Hi RUsers,
I am currently tying to use RExcel via R installed in my remote server
(CentOS 5.5).
Could you please help me set up a R remote server?
All I know about the server is its IPAddress.
Apologies for my fundamental question.
Thank you so much!
Taka
It likely means that your x and y are differently long. That is, affect1 and
adh1scr do not contain the same number of values in that instance. That
precludes them from being plotted against each other. abline and lowess
would fail for the same reason.
x-c(1,2,3)
y-c(2,4)
plot(y~x)
Note what are supposed to be quotation marks around ID in your post are a
circumflex instead. Maybe the problem is that the quotation marks that are
supposed to be around ID are not recognized by your R version. May you do
not use the proper font encoding. Use straight quotation marks or
Thanks for the advice. I tried to install the package as instructed. Command
and results below:
install.packages(patchDVI, repos=http://R-Forge.R-project.org;)
Installing package(s) into C:/Users/nova/Documents/R/win-library/2.13
(as lib is unspecified)
Warning message:
In
Dear R-Users,
I am currently trying my luck with Hidden Markov Chains and use the
package mhsmm. I was very shocked to see how the initial values for
the transition matrix and probability distributions affected the
outcomes. The results were extremely different even for small changes
in the
Hello Everybody.
i am installing rgeo in R-2.13 on debian lenny. i am getting following
error. Can anybody help me why i m not able to compile rgeos.
rgeos_misc.c: In function rgeos_hausdorffdistance:
rgeos_misc.c:55: error: GEOSHausdorffDistance_r undeclared (first use in
this function)
A function for non-parametric multivariate analysis of variance (should do
univariate, too, I guess) allowing for interactions (as far as I can tell)
is implemented in the anosim() function of the vegan package.
See also:
http://cc.oulu.fi/~jarioksa/opetus/metodi/vegantutor.pdf
Thank you for your post Greg.
Do you have any useful references regarding this variability (papers etc)?
Many thanks.
Dave
From: Greg Snow greg.s...@imail.org
To: Dave Evens daveeve...@yahoo.co.uk; r-help@r-project.org
r-help@r-project.org
Sent: Thursday, 16 June 2011, 21:32
Subject: RE: [R]
Thank you for all the previous support. I have another question:
I am using eRm to run a partial credit model. Is there a way to ask eRm
to compute and present to me the probability of a person with a specific
ability to achieve each one of the scores 0, 1, 2, 3, or 4 on a specific
question with
On 16.06.2011 16:19, Eric Elguero wrote:
Hi everybody,
I just tried to run R on one of my projects
but it did not want to run:
R version 2.12.1 (2010-12-16)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-pc-linux-gnu (64-bit)
some lines
Since this example is not reproducible (and you have not quuoted any
former code) I can only give advice in principle:
1. Never use 1:length(x) since this will seriously fail if x is a length
0 object. Instead, use seq_along(x)
2. If k is a list, then you probably want to use doubled brackets
On 17.06.2011 09:24, christiaan pauw wrote:
Thanks for the advice. I tried to install the package as instructed. Command
and results below:
install.packages(patchDVI, repos=http://R-Forge.R-project.org;)
Installing package(s) into ‘C:/Users/nova/Documents/R/win-library/2.13’
(as ‘lib’ is
Well you need to recalculate the x values and need to interpolate for
the position where you lines cross the m lines
Uwe Ligges
On 16.06.2011 23:35, Muhammad Rahiz wrote:
Hi all,
I have the following script which fills the values which are less than
the mean of a given timeseries.
Hi
Hi,
If you truly have an array, this is option that should be much faster
than a loop:
index - which(is.na(dat))
dat[index] - dat[index - 1]
the only catch is that when there previous value is NA, you may have
to go through the process a few times to get them all. One way to
- Original Message
From: Uwe Ligges lig...@statistik.tu-dortmund.de
To: Ben Rhelp benrh...@yahoo.co.uk
Cc: r-help@r-project.org
Sent: Thu, 16 June, 2011 14:38:12
Subject: Re: [R] Porting unmaintained packages to post R 2.10.0 era
[...]
What about --binary is deprecated? What
Hi Prof Brian,
Thank you for your email and for writing MASS. This book is brilliant.
- Original Message
From: Prof Brian Ripley rip...@stats.ox.ac.uk
To: Ben Rhelp benrh...@yahoo.co.uk
Cc: r-help@r-project.org
Sent: Thu, 16 June, 2011 14:48:00
Subject: Re: [R] Porting
On 17.06.2011 12:04, Ben Rhelp wrote:
- Original Message
From: Uwe Liggeslig...@statistik.tu-dortmund.de
To: Ben Rhelpbenrh...@yahoo.co.uk
Cc: r-help@r-project.org
Sent: Thu, 16 June, 2011 14:38:12
Subject: Re: [R] Porting unmaintained packages to post R 2.10.0 era
[...]
