Hi there,
I'm not sure whether this is the right mailing list to put this question... if
not, please tell me where I can find help.
I have a spare dual-core machine which run Windows XP Professional and is
nearly unused (nearly! Unfortunately, I cannot change OS).
And I'm investigating Knime
I know that Ryacas is promoted here whenever requests about symbolic algebra
or calculus appear on the R-help list. But to say the truth, Yacas itself is
a very very limited Computer Algebra System and looking onto its home page
it appears the development will stop or has stopped anyway.
It would
I am getting the following error on iMac running Mac OS 10.5.6; this occurs
whenever I try to import a text file into R Commander:
Error in splitCmd(command) : unbalanced quotes
Thus, I am unable to import data into R Commander. Any suggestions as to how I
might fix this error? I have gotten i
I am getting the following error on iMac running Mac OS 10.5.6; this occurs
whenever I try to import a text file into R Commander:
Error in splitCmd(command) : unbalanced quotes
Thus, I am unable to import data into R Commander. Any suggestions as to how I
might fix this error. I have gotten i
A recent thread on summary statistics, got me thinking. (Note this may not
happen often.) A function that would do summaries as describe below (similar
to SAS PROC UNIVARIATE) might be a nice addition to the main R system. Is
there a process by which functions, from packages can eventually be
inco
See ?"[" and its examples
Also, section 2.7 of An Introduction to R is a good place to start:
http://cran.r-project.org/doc/manuals/R-intro.html#Index-vectors
hth,
Kingsford Jones
On Mon, Feb 9, 2009 at 5:27 PM, jjh21 wrote:
>
> Hello,
>
> I am trying to do some data cleaning in R. I need to
It depends on how the data is set up (I am not an expert), but I have
had good results with the subset function. subset(x, var!=3 & var!=4)
this will take the subset of the dataframe x where var is not equal
to 3 or 4.
a <- rnorm(25)
var <- rep(c(1:5), 5)
x <- data.frame(a, var)
subset(x, var!=3
Hello,
I am trying to do some data cleaning in R. I need to drop observations that
take on certain values of a variable. In STATA I might type something like:
drop if == 3
drop if == 4
Is there an R equivalent of this? I have tried playing around with the
subset command, but it seems a bit cl
Dmitriy Verkhoturov writes:
> Hello listmemebers,
>
> I have data from two-color microarray expression profiling experiments
> where 3 whole brain (WB) samples were compared to 3 Mauthner Cells
> (MC) in a loop design (-> MC #1 -> WB #1 -> MC #2 -> WB #2 -> MC #3 ->
> WB #3 -> MC #1 ->). In addit
Dear fellows: This is the problem: I have 5 variables A, B, C, D and E
with a range from 1 to 100 with 0.1 steps. Depending on the different
values these have, the results of the formula change:
alitemp <- ((Abase/llmcc$Clase)*PClase)+(((1/llmcc
$Categoria)*Abase)*PCategoria)+((Abase*llmcc$P
The forms of equations are limited but its not limited to just one:
> library(Ryacas)
Loading required package: XML
> x <- Sym("x")
> y <- Sym("y")
> Solve(List(x+y == 2, x-y == 0), List(x, y))
[1] "Starting Yacas!"
expression(list(list(x == 2 - y, y == 1)))
On Mon, Feb 9, 2009 at 7:45 PM, Carl
rkevinbur...@charter.net wrote:
I am new to 'R' and also new to the concept of a 'Hessian' with non-linear
optimization. I would like to avoid going through all of the reference articles
given with ?optim as access to a library is not handy. Would someone be able to
elighten me on what is in t
At 6:41 PM -0500 2/9/09, David Winsemius wrote:
describe() in Hmisc provides much of the rest of what you asked for:
describe(pref900$TCHDL)
pref900$TCHDL
n missing uniqueMean .05 .10 .25 .50
.75 .90 .95
9061904469 16051 4.123 2.320 2.557 3.0
Hi,
I'm trying to create a connectivity diagram using
RgraphViz library. I want to increase the edge's label fontsize, but
the size did not change.
