On 30/04/11 01:14, Nick Sabbe wrote:
Hi R people.
I ran into this problem: I created a plot with errbars, like this:
errbar(x=c(1,2,3,4), y=c(2,1,3,3), yminus=c(1.5,0.5,2.5,2.5),
yplus=c(2.5,1.5,3.5,3.5))
Next, I wanted to accentuate some x value with an abline, like this:
abline(v=2)
That works like a charm! Thanks so much Duncan.
On Fri, Apr 29, 2011 at 6:37 PM, Duncan Murdoch wrote:
> On 29/04/2011 9:34 PM, Miao wrote:
>
>> Thanks Duncan for clarifying this. I'm pretty a newbie to such type of
>> characters and special characters. In R's gsub() what regular
>> expression
Two additions to Michael Friendly's comments:
1. Have you considered putting your package on R-Forge
(r-forge.r-project.org)? They have a facility (which has been broken
for several months now but will likely be fixed again at some time in
the future) to test your package more or less
On 4/28/2011 8:43 AM, Singmann wrote:
Dear all,
I (and a colleague) have been working on our first package (for fitting a
certain type of cognitive models:
http://www.psychologie.uni-freiburg.de/Members/singmann/R/mptinr) for quite
a while now and have the feeling, that it is "good to go". That
On Fri, 29 Apr 2011, Duncan Murdoch wrote:
On 29/04/2011 7:41 PM, Miao wrote:
Can anyone help on gsub() in R? I have a string like something below, and
wanted to delete all the strings with leading backslash, including
"\xa0On",
"\023, "\xab", and many others. How should I write a regular
On 29/04/2011 9:34 PM, Miao wrote:
Thanks Duncan for clarifying this. I'm pretty a newbie to such type of
characters and special characters. In R's gsub() what regular
expressions shall I use to handle all these situations?
I don't know. This might work:
gsub("[\x01-\x1f\x7f-\xff]", "", x)
Thanks Duncan for clarifying this. I'm pretty a newbie to such type of
characters and special characters. In R's gsub() what regular expressions
shall I use to handle all these situations?
On Fri, Apr 29, 2011 at 6:07 PM, Duncan Murdoch wrote:
> On 29/04/2011 7:41 PM, Miao wrote:
>
>> Hello,
>
On 30/04/11 02:44, Alaios wrote:
Thanks a lot.
I finally used
M2<- M
M2[M< thresh]<- 0
M2[M>= thresh]<- 1
as I noticed that this one line
M2<- as.numeric( M[]< thresh )
vectorizes my matrix.
One more question I have two matrices that only differ slightly. What will be
the easiest way to co
On 29/04/2011 7:41 PM, Miao wrote:
Hello,
Can anyone help on gsub() in R? I have a string like something below, and
wanted to delete all the strings with leading backslash, including "\xa0On",
"\023, "\xab", and many others. How should I write a regular expression
pattern in gsub()? I don't
Unless I misunderstand, I think you might want the nlsList function from the
nlme package. It will let you fit a nonlinear model for grouped data. You use
"|" to separate the model formula from the grouping factor.
Rick
From: r-help-boun...@r-project.org
Putting SQL columns/variables into square brackets is valid syntax for sqlite.
Expecting sqlite to share variables with R is not, so there was only one
necessary change.
---
Jeff Newmiller The . . Go Live...
DCN: Basi
hi Christiano,
the error is that FUN is not a function. That is true, the argument
that you are passing to FUN is a different class. Instead of fx, for
example, where fx is your model code below, try to write it as a
function of the arguments that you want to split by Cerca.
You might try to con
Hello,
Can anyone help on gsub() in R? I have a string like something below, and
wanted to delete all the strings with leading backslash, including "\xa0On",
"\023, "\xab", and many others. How should I write a regular expression
pattern in gsub()? I don't care how many characters following ba
On Apr 29, 2011, at 4:27 PM, mathijsdevaan wrote:
Hi list,
Can anyone tell my why the following does not work? Thanks a lot!
Your help
is very much appreciated.
DF = data.frame(read.table(textConnection("B C D E F G
8025 1995 0 4 1 2
8025 1997 1 1 3 4
8026 1995 0 7 0
Hi, Dennis,
Thanks for the reply. I tried to upgrade to R 2.13.0. Then when I tried to
load the library(nparcomp), I got an error
Error: package 'mvtnorm' is not installed for 'arch=i386'
What does that mean? Thanks.