What
Juergen,
I try to install Rmpi as root with install.packages(Rmpi).
It fails with:
...
I encountered a similar problem with openmpi-1.5.3 and
Rmpi_0.5-9.tar.gz. The reason was an unchecked call to
dlopen(libmpi.so.0). At least openmpi-1.5.3 does not provide
libmpi.so.0 anymore. The patch
Hi All,
I am trying to trace the origin of the current loess implementation in
R. The reference mentions that Prof Ripley based it on the 1998 version
of dloess. When I look at dloess in http://www.netlib.org/a, the file
changes mentions dloess was made available in 1992 and that a memory
Dube, Jean-Pierre Jean-Pierre.Dube at chicagobooth.edu writes:
To whom it may concern,
I am trying to maximize a log-likelihood function using optim.
This is a simple problem with only 18
parameters. To conserve memory, I am using sparse matrices
(SLAM) for some of the data matrices
Dear R-Users,
I am currently trying my luck with Hidden Markov Chains and use the
package mhsmm. I was very shocked to see how the initial values for
the transition matrix and probability distributions affected the
outcomes. The results were extremely different even for small changes
in the
Hello all,
I have a php script that makes a call
system(PATH_PROJECT_ROOT . '/sh/combineDays.sh ' . $dateString . .
PATH_PROJECT_ROOT);
and within combineDays.sh I run an R script:
R --slave --vanilla --args $testDates $BASEDIR $defaultSearchVersion
$BASEDIRR/combineDays.R
At
It is very likely that your GEOS installation is too old. Are you trying to
install rgeos 0.1-8, available since yesterday from CRAN? This should check
that GEOS is = 3.2.2, so perhaps you are trying to install an earlier
version of rgeos? You have not provided your full configure report, so it is
Sorry, forgot to quote:
Hi,
I am trying to use the objects from the list below to create more objects.
For each year in h I am trying to create as many objects as there are B's
keeping only the values of B. Example for 1999:
$`1999`$`8025`
B
B 8025 8026 8027 8028 8029
80251
Thanks! na.locf was spot on
Very fast approach to this type of problem.
--
View this message in context:
http://r.789695.n4.nabble.com/Replacing-values-without-looping-tp3602247p3605203.html
Sent from the R help mailing list archive at Nabble.com.
__
Dear R-users
I seem to be stumped on something simple. I want to split a data frame
by factor levels given in one or more columns e.g. given
dat - data.frame(x = runif(100),
fac1 = rep(c(a, b, c, d), each = 25),
fac2 = rep(c(A, B), 50))
I know I can split
try this:
dat - data.frame(x = runif(100),
fac1 = rep(c(a, b, c, d), each = 25),
fac2 = rep(c(A, B), 50))
splits - c(fac1, fac2)
split(dat, dat[splits])
I hope it helps.
Best,
Dimitris
On 6/17/2011 2:34 PM, carslaw wrote:
Dear R-users
I seem to be stumped on something simple. I
Dear Erin,
On Fri, Jun 17, 2011 at 1:45 AM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Dear R People: (particularly those who have build Rcmdr Plugin packages):
I'm building a new Plugin and keep getting the following error:
Error in if (is.null(where) || where = n) rbind(object1, object2)
On 17.06.2011 14:39, Stephen Ellison wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Uwe Ligges
Sent: 17 June 2011 12:04
See the manual. It tells you
R CMD INSTALL --build
will generate a binary package.
...
it
Thanks Uwe
I have installed R in a path without spaces ( c:\R\R-2.13.0 ..) and ten
installed patchDVI
I Then rAn the command provided by Duncan on a file known to be correct from
the Sweave examples:
example(Sweave)
(lots of output)
and then
patchDVI::SweavePDF(Sweave-test-1.tex)
I still do not get the point for what task this expansion of data may be
useful, by I guess you want (in this case probably very inefficient, but
other can work out how to improve if interested) to insert after
k - lapply(h, function (x) x*0)
the lines:
for(i in seq_along(k)){
temp -
I am looking for a discrete test for testing SNP data. Is ther any common
ones on the market for the 2 x 3 SNP table? I don't want to use the
Chi-square test.
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Hi everyone,
I have bunch of date and time observations in the format %Y-%m-%d %I %M %S
%p. I used strptime() to read this format. But the problem is some of the
times are in the format of %I %M %p, so for those times, strptime is giving
me NA values.