Did I do something wrong ?
M <- matrix(nrow=5,ncol=5,byrow=TRUE,data=mytable)
colnames(M) <- levels(pf$agent)
A <- new("graphAM", M, "directed", va
Try the functions Anova() or linear.hypthesis() in the package "car". If
using Anova(), you probably want the type II table.
Ian Fiske
dl7631 wrote:
>
> Hello!
>
> I've run a simple linear model: result<-lm(DV~A+B+C,data=Data)
>
> My Data$A,Data$B, and Data$C are factors. So, lm automatical
> Thank you very much Jorge, Phil and David: I was finally able to
> perform the operations I needed. I changed the function in order to
> adapt it to the simplest form like the following:
> ali <- function(Abase) {
> alitemp <- ((Abase/llmcc$Clase)*PClase)+(((1/llmcc
> $Categoria)*Ab
To bring closure to this thread, we found that the following simple patch to
Rmpi/src/Rmpi.c fixes the problem:
--- rmpi-0.5-6.orig/src/Rmpi.c
+++ rmpi-0.5-6/src/Rmpi.c
@@ -63,7 +63,7 @@
else {
#ifdef OPENMPI
- dlopen("libmpi.so.0", RTLD_GLOBAL);
+ dlopen("libmpi.so.0", RT
Gabor G a ecrit:
Check out the Ryacas package. There is a vignette with some
examples.
Which led me to the manuals for yacas itself. I'm guessing there may be
a way to use yacas' "AND" construct to combine a few equations and then
hope the Newton Solver can work with that, but it's not
The binary in R.app starts the GUI. I suppose I would have to install
the command-line version separately.
On Feb 9, 2009, at 7:12 PM, Rolf Turner wrote:
On 10/02/2009, at 12:55 PM, Duncan Murdoch wrote:
On 09/02/2009 6:27 PM, Fuchs Ira wrote:
a=function() {
print("start")
print(Sys.time(
Just jumping to conclusions since the site is down as I type (7PM EST).
What's the news from that part of Australia -- thus showing my complete
ignorance of DownUnder geography.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listin
Just to play a little:
mystats<- function(x)
{
xvalid<-x[!is.na(x)]
XNtotal<-length(x)
XNnull<-length(xvalid)
XNnotNull<-XNtotal-XNnull
Xmean<-mean(xvalid)
Xmedian<-median(xvalid)
Xsd<-sd(xvalid)
Xrange<-range(xvalid)
my.results<-cbind(XNtotal, XNnull, XNnotNull, Xmean, Xmedian, Xsd,
Xmin=Xrange[1
Hi all,
I have modeled my observations/data,say, D=y(i) (i=1,2,...,2), as a
univariate non-gausian distribution, gamma distribution y(i)~gamma(a,b) for
example, and I have also obtained N samples of a, b via WinBUGS(Gibbs
sampling), my question is: how to calculate the marginal density of D,
m
On 10/02/2009, at 12:55 PM, Duncan Murdoch wrote:
On 09/02/2009 6:27 PM, Fuchs Ira wrote:
a=function() {
print("start")
print(Sys.time())
Sys.sleep(5)
print(Sys.time())
print("end") }
a()
results in:
[1] "start"
[1] "2009-02-09 18:25:41 EST"
[1] "2009-02-09 18:25:46 EST"
[1] "end"
all prin
One way. there may be better. The apply function will work with just
one row (or one column) at a time.