Jun
On Fri, Apr 29, 2011 at 5:49 PM, Dennis Murphy wrote:
> Hi:
>
> Is this
Hi:
Is this the function nparcomp() in the nparcomp package or the one
from the mutoss package? When using functions from packages, it is
useful to indicate the package name. I'm assuming you're using the
nparcomp package, because your code worked for me when that package
was loaded:
> library(np
Hi Ryan
postForm() is using a different style (or specifically Content-Type) of
submitting the form than the curl -d command.
Switching the style = 'POST' uses the same type, but at a quick guess, the
parameter name 'a' is causing confusion
and the result is the empty JSON array - "[]".
A qui
Hi Tal
You can add
ssl.verifypeer = FALSE
in the .opts list so that the certificate is simply accepted.
Alternatively, you can tell libcurl where to find the certification
authority file containing signatures. This can be done via the cainfo
option, e.g.
cainfo = system.file("CurlSSL", "c
Hi:
Try
split(DF, DF$C)
Does that work?
Dennis
On Fri, Apr 29, 2011 at 1:27 PM, mathijsdevaan wrote:
> Hi list,
>
> Can anyone tell my why the following does not work? Thanks a lot! Your help
> is very much appreciated.
>
> DF = data.frame(read.table(textConnection(" B C D E F G
> 802
Dear list,
I tried to use the nparcomp to run some post hoc non-parametric comparison
and got and error.
Error in uniroot(pfct, interval = interval) :
f() values at end points not of opposite sign
Appreciate any comments.
the command line:
>nparcomp(Ulceration~Group,data=test,type='Dunnett'
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of
> luke-tier...@uiowa.edu
> Sent: Friday, April 29, 2011 9:35 AM
> To: Jonathan Daily
> Cc: r-help@r-project.org; Hadley Wickham; Barry Rowlingson
> Subject: Re: [R] setting optio
Hello Duncan,
Thank you for having a look at this.
I tried the code you provided but it failed in the getForm stage. running
this:
> tt = getForm("http://spreadsheets0.google.com/spreadsheet/pub";,
+ hl ="en", key =
"0AgMhDTVek_sDdGI2YzY2R1ZESDlmZS1VYUxvblQ0REE",
+ sing
Hello expeRts,
here is something which strikes me as kind of odd and I would like to ask for
some enlightenment:
First let's do this:
tkern <- kernel("modified.daniell", c(5,5))
test <- rep(1,100)
system.time(kernapply(test,tkern))
User System verstrichen
1.100 0.0
The general idea of the KS test (and others) can be applied to discrete data,
but the implementation in R assumes continuous data (does not have the needed
adjustments to deal with ties). The chi-square and other tests suffer from the
same problems in your case. In all cases the null hypothesi
On 29 April 2011 08:43, viostorm wrote:
>
> After I shared comments form the forum yesterday with the biostatistician he
> indicated this:
>
> "Fisher's exact test is the non-parametric analog for the Chi-square
> test for 2x2 comparisons. A version (or extension) of the Fisher's Exact
> test, kno
Rob--
Your biostatistician has not disagreed with the rest of us about anything
except for his preferred name for the test. He wants to call it the
Freeman-Halton test, some people call it the Fisher-Freeman-Halton test,
but most people call it the Fisher Exact test -- all are the same test.
Hi:
Does this work?
library(plyr)
ddply(tmp, .(trial, Gender), transform, rankscore = rank(score))
score trial Gender rankscore
1 1 1 M 1
2 2 1 M 2
3 3 1 F 1
4 4 1 F 2
5 4 2 M 2
6 3
Yes, I wrote also in the other forum because here I didn't take an answer.
Thanks for your reply
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-qualitative-variables-in-anova-tp3483845p3484599.html
Sent from the R help mailing list archive at Nabble.com.
_
Hi,
I am calculation pairwise correlation coefficient for a matrix of 234 X
3.
I am getting the following error,
Error in cbind(as.vector(row(cl)), as.vector(col(cl)), as.vector(cl)) :
allocMatrix: too many elements specified
In addition: There were 50 or more warnings (use warnings() to se
Hi list,
Can anyone tell my why the following does not work? Thanks a lot! Your help
is very much appreciated.