For example,
You could loop over the list of reasonable formats,
using the first one that works (does not return NA):
f - function (strings, formats) {
times - strptime(strings, formats[1])
i - 1
while ((i length(formats)) any(isBad - is.na(times))) {
i - i + 1
times[isBad] -
I am sure there is a more elegant version of doing this. But this works:
x-rnorm(20)
y-matrix(1:5) #number of points to sample
f-function(z){sample(x,z)}
apply(y,1,f)
Just adjust y to your liking.
HTH,
Daniel
alfredo wrote:
Hi All,
I'd like to randomly sample a vector N times, where
Just use lapply():
set.seed(1)
x - rnorm(20)
lapply(1:3, function(i) sample(x, i))
[[1]]
[1] -0.6264538
[[2]]
[1] -0.3053884 -2.2146999
[[3]]
[1] -2.21469989 -0.30538839 -0.04493361
HTH,
Marc Schwartz
On Jun 17, 2011, at 12:53 PM, Daniel Malter wrote:
I am sure there is a more elegant
You don't say what happens if both arrays have non-missing entries, but
assuming that doesn't happen:
ifelse(is.na(xf),xg,xf)
[1] W k h NA g r j NA v d NA v NA z r r i
--
David L Carlson
Associate Professor of Anthropology
Texas AM University
getting errors when running this file
http://r.789695.n4.nabble.com/file/n3605311/il1ra_status.txt
il1ra_status.txt
Age.co-c(dat2b$Dr_Age)
Sex.co-as.factor(dat2b$Sex)
casecont.co-as.factor(dat2b$Self_T1D)
stat.co-as.factor(dat2b$status)
m -
Please, can you help me with the following? I want to include a Greek letter
as part of a factor label that I need to use as factor on a xyplot. Probably
this was already solved so can I get the place where I can look for?
Thanks,
Hugo
[[alternative HTML version deleted]]
Dear all,
I would like to fit an ARMA model, but I'm not sure exactly how to fit it.
Here's an example of the problem.
This is my time variable, hourly data
t - seq(as.POSIXct(2011-01-01 00:00:00), as.POSIXct(2011-12-31 23:00:00),
by=hour)
my response
y - rnorm(length(t), 1000, 500)
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Uwe Ligges
Sent: 17 June 2011 12:04
See the manual. It tells you
R CMD INSTALL --build
will generate a binary package.
...
it seems I am not the only one to make
If the problem is that in the history R adds a new newline avery time, you
should go to Options and activate clear input lines (or something
similar).
marco t.
On Thu, Jun 16, 2011 at 7:38 PM, Jonathan Daily biomathjda...@gmail.comwrote:
Do you mean prompts (the that indicates R is waiting
Hi,
I used R CMD build to build a package but realized that the .Rnw file was
deleted after the .tar.gz file was generated. Is it supposed to be the case? If
not, any idea as to why it happened?
Thanks,
YL
__
R-help@r-project.org mailing list
Hi All,
I'd like to randomly sample a vector N times, where each successive random
sample increases in size. I have realised that the function sample does not
take vectors for arguments. For example,
x-rnorm(20,0,1)
sample(x,c(1,2,3)) ## will only return one random sample of size 1.
The trick
Hi,
Does anyone know how i get rid of the marker point in my error bars? The
bars default function (it seems) is to have a dot in the middle of the bar,
however i don't want it there.
I am using Hmis to draw my errorbars and code is
a-as.vector(tapply(Sporangia,list(Host,Isolate),mean))
-Original Message-
You can install to any library, even a temporary one that
does not need to be in .libPaths() of your regular R
installation. So you won't do any harm to your sytsem.
The principal point I was making is that the manuals are not very informative
on the
Hi
A client of mine has asked me to investigate the installation of R-software.
Could anyone tell me whether the software works only on a client machine or
whether it sits on a server with clients attaching to it?
Not immediately clear from the docs.
Best
Oliver
--
Oliver
Hello,
I am working with two datasets with ~5,200 observations. Apparently,
neither the correlog function (ncf package) nor the mantel function (ecodist
package) can handle these datasets. Does anyone have any experience about
the maximum number of observations these functions
You need to use hhid as the rownames for housing.cluster rather than
including it as a variable in the data.frame:
housing.cluster -data.frame(htypec1, afforcr1, resyrc1, crowdcc1, chprbos1)
rownames(housing.cluster) - hhid
Then it will not be included in the cluster analysis but will be used to
Hello,
I need to create a scatterplot where the y-axis is upside down. If I have
non-negative bivariate data in objects x and y, then the operation
plot(x, -y)
gives me the figure I want -- a mirror image of plot(x, y) -- except that the
y-values (coordinates) are negative, which I don't
?plotmath
On Fri, Jun 17, 2011 at 6:32 AM, Hugo Loyola quinc...@gmail.com wrote:
Please, can you help me with the following? I want to include a Greek letter
as part of a factor label that I need to use as factor on a xyplot. Probably
this was already solved so can I get the place where I can
Hi friends,
I have a matrix with following format.
group var1 var2 ...varN
c1 group1 1.2399 1.4990-1.4829
c2 group4 0.8989 0.7849.1.8933
...