> DF
Month Week Estpassage MedFL
1 July 2766534
2 July 28 223235
3 July 29 924135
4 July 30 2846435
5Aug 31 4104935
You can use 'ifelse':
> x <- data.frame(id=sample(1:4,20,TRUE))
> # use ifelse to do the calculations
> x$cal <- ifelse(x$id == 1, 21,
+ifelse(x$id == 2, 221,
+ ifelse(x$id == 3, 2221, 1)))
> x
id cal
1 4 1
2 121
3 3 2221
4 121
5 2 221
6
On 09/02/2009 6:27 PM, Fuchs Ira wrote:
a=function() {
print("start")
print(Sys.time())
Sys.sleep(5)
print(Sys.time())
print("end") }
a()
results in:
[1] "start"
[1] "2009-02-09 18:25:41 EST"
[1] "2009-02-09 18:25:46 EST"
[1] "end"
all printed together after 5 seconds.
The reason for this m
Dear SY,
Also take a look at basicStats in the fBasics package.
# fBasics package
install.packages('fBasics')
require(fBasics)
# For reproducibility
set.seed(123)
# Some data
x<-c(NA,rnorm(10),NA)
basicStats(x)
HTH,
Jorge
On Mon, Feb 9, 2009 at 6:04 PM, phoebe kong wrote:
> Hi all,
>
> I
describe() in Hmisc provides much of the rest of what you asked for:
> describe(pref900$TCHDL)
pref900$TCHDL
n missing uniqueMean .05 .10 .25 .50 .
75 .90 .95
9061904469 16051 4.123 2.320 2.557 3.061 3.841
4.886 6.054 6.867
lowest
On 10/02/2009, at 12:09 PM, Fuchs Ira wrote:
Yes, Windows has that in the Misc menu, but I don't see a way to do
this on the Mac.
In my experience output is not buffered on a Mac.
Are you sure you're *printing* the output you want to see?
Can you provide, as the postin
a=function() {
print("start")
print(Sys.time())
Sys.sleep(5)
print(Sys.time())
print("end") }
a()
results in:
[1] "start"
[1] "2009-02-09 18:25:41 EST"
[1] "2009-02-09 18:25:46 EST"
[1] "end"
all printed together after 5 seconds.
On Feb 9, 2009, at 6:21 PM, Rolf Turner wrote:
>
> On 10/02/2
Have you tried flush.console() ?
r
-
Remko Duursma
Post-Doctoral Fellow
Centre for Plant and Food Science
University of Western Sydney
Hawkesbury Campus
Richmond NSW 2753
Dept of Biological Science
Macquarie University
North Ryde NSW 2109
Austra
On 10/02/2009, at 12:09 PM, Fuchs Ira wrote:
Yes, Windows has that in the Misc menu, but I don't see a way to do
this on the Mac.
In my experience output is not buffered on a Mac.
Are you actually *printing* out the results you want to see?
Could you, as the posting g
Simon,
thank you very much for your very instructive response!
Your explanation has not only justified my intuitive solution to just
delete the corresponding row & column of the penalty matrix, when
choosing a reference category for dummy coding, but leads to a much
better understanding of this i
For a simulation study, I need to fit a GEE with a IG distribution and using a
log link function. As far as I know, there are two GEE packages available
('gee' and 'geepack') but none of them supports IG. I've also tried using
family=quasi("log","mu^3") (without luck!).'
Any guidance is highly a
I wonder if there are any useRs sharing day-to-day realizations/tricks
on twitter...
Seems like a good place for those things that are good findings, but
one is too lazy to blog about them...
--
Jose Quesada, PhD.
Max Planck Institute, Human Development, Berlin
http://www.josequesada.name/
_
Dear list,
I will be working with RDF in the future, and was wondering if there's
anyone else in this list doing this.
Any libraries/resources to check?
Thanks,
-Jose
--
Jose Quesada, PhD.
Max Planck Institute, Human Development, Berlin
http://www.josequesada.name/
Yes, Windows has that in the Misc menu, but I don't see a way to do
this on the Mac.
On Feb 9, 2009, at 5:51 PM, Duncan Murdoch wrote:
On 09/02/2009 5:07 PM, Fuchs Ira wrote:
I would like to have a function which gets data, does a
calculation and prints a result and then waits some number
Hi all,
I'm wondering if there is a function that can return summary statistics:
N=total number of observation, # missing, mean, median, range, standard
deviation.
