
have a look at ?plotmath and try
mtext(expression(paste("Total Landings of"~~italic("Pecten
maximus"),"(tonnes)")), side=2,line=2)
"~~" is used for extra space between plain and italic font, but that
might be a matter of taste.
hth.
Am 29.07.2011 15:34, schrieb Philip Boulcott:
> I w
Dear Eik,
Am Freitag, den 29.07.2011, 15:46 +0200 schrieb Eik Vettorazzi:
> how about this
> for (i in 1:2) { print( do.call(paste("f",i,sep=""),list(2, 3) )) }
>
> or using get
> for (i in 1:2) { print( get(paste("f",i,sep=""))(2, 3) ) }
works great for me. Thank you very much. Regarding searc
On Jul 29, 2011, at 9:28 AM, Paul Menzel wrote:
Dear R folks,
wanting to compare different implementations of a solution I want to
script it to iterate over the different implementations. Is there a
way
to do this in the R shell/command line?
You might consider examining the short and se
Martin,
I don't know of an easy way to make that change, but do note:
The two ‘hinges’ are versions of the first and third quartile,
i.e., close to ‘quantile(x, c(1,3)/4)’. The hinges equal the
quartiles for odd n (where ‘n <- length(x)’) and differ for even
n. Whereas the q
Hi Paul,
how about this
for (i in 1:2) { print( do.call(paste("f",i,sep=""),list(2, 3) )) }
or using get
for (i in 1:2) { print( get(paste("f",i,sep=""))(2, 3) ) }
cheers
Am 29.07.2011 15:28, schrieb Paul Menzel:
> Dear R folks,
>
>
> wanting to compare different implementations of a solution
Hi:
The argument stringsAsFactors = FALSE will suppress character data
from being converted to factors. For more control, see the arguments
colClasses and as.is in ?read.table - they should carry over to
read.delim[2] but don't quote me on that :)
HTH,
Dennis
On Fri, Jul 29, 2011 at 1:45 AM, ang
1 - 100 of 147 matches
Mail list logo