...
c100 group10 .
I want to draw a profile plot
of each condition c1 to c100, which rows in above matrix and each line
On Jun 17, 2011, at 2:31 PM, Owen, Jason wrote:
Hello,
I need to create a scatterplot where the y-axis is upside down. If I have
non-negative bivariate data in objects x and y, then the operation
plot(x, -y)
gives me the figure I want -- a mirror image of plot(x, y) -- except that
That's fantastic! Thanks.
-Original Message-
From: Marc Schwartz [mailto:marc_schwa...@me.com]
Sent: Friday, June 17, 2011 3:47 PM
To: Owen, Jason
Cc: 'r-help@r-project.org'
Subject: Re: [R] plot the y-axis upside down
On Jun 17, 2011, at 2:31 PM, Owen, Jason wrote:
Hello,
I am not an expert in time series (that is why I referred you to the task view
rather than give my own inexpert opinion). I do remember from a textbook that
covered the basics of time series that prediction 2 time points ahead was
different from plugging in the next time estimate and
Does this do what you want?
x - abs(rnorm(100))
tt - 1:100
m - mean(x)
par(mfrow=c(2,1))
yy - c(0,3)# y-limit
plot(tt,x,type=l,ylim=yy)
abline(h=m)
clip(0,100,0,m)
polygon( c(1,tt,100), c(m,x,m), col='red' )
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
Hello,
Is the following a bug? I always thought that df$varname - does the same as
df[varname] -
df - data.frame(weight=round(runif(10, 10, 100)), sex=round(runif(100, 0,
1)))
df$pct - df[weight] / ave(df[weight], df[sex], FUN=sum)*100
names(df)
[1] weight sexpct ### -- ok
What do you want to happen when both are NA? what do you want to happen if both
have values?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org
df$varname is a column of df.
df[varname] is a one-column df containing that column.
df[[varname]] is a column of df (same as df$varname).
df[,varname] is a column of df (same as df$varname).
df[,varname,drop=FALSE] is a one-column df (same as df$varname).
df$newVarname - df[varname] inserts
And the extra twist in the tale is exemplified by this
mini-version of Albert-Jan's first example:
DF - data.frame(A=c(1,2,3))
DF$B - c(4,5,6)
DF$C - c(7,8,9)
DF
# A B C
# 1 1 4 7
# 2 2 5 8
# 3 3 6 9
DF$D - DF[A]/DF[B]
DF
# A B CA
# 1 1 4 7 0.25
# 2 2 5 8 0.40
On 6/17/2011 2:24 PM, (Ted Harding) wrote:
And the extra twist in the tale is exemplified by this
mini-version of Albert-Jan's first example:
DF- data.frame(A=c(1,2,3))
DF$B- c(4,5,6)
DF$C- c(7,8,9)
DF
# A B C
# 1 1 4 7
# 2 2 5 8
# 3 3 6 9
DF$D- DF[A]/DF[B]
DF
Hi,
There is nothing sacrosanct about taking a random sample of size 1 then 2 then
3 versus just taking one random sample of size 6 (with replacement, if you
like) and then just use the first element, elements 2:3 and elements 4:6.
HTH,
Josh
On Jun 17, 2011, at 6:59, alfredo
I think need to do something like this:
dat-data.frame(state=sample(id=rep(1:5,each=200),1:3, 1000,
replace=T,prob=c(0.7,0.05,0.25)),V1=runif(1,10,1000),V2=rnorm(1000))
rle.dat-rle(dat$state)
temp-1
out-data.frame(id=1:length(rle.dat$length))
for(i in 1:length(rle.dat$length)){
Justin Haynes jtor14 at gmail.com writes:
I think need to do something like this:
dat-data.frame(state=sample(id=rep(1:5,each=200),1:3, 1000,
replace=T,prob=c(0.7,0.05,0.25)),V1=runif(1,10,1000),V2=rnorm(1000))
brown bag...
dat-data.frame(id=rep(1:5,each=200),state=sample(1:3, 1000,
Hello Dear R user,
I want to conduct a Principal components analysis and I need to run two
tests to check whether I can do it or not. I found how to run the KMO
test, however i cannot find an R fonction for the Bartlett's test of
sphericity. Does somebody know if it exists?