As I know, summary() returns some of info I've mentioned above.
Thanks,
SY
[[alternative HTML version deleted]]
__
Dear Christian,
Every single time check update package, “cwhmisc” always requests updating.
I’m aware that the package was latest updated in CRAN on 20Nov2008.
Is there anything wrong with my R library or somethingelse?
I use R 2.8.1 on Window XP service pack 2
Regards
Nguyen
Garvan Institute of Me
On 09/02/2009 5:07 PM, Fuchs Ira wrote:
I would like to have a function which gets data, does a calculation
and prints a result and then waits some number of seconds and
repeats. If I use Sys.sleep, the execution is pausing but the
function output is buffered so that it all comes out when t
I am not too sure what your question is, but try
?debug
or take a look at http://www.ats.ucla.edu/stat/r/library/R-debug-tools.pdf
On Feb 10, 9:07 am, Fuchs Ira wrote:
> I would like to have a function which gets data, does a calculation
> and prints a result and then waits some number of sec
First of all, this is not a list, this is a vector. Second, see ?cumsum.
Best,
Gabor
On Mon, Feb 9, 2009 at 11:13 PM, glenn wrote:
>very simple question I am sure sorry:
>
>
>
> for a list;
>
>
>
> test <-c(1,2,3)
>
>
>
> how do I total them up please to return the result
>
>
>
>
>
>
very simple question I am sure sorry:
for a list;
test <-c(1,2,3)
how do I total them up please to return the result
1,3,6
Regards
Glenn
[[alternative HTML version deleted]]
__
R-help@r-project.or
Dear Sirs: I've been working with several variables in a dataframe
that serve as part of a calculation that I need to perform in a
different way depending on its value. Let me explain:
The main dataframe is called llmcc
llmcc : 'data.frame': 283 obs. of 11 variables:
$ Area : num 30
I would like to have a function which gets data, does a calculation
and prints a result and then waits some number of seconds and
repeats. If I use Sys.sleep, the execution is pausing but the
function output is buffered so that it all comes out when the function
terminates. How can I get
Kia ora Gina
If I understand you correctly, you need to specify mfrow argument as c(1,2) and
then make your two calls to plot(). HTH ...
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of gina patel
> Sent: Tues
Try something like this where we use the built in data frame anscombe:
opar <- par(mfrow = c(1, 2), no.readonly = TRUE)
matplot(1:11, anscombe[1:4], pch = 1:4, col = 1:4, log = "x", ylab = "Y")
matplot(1:11, anscombe[5:8], pch = 1:4, col = 1:4, log = "x", ylab = "Y")
par(opar)
On Mon, Feb 9, 200
on 02/09/2009 03:21 PM Steve Sidney wrote:
> Dear all
>
> As a new user of R, can someone please help me with the following
>
> I have created a programme to analyse laboratory data and one of the
> graphs is a bar plot of 'Z' scores.
>
> On the bar plot I am using the following line to plot som
Hello!
I've run a simple linear model: result<-lm(DV~A+B+C,data=Data)
My Data$A,Data$B, and Data$C are factors. So, lm automatically recoded
them into dummy variables. I have all the results I need but one.
Question: Where could I see the variance explained by all A dummy
variables together, the
I am new to R and have a problem that I haven't been able to find the answer to
in the guides or online.
I have multiple datasets, D1, D2, D3, D4, D5, D6, D7 and D8, and I would like
to produce two plots side by side using mfrow. The first plot should contain
data from D1-D4, the second should
Dear all
As a new user of R, can someone please help me with the following
I have created a programme to analyse laboratory data and one of the graphs is
a bar plot of 'Z' scores.
On the bar plot I am using the following line to plot some results
barplot (zb[,c("ZBW")], ylim = c(-6,6), names.