DF = data.frame(read.table(textConnection("B C D E F G
8025 1995 0 4 1 2
8025 1997 1 1 3 4
8026 1995 0 7 0 0
8026 1996 1 2 3 0
8026 1997 1 2 3 1
8026 1
Rob: Fisher's exact test is conceptually possible for any r x c
contingency table problem and uses the observed multinomial table
probability as the test statistic. Other tests for r x c contingency
tables use a different test statistic (Chi-squared, likelihood ratio,
Zelterman's). It is p
Barth sent me a very good code and I modified it a bit. Have a look:
Error<-rnorm(1000, mean=0, sd=0.05)
estimate<-(log(1+0.10)+Error)
DCF_korrigiert<-(1/(exp(1/(exp(0.5*(-estimate)^2/(0.05^2))*sqrt(2*pi/(0.05^2
))*(1-pnorm(0,((-estimate)/(0.05^2)),sqrt(1/(0.05^2))-1))
DCF_verzerrt<-(1/(e
Are you working on the same homework problem as user494766?
http://stackoverflow.com/questions/5835605/1-way-anova-in-r-help
--
David.
On Apr 29, 2011, at 10:43 AM, katerinaaa wrote:
Hi,
I am newbie in R programming and I need some help.
I have two columns the first has 1000 values Y/N/U
On Fri, 29 Apr 2011, David Winsemius wrote:
On Apr 29, 2011, at 1:29 PM, Mike Miller wrote:
On Fri, 29 Apr 2011, Giovanni Petris wrote:
Well, but the original poster also refers to 0.2 and 0.8 as "expected min
and max", in which case we are back to a joke...
Well, he is a lot better with E
-- snip --
It clogs up my email, takes a long
time to delete, and is hard to be selective enough to not
delete some of my other important email.
-- snip --
If you don't care about contributing to the R listserve community, it's hard
to imagine why that community should care about you.
Some p
Hi:
I'm starting a research of Support Vector Regression. I want to obtain a
model to predict a property A with
a set of property B, C, D, ... This problem is very common for example in
QSAR models. I want to know
some examples and package that could help me in this way. I know about
caret
Hi,
I've been able to use R in C# on my machine by following the steps on the
http://www.codeproject.com/KB/cs/RtoCSharp.aspx.
This works locally, i.e. if R is running on my box. I was wondering if its
possible to change it so that I can connect to another machine that is running
R (and has rs
Hi everybody,
I think that I am missing something fundamental in how strings are passed from
a postForm() call in R to the curl or libcurl functions underneath. For
example, I can do the following using curl from the command line:
$ curl -d "Archbishop Huxley" "http://www.datasciencetoolkit.or
Hello,
This is my first post in this e-mail list and I hope it's enough to justify
calling for help. In case it's not, sorry.
I'm trying to do analysis and graphics using a factor as a criteria to split
data and do the analysis/graphics for each subset of data.
Right now what I'm trying to do is
Thierry,
The first suggestion worked. Thank you very much.
*Ben Caldwell*
University of California, Berkeley
137 Mulford Hall #3114
Berkeley, CA 94720
Office 223 Mulford Hall
(510)859-3358
On Fri, Apr 29, 2011 at 1:52 AM, ONKELINX, Thierry wrote:
> Dear Ben,
>
> Are site, transect and plot fa
Suppose I have data such as
tmp <- data.frame(score = c(1,2,3,4, 4,3,2,1), trial = gl(2,4), Gender =
gl(2,2,8, labels=c('M', 'F')))
Now I would like to compute a rank on the variable score conditional on trial
and gender. I could do
res <- with(tmp, tapply(score, list(Gender, trial), rank))
re
Ryan -
summary expects an lm object, and fit is a list. So
you need to use something like
lapply(fit,summary)
to pass each list element to the summary function.
- Phil Spector
Statistical Computing Facility
Hi Ben,
Thank you for your help.