Thanks for your
you can do something like this
x-rnorm(20,0,1)
time=c(1,2,3)
sapply(1:length(time),function(t) sample(x,time[t]))
Weidong Gu
On Fri, Jun 17, 2011 at 9:59 AM, alfredo alfredote...@gmail.com wrote:
Hi All,
I'd like to randomly sample a vector N times, where each successive random
sample
I am trying to fit a curve to a cumulative mortality curve (logistic) where y
is the cumulative proportion of mortalities, and t is the time in hours (see
below). Asym. at 0 and 1
y
[1] 0. 0.04853859 0.08303777 0.15201970 0.40995074 0.46444992
0.62862069 0.95885057 1.
[10]
R Experts
I'm currently using an S+ script of the following format and would
like to convert it to R. The script opens a graphsheet with an
associated name, plots something (in this case a boxplot) and then
exports the contents of the graphsheet of the assigned name to an EMF
file. I've
Hi everyone,
Apologies if this is a silly question but I am a student and this is my
first time using R so I am still trying to educate myself on commands,
models e.t.c
I have a mixed model with four dichotomous fixed factors and subject as a
random factor (as each person completed four
Hello -
I am having trouble extracting data from NetCDF data files using
RNetCDF. The data files each have 3 dimensions (longitude, latitude,
and a date) and 3 variables (latitude, longitude, and a climate
variable).
Here is some of the output from print.nc for clarity:
-
dimensions:
Justin jtor14 at gmail.com writes:
I think need to do something like this:
dat-data.frame(state=sample(id=rep(1:5,each=200),1:3, 1000,
replace=T,prob=c(0.7,0.05,0.25)),V1=runif(1,10,1000),V2=rnorm(1000))
brown bag...
... its friday and im sleepy!...
chmod 777 on the destination folder fixed the sink() issue. Still don't know
why library() was causing the script to stop though.
- Original Message -
From: Yoni Teitelbaum yoni.teitelb...@escapemg.com
To: r-help@r-project.org
Sent: Friday, June 17, 2011 5:41:33 AM
Subject: [R]
I am trying to get the
Vim-R-Pluginhttp://www.vim.org/scripts/script.php?script_id=2628 to
work with gvim and R on Windows 7.
When I open a .R file in VIM, it complains and says Python interface must
be enabled to run Vim-R-Plugin. I have installed pywin32 for python 2.7,
and added the following
Hi,
How can I accomplish this in R. Example:
I have the following data.frame:
data -
data.frame(x=c(1,2,3,4,5,6,5,3,7,1,0,4,8),y=c(1,2,1,2,2,2,1,1,1,2,2,2,2),z=c(5,8,4,3,4,1,6,3,3,6,3,5,7))
Supposing that data$y is a factor, I would like to find the Spearman
correlation between data$x and
This question was just answered yesterday in this post:
http://r.789695.n4.nabble.com/Correlations-by-subgroups-td3599548.html#a3600553
One solution:
x-c(1,1,1,1,1,2,2,2,2,2)
y-rnorm(10)
z-y+rnorm(10)
by(data.frame(y,z),factor(x),cor)
HTH,
Daniel
Mateus Rabello wrote:
Hi,
How can I
Can this be used in matplot(x,y,)? where x and y have matching rows.
However, each column may have different rows or length. Thanks. Sheng
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-join-matrices-of-different-row-length-from-a-list-tp3177212p3607158.html
Sent from
The formula for the chi-square value is:
-( (n-1) - (2*p-5)/6 )* log(det(R))
where n is the number of observations, p is the number of variables, and R
is the correlation matrix. The chi square test is then performed on
(p^2-p)/2 degrees of freedom. So you can compute it by hand. Or you can use
I have 'x' variables that I need to find the optimum combination of, with the
constraint that the sum of all x variables needs to be exactly 100. I need
to test all combinations to get the optimal mix.
This is easy if I know how many variables I have - I can hard code as below.
But what if I
No.
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers)
Here ia an idea that might be useful to adapt
fixedSumCombinations - function(N, terms)
if(terms == 1) return(N) else
if(terms == 2) return(cbind(0:N, N:0)) else {
X - NULL
for(i in 0:N)
X - rbind(X, cbind(i, Recall(N-i, terms-1)))
X
}
On Fri, 17 Jun 2011, Michael Karol wrote:
R Experts
I'm currently using an S+ script of the following format and would
like to convert it to R. The script opens a graphsheet with an
associated name, plots something (in this case a boxplot) and then
exports the contents of the graphsheet of
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