On Mon, Feb 9, 2009 at 4:46 PM, Giovanni Petris wrote:
>
> As suggested by somebody else, uniroot() can be used to solve
> equations in one unknown variable. However, you will never get the
> "exact x" in R, only an approximate root...
>
> Best,
> Giovanni
>
>> Date: Sat, 07 Feb 2009 02:20:10 -080
Hi Frank,
Hello, everyone!
I have a set of proteomic data .And I do a solexa sequencing in the
corresponding sample. So I get much mass sequencing data. How can I using R to
integrate those two set data. I wonder if some tool or R package would help me?
You are more likely to receive a reply
Hi all!!
I would like to know if anyone has experienced this behaviour with
ROracle package. I'm attaching information to reproduce the issue.
Bug maybe?
Thank you very much for your attention.
Ramon.
##
##
on 02/09/2009 01:30 PM Josip Dasovic wrote:
> Dear R Help-Listers:
>
> I have a problem that seems like it should have a simple solution, but I've
> spent hours on it (and searching the r-help archives) to no avail. What I'd
> like to do is to generate a new variable within a data frame, the val
One way to do this is through transform, assuming that there is one-to-one
correspondence between regions and elements:
mydf <- data.frame(region=c(rep("North", 5), rep("East", 5), rep("South",
5), rep("West", 5)))
elements <- c("earth", "water", "air", "fire")
transform(mydf, element = factor(reg
Hi Vie,
Something like the following should be fine:
## R Start...
> n<-1
> my.min <- 0
> my.max <- 0.7
> runif(n, my.min, my.max)
[1] 0.01145260
## R end.
see ?runif for details. Hope that helps a little,
Tony Breyal
On 9 Feb, 14:40, Vie wrote:
> Hi,
>
> Ive been trying to find a function tha
Ram Pandit gmail.com> writes:
> I am working in a country level data. After running the regression, I would
> like to plot the residuals of each observation based on the group created
> for a particular variable.
>
> For example, one of my independent variable is "Income", I would like to
> plot
XPath is your friend here.
getNodeSet(mf,
'//characterist...@type="File" and @eName="FileTypeId"
and @eValue="10"]/parent::File
/characterist...@type="Patient"
and @eName="PatientReference"]/@eValue')
I have broken the XPath expressi
If you want to chose numbers from your range with uniform probability
runif(1, 0, 0.5)
See ?runif
-Christos
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Vie
> Sent: Monday, February 09, 2009 9:41 AM
> To: r-help@r-proje
Dear R Help-Listers:
I have a problem that seems like it should have a simple solution, but I've
spent hours on it (and searching the r-help archives) to no avail. What I'd
like to do is to generate a new variable within a data frame, the values of
which are dependent upon the values of an exis
Hi Vie,
Take a look at ?runif.
set.seed(123)
> runif(1,0,0.5)
[1] 0.1437888
> runif(1,0,0.7)
[1] 0.5518136
HTH,
Jorge
On Mon, Feb 9, 2009 at 9:40 AM, Vie wrote:
>
> Hi,
>
> Ive been trying to find a function that will allow me to pull out a number
> between a minimum and maximum threshold.
>
on 02/09/2009 08:40 AM Vie wrote:
> Hi,
>
> Ive been trying to find a function that will allow me to pull out a number
> between a minimum and maximum threshold.
>
> I want a random decimal number between, for example, 0 and 0.5 or 0 and 0.7.
> I've been searching everywhere for a function that w
Vie bham.ac.uk> writes:
>
> Ive been trying to find a function that will allow me to pull out a number
> between a minimum and maximum threshold.
>
> I want a random decimal number between, for example, 0 and 0.5 or 0 and 0.7.