I did the same question in the r-sig-phylo mailing list. Liam Revell gave
the following solution:
temp<-prop.part(tree)
X<-matrix(0,nrow=length(tree$tip),ncol=length(temp),dimnames=list(tree$tip.label,tree$node.label))
for(i in 1:ncol(X)) X[temp[[i]],i]<-1
Vand
After I shared comments form the forum yesterday with the biostatistician he
indicated this:
"Fisher's exact test is the non-parametric analog for the Chi-square
test for 2x2 comparisons. A version (or extension) of the Fisher's Exact
test, known as the Freeman-Halton test applies to comparison
Folks,
I'm new to R and would like to use it to analyze web server performance data.
I collect the data in this CSV format:
1304083104.41,Y,668.856249809
1304083104.41,Y,348.143193007
First column is a timestamp, rows with N instead of Y
need to be skipped and the last column has the same fo
(11/04/29 22:09), Frank Harrell wrote:
Yes I would select that as the final model.
Thank you for your comment. I am able to be confident about my model now.
The difference you saw is caused
by different treatment of penalization of factor variables, related to the
use of the sum squared diffe
Hi,
I am newbie in R programming and I need some help.
I have two columns the first has 1000 values Y/N/U and the other has f/m.
Like that :
q7 sex
==
Um
U f
Um
Nf
I want to do one way anova parametric and no parametric.
But I have some problems.
Code:
frameq7 <- data.fram
Hi!
I was trying to install RWeka to be able to use SnowballStemmer in a Mac OS
X 10.6.7 environment... but coudn't do it... I get error messages after:
> library(RWeka);
> install(Snowball);
> ## Test the supplied vocabulary for the default stemmer ('porter'):
> source <- readLines(system.file("
Hi,
I am trying to run a regression on two matrices with 10 columns. I have
been able to run the regression with the following code:
fit=list()
for(i in 1:10) {
fit[[i]]=lm(monret[,i]~janret[,i])
}
However, I can't get the regression to spit out more than the coefficients
(summary(fit) does not
Thanks a lot Duncan, this is what I was looking for!!Thank you,Nandini
> Date: Fri, 29 Apr 2011 09:53:06 -0400
> From: murdoch.dun...@gmail.com
> To: nandini...@hotmail.com
> CC: r-help@r-project.org
> Subject: Re: [R] replace non numeric with "NA"
>
> On 29/04/2011 6:45 AM, Nandini B wrote:
Hi there,
I have the problem, that I'm not able to reproduce the SPSS residual
statistics (dfbeta and cook's distance) with a simple binary logistic
regression model obtained in R via the glm-function.
I tried the following:
fit <- glm(y ~ x1 + x2 + x3, data, family=binomial)
cooks.distance
I am using the mlogit packages and get a data problem, for which I
can't find any clue from R archive.
code below shows my related code all the way to the error
#---
mydata <- data.frame(dependent,x,y,z)
mydata$dependent<-as
On Fri, Apr 29, 2011 at 06:19:24PM +0300, Tal Galili wrote:
>
> data_url <- "
> http://spreadsheets0.google.com/spreadsheet/pub?hl=en&hl=en&key=0AgMhDTVek_sDdGI2YzY2R1ZESDlmZS1VYUxvblQ0REE&single=true&gid=0&output=csv
> "
> read.csv(data_url)
> Error in file(file, "rt") : cannot open the connection
Thanks David for fixing the early issues.
The reason for the failure is that the response
from the Web server is a to redirect the requester
to another page, specifically
https://spreadsheets0.google.com/spreadsheet/pub?hl=en&hl=en&key=0AgMhDTVek_sDdGI2YzY2R1ZESDlmZS1VYUxvblQ0REE&single=true&g
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius
> Sent: Friday, April 29, 2011 10:36 AM
> To: Tal Galili
> Cc: r-help@r-project.org
> Subject: Re: [R] read.csv fails to read a CSV file from google docs
>
>
> O
Hello,
I notice that e.g /home/sguha/lib64 is hard coded into the /bin/R file .
I nstalled R as ./configure --prefix=$HOME ...
What i need to do is ship the entire R distribution to remote nodes,
and run R. These are shipped to ephemeral directories
so I dont know the path ahead of time.
R_HOME
On Apr 29, 2011, at 1:29 PM, Mike Miller wrote:
On Fri, 29 Apr 2011, Giovanni Petris wrote:
Well, but the original poster also refers to 0.2 and 0.8 as
"expected min and max", in which case we are back to a joke...