> I've been searching everywhere for a function that will allow me t
How about this:
> runif(1,0,0.5)
[1] 0.4806179
> runif(1,0,0.7)
[1] 0.1789742
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Vie
Sent: Monday, February 09, 2009 9:41 AM
To: r-help@r-project.org
Subject: [R] Choosing a random numb
See ?runif
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Vie
> Sent: Monday, February 09, 2009 9:41 AM
> To: r-help@r-project.org
> Subject: [R] Choosing a random number between x and y
>
>
> Hi,
>
> Ive been trying to f
The runif function meets the stated criteria, if that is not good enough, then
give us more detail.
--
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 [mailto:r-help
Vie wrote:
>
> Hi,
>
> Ive been trying to find a function that will allow me to pull out a number
> between a minimum and maximum threshold.
>
> I want a random decimal number between, for example, 0 and 0.5 or 0 and
> 0.7.
>
I'm no R expert, but this should give you n uniformly distributed
Hello, everyone!
I have a set of proteomic data .And I do a solexa sequencing in the
corresponding sample. So I get much mass sequencing data. How can I using R to
integrate those two set data. I wonder if some tool or R package would help me?
Thank you !
[[alternative HTML version dele
Dear R users,
I want to run nested fixed-factor Anova in R on different experiments.
In this toy example I have 3 levels of the main factor x1 and 7 levels
of the nested factor z1
x1 and continuous response variable y1.
x1
[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
Hi,
Ive been trying to find a function that will allow me to pull out a number
between a minimum and maximum threshold.
I want a random decimal number between, for example, 0 and 0.5 or 0 and 0.7.
I've been searching everywhere for a function that will allow me to do this
in R, but I have yet to
"Without a full example (see the footer of this message) we cannot tell
what you (or a package you are using) did wrong. "
That's the main point. But one possibility might be that if they're using
TINN-R on Windows, this warning occurs all the time as Tinn-R opens
connections for sourcing data in
On Mon, Feb 9, 2009 at 12:36 PM, Neotropical bat risk assessments <
neotropical.b...@gmail.com> wrote:
> Read a string of data and had this message during a plot run.
>
> Warning message:
> closing unused connection 3 (Lines)
>
> Not sure what this means or if it should be of concern.
>
This simp
I am using AOV and I am not sure if my model structure is correct for the
repeated measures.
I want to verify if there is a interaction of a random factor(block) with
fixed factors(veget,fruit, time,)
> model<-aov(trackarcsin~veget*fruit*time*block
> +Error(block/(veget*fruit)),data=mice)
Becau
Hello listmemebers,
I have data from two-color microarray expression profiling experiments where 3
whole brain (WB) samples were compared to 3 Mauthner Cells (MC) in a loop
design (-> MC #1 -> WB #1 -> MC #2 -> WB #2 -> MC #3 -> WB #3 -> MC #1 ->). In
addition to phenotype analysis I would also
Hi,
I have a rather complex xml document that I am attempting to parse based on
attributes:
http://www.w3.org/2001/XMLSchema-instance";>
D:\CN_data\Agilent\Results\
Dear all,
I am working in a country level data. After running the regression, I would
like to plot the residuals of each observation based on the group created
for a particular variable.
For example, one of my independent variable is "Income", I would like to
plot the residual based on income cat
How about 'cmh_test' in the coin package?
>From the PDF: The null hypothesis of the independence of y and x is tested,
block defines an optional factor for stratification. chisq_test implements
Pearson’s chi-squared test, cmh_test the Cochran-Mantel-Haenzsel test and
lbl_test the linear-by-linear
On Mon, 9 Feb 2009, Neotropical bat risk assessments wrote:
Hi all,
Read a string of data and had this message during a plot run.
Warning message:
closing unused connection 3 (Lines)
Not sure what this means or if it should be of concern.
It means R tidied up after you. But it may have tid
Thanks Jim. That was the first thing I tried. I should have mentioned
that I was also trying to write the three columns of the legend using
three different colors. So I want the "blah" column to be black, second
column to be red and the third column to be blue.
I would appreciate if you can share
Hi all,
Read a string of data and had this message during a plot run.
Warning message:
closing unused connection 3 (Lines)
Not sure what this means or if it should be of concern.
Tnx.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman
Dear Patricia,
Take a look at this post:
http://www.nabble.com/Re:-applying-cor.test-to-a-(m,-n)-matrix---SUMMARY-to17150239.html#a17150239
The easiest way is using cor. See ?cor for details. Here is an example:
set.seed(12)
A<-matrix(rnorm(30),ncol=3)
cor(A)
# [,1] [,2] [,3]
# [1,
'combn' will give you the combinations that you can then use as
parameters in the function:
> combn(c('a', 'b', 'c'), 2)
[,1] [,2] [,3]
[1,] "a" "a" "b"
[2,] "b" "c" "c"
>
On Mon, Feb 9, 2009 at 11:44 AM, patricia garcía gonzález
wrote:
>
> Hi all,
>
> Having a matrix A formed by n ve
As suggested by somebody else, uniroot() can be used to solve
equations in one unknown variable. However, you will never get the
"exact x" in R, only an approximate root...
Best,
Giovanni
> Date: Sat, 07 Feb 2009 02:20:10 -0800 (PST)
> From: oryie <43248...@qq.com>
> Sender: r-help-boun...@r-pr
Hi all,
Having a matrix A formed by n vectors as columns. Is there anything to
calculate a determined function to all combination of vectors?
For example imagine A matrix is compose by vectors a, b and c. And the function
to perform is correlation, so I would like to obtain cor(a, b), cor(a,
Dear Vito,
Yes, these tests are *similar* in spirit to loglinear models using
either row/col/both scores for the association.
But I'm still looking for something equivalent to give the same results
as SAS with CMH for
non-parametric tests. One advantage of the CMH tests is that for
stratified
On Mon, Feb 9, 2009 at 4:51 AM, Neil Shephard wrote:
>
> The preceived "difficulty" of installing R under whatever flavour of
> GNU/Linux in this thread stems from being unfamiliar with the process of the
> package management of the flavour of GNU/Linux you use (and in part by the
> various distro
Dear all,
I need to fit a gee model with an auto-regressive correlation structure and I
faced some problems.
I attach a simple example:
###
library(gee)
library(geepack)
# I SIMULATE DATA FROM POISSON DISTRIBUTION, 10 OBS FOR EACH OF 50 GROU
On Mon, Feb 9, 2009 at 4:57 PM, wrote:
>
> hi,
>
> I have some session data in a dataframe, where each session is recorded with
> a start and a stop date. Like this:
>
> session_start session_stop
> ===
> 2009-01-03 2009-01-04
> 2009-01-01 2009-01-05
> 2009-01-02
Try this:
> dateseq <- function(i) seq(DF[i, 1], DF[i, 2], 1)
> table(as.Date(unlist(lapply(1:nrow(DF), dateseq)), origin = "1970-01-01"))
2009-01-01 2009-01-02 2009-01-03 2009-01-04 2009-01-05 2009-01-06 2009-01-07
1 2 3 3 2 1 1
2009
hi,
I have some session data in a dataframe, where each session is recorded with a
start and a stop date. Like this:
session_start session_stop
===
2009-01-03 2009-01-04
2009-01-01 2009-01-05
2009-01-02 2009-01-09
A session is at least one day long. Now I want
Dear Michael,
It sounds as a linear-by-linear loglinear model (and its variants) which
uses scores for one or more variables in the table.. (see Agresti, 1990,
Categorical Data Analysis. I do remember the pages and I have not the
book here..)
If this is the case, you can use standard call to
Try this:
a <- structure(1:3, x = 3)
b <- "attributes<-"(11:15, attributes(a))
dput(b)
On Mon, Feb 9, 2009 at 7:09 AM, Alon Wasserman wrote:
> Hi,
> I would like to know how to assign values to a whole vector while keeping
> its attributes. For example, say I have
> a <- structure(1:3,x=3)
> and
I still remember my