Well, he is a lot better with English than I am with Mandarin. He
seemed
Dear Rxperts,
Can "Jordan decomposition" of submatrices be useful to determine size of sub
blocks? "http://en.wikipedia.org/wiki/Jordan_normal_form";..
Thanks for the ideas/suggestions.
.
I have another similar situation, where at least one of the off diagonal
elements of the lower triangle subma
On Apr 29, 2011, at 11:19 AM, Tal Galili wrote:
Hello all,
I wish to use read.csv to read a google doc spreadsheet.
I try using the following code:
data_url <- "
http://spreadsheets0.google.com/spreadsheet/pub?hl=en&hl=en&key=0AgMhDTVek_sDdGI2YzY2R1ZESDlmZS1VYUxvblQ0REE&single=true&gid=0&outp
On Fri, 29 Apr 2011, Giovanni Petris wrote:
Well, but the original poster also refers to 0.2 and 0.8 as "expected
min and max", in which case we are back to a joke...
Well, he is a lot better with English than I am with Mandarin. He seemed
to like the truncated normal answers, so we'll let t
In python, opening a connection using with allows for a temporary
assignment using "as". So:
with file("/path/to/file") as con:
permanent_object = function(con)
would provide the return of function(con) globally, but close con. If
function(con) causes an error, con is still closed.
I agree w
On Apr 29, 2011, at 10:44 AM, Alaios wrote:
Thanks a lot.
I finally used
M2 <- M
M2[M < thresh] <- 0
M2[M >= thresh] <- 1
as I noticed that this one line
M2 <- as.numeric( M[] < thresh )
vectorizes my matrix.
One more question I have two matrices that only differ slightly.
What will be th
The Python solution does not extend, at least not cleanly, to things
like dev on/ dev off or to Hadley's locale example. In any case if I
am reading the Python source correctly on how they handle user
interrupts this solution has the same non-robusness to user interrupts
issue that Bill's initial
Thanks a lot Jim, this is perfect!!
Thank you,
Nandini Badarinarayan
> Date: Fri, 29 Apr 2011 09:49:26 -0400
> From: jmac...@med.umich.edu
> To: nandini...@hotmail.com
> CC: r-help@r-project.org
> Subject: Re: [R] replace non numeric with "NA"
>
> Hi Nandini,
>
> On 4/29/2011 6:45 AM, Nandi
Hello all,
I wish to use read.csv to read a google doc spreadsheet.
I try using the following code:
data_url <- "
http://spreadsheets0.google.com/spreadsheet/pub?hl=en&hl=en&key=0AgMhDTVek_sDdGI2YzY2R1ZESDlmZS1VYUxvblQ0REE&single=true&gid=0&output=csv
"
read.csv(data_url)
Which results in the fo
Thanks, to all. I didn't know about either *methods( ) *or the package *
lubridate*, which seems like a very nice *Date *package.
*-- Russ *
On Fri, Apr 29, 2011 at 1:35 AM, Kenn Konstabel wrote:
> The function for getting the year from date is there in package
> lubridate (as well as many o
Dear R Users,
I am doing stats::decompose() on 4 different time series. When I issue
csdA <- decompose(tsA)
plot(csdA)
I get a summary plot for observed, trend, seasonal and random components
of decomposed time series tsA. As I understand it, the object returned
by decompose() has it's own plo
On Fri, Apr 29, 2011 at 07:44:59AM -0700, Alaios wrote:
> Thanks a lot.
> I finally used
>
> M2 <- M
> M2[M < thresh] <- 0
> M2[M >= thresh] <- 1
>
> as I noticed that this one line
>
> M2 <- as.numeric( M[] < thresh )
> vectorizes my matrix.
Hi.
This may be avoided, for example
M2 <- M
M
H do I obtain a strictly rectangular
type-double array (converted to an R 2-dimensional array) from a Java class?
I can obtain a 1-dimensional type-double array (vector) or a scalar,
but I cannot figure out the two-dimensional from the instructions.
Is .jevalArray also involved?
My simple Java te
Thanks a lot.
I finally used
M2 <- M
M2[M < thresh] <- 0
M2[M >= thresh] <- 1
as I noticed that this one line
M2 <- as.numeric( M[] < thresh )
vectorizes my matrix.
One more question I have two matrices that only differ slightly. What will be
the easiest way to compare and find the cells that
Well, but the original poster also refers to 0.2 and 0.8 as "expected
min and max", in which case we are back to a joke...
Giovanni
On Thu, 2011-04-28 at 13:06 -0400, David Winsemius wrote:
> On Apr 28, 2011, at 12:09 PM, Ravi Varadhan wrote:
>
> > Surely you must be joking, Mr. Jianfeng.
> >
David Winsemius wrote:
>
> On Apr 29, 2011, at 4:27 AM, ivan wrote:
>
>> Hi All,
>>
>> I am trying to create a function which evaluates whether the values
>> (which
>> are equal to one) of a matrix are the same as their mirror values.
>> Consider
>> the following matrix:
>>
>>> n<-matrix(cbi
On Apr 29, 2011, at 9:37 AM, Alaios wrote:
Dear all,
I have a quite big matrix which I would like to threshold.
If the value is below threshold the cell should be zero
and
if the value is over threshold the cell should be one
M2 <- M
M2[M < thresh] <- 0
M2[M >= thresh] <- 1
or perhaps simply
On 2011-04-29 06:14, Nick Sabbe wrote:
Hi R people.
I ran into this problem: I created a plot with errbars, like this:
errbar(x=c(1,2,3,4), y=c(2,1,3,3), yminus=c(1.5,0.5,2.5,2.5),
yplus=c(2.5,1.5,3.5,3.5))
Next, I wanted to accentuate some x value with an abline, like this:
abline(v=2)
If you are plotting that many data points, you might want to look at
'hexbin' as a way of aggregating the values to a different
presentation. It is especially nice if you are doing a scatter plot
with a lot of data points and trying to make sense out of it.
On Wed, Apr 27, 2011 at 5:16 AM, Jonat
On 29/04/2011 6:45 AM, Nandini B wrote:
Hello,
I have a sample data frame which looks like this
day od month
1 1 0.12
2 3 #VALUE! 1
3 5 0.4 12
4 7 0.8 10
5 11 - 3
6 14 s 7
7 18
On Apr 29, 2011, at 6:47 AM, Mathias Walter wrote:
Hi,
I have large data frame with many columns. A short example is given
below:
dataH
host ms01 ms31 ms33 ms34
1 cattle4 2096
2 sheep4345
3 cattle4345
4 cattle4345
5
Hi Nandini,
On 4/29/2011 6:45 AM, Nandini B wrote:
Hello,
I have a sample data frame which looks like this
day od month
1 1 0.12
2 3 #VALUE! 1
3 5 0.4 12
4 7 0.8 10
5 11 - 3
6 14 s
Dear all,
I have a quite big matrix which I would like to threshold.
If the value is below threshold the cell should be zero
and
if the value is over threshold the cell should be one
One really simple way to do that is two have a nested loop and check cell by
cell.
The problem is that this seem
Yes I would select that as the final model. The difference you saw is caused
by different treatment of penalization of factor variables, related to the
use of the sum squared differences between the estimate at one category from
the average over all categories. I think that as long as you code it
Hi, Jerome and Phil,
Thank you for your solutions and I have studied carefully your codes but I have
further questions (since I guess the simple lines of codes may not do the real
job I am going to describe to you. Please forgive me for my shallowness!)
I guess I over-simplified my question, ba
The rgdal package is not a dependency of sp, only suggested. In addition, you
are trying to install source packages, but should (probably) be installing
binaries, with type="mac.binary.leopard" the most likely. If you need the
OSX rgdal binary, make sure that CRAN extras is on your repository path
Hello,
I have a sample data frame which looks like this
day od month
1 1 0.12
2 3 #VALUE! 1
3 5 0.4 12
4 7 0.8 10
5 11 - 3
6 14 s 7
7 18 -- 12
8 27 19
Dear all:
Is there a way one could plot two conditional inference trees (party
package, ctree) in a figure specified by layout? My attempts failed as
plot.party seemed to take over the layout functionality and forced a single
ctree plot to be displayed. A brief (non reproducible) example togethe
thanks everyone for the help.
I ended up copying and pasting the legend function from the R source files.
I changed it so that the title.cex is not set by default to cex and so
that this title.cex can be given as a parameter.
It works fine for me.
Note that if you make the title too big it goe
Dear Colleague
I am trying to figure out how to use R to do OLS restricted VECM regression.
However, there are some notation I cannot understand.
Please tell me what is 'ect', 'sd' and 'LRM.dl1 in the following practice:
#OLS retricted VECM regression
data(denmark)
sjd <- denmark[, c("LRM",
Hi,
I have large data frame with many columns. A short example is given below:
> dataH
host ms01 ms31 ms33 ms34
1 cattle4 2096
2 sheep4345
3 cattle4345
4 cattle4345
5 sheep4355
6goat4345
7
Hi R people.
I ran into this problem: I created a plot with errbars, like this:
> errbar(x=c(1,2,3,4), y=c(2,1,3,3), yminus=c(1.5,0.5,2.5,2.5),
yplus=c(2.5,1.5,3.5,3.5))
Next, I wanted to accentuate some x value with an abline, like this:
> abline(v=2)
In one of my R sessions (which admi
... is the apply function what you are looking for?
A=matrix(1,2,4)
apply(A,1,sum)
HTH
Pete
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Hi Rob,
fastbw does not try to produce a full fit object. You have to re-run the
fit manually based on what you (sometimes dangerously) learn from fastbw.
If I can find a way to add a 'formula' component to the fastbw result then
you could do something like lrm(fastbw(fit)$formula, ...).
Frank
On 04/29/2011 08:35 PM, hck wrote:
Dear all
Problem: hist()-function, scale = “percent”
I want to generate histograms for changing underlying data. In order to make
them comparable, I want to fix the y-axis (vertical-axis) to, e.g., 0%, 10%,
20%, 30% as well as to fix the spaces, too. So the y-
On Apr 29, 2011, at 4:27 AM, ivan wrote:
Hi All,
I am trying to create a function which evaluates whether the values
(which
are equal to one) of a matrix are the same as their mirror values.
Consider
the following matrix:
n<-matrix(cbind(c(0,1,1),c(1,0,0),c(0,1,0)),3,3)
colnames(n)<-cbi
Kenn Konstabel writes:
> Another way (not elegant but better and shorter than the eval-parse
> way) is to use get. ?get
This one is handy for interactive use, thanks for the hint.
Kind Regards,
Michael Bach
__
R-help@r-project.org mailing list
https:
"Nick Sabbe" writes:
> ObjectsOfInterest<- list(one_df, two_df, three_df)
> for(namedf in ObjectsOfInterest){...}
I see. This is also more readable and traceable for others.
> or probably even better
> sapply(ObjectsOfInterest, function(namedf){...})
I like this one for its functional style.
On 04/29/2011 04:09 AM, Bogaso Christofer wrote:
Hi all, please consider this plot:
xx<- seq(4, 0.01, by = -0.04)
yy<- rnorm(xx)
plot(xx, yy, type="l")
Here you see my original 'xx' was in decreasing order, however R puts it in
the increasing order. I understand that in any plot x and y a
Thanks for the note: Indeed, the function is the hist() function not Hist()
with capital letter.
I use the standard R hist()-function with the lower case only. Nevertheless,
the ylim does not work as supposed to.
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On Fri, Apr 29, 2011 at 03:35:41AM -0700, hck wrote:
> Problem: hist()-function, scale = “percent”
[...]
> ="Hist(na.exclude("&AA3&"), breaks=50, col=""seashell3"",
> scale=""percent"",xlim=c(-1, 1), xlab=""Bewertungsfehler"",
> ylab=""Haeufigkeit (in %)"", main=""KBV"", border=""white"")"
Before
Dear R-lings,
I did not know which list to post to, because it is a studentship so not
really a job, so it did not fit the r-sig-jobs list and it is about
devloping an extension package interfaced with R I hope I did not upset
anyone. If so apologies.
The Centre For Complex systems A
I would like to generate some noisy time series. I know that it is possible to
"classify" noise by looking at the exponent (beta) of the relationship between
the spectrum of the time series and the frequencies (i.e. spectrum ~ frequency
^ beta ).
Is there a way to generate White (beta=0), Pink
Hi Michael.
This is a classic :-)
ObjectsOfInterest<- list(one_df, two_df, three_df)
for(namedf in ObjectsOfInterest){...}
or probably even better
sapply(ObjectsOfInterest, function(namedf){...})
hth.
Nick Sabbe
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link: http://biomath.ugent.be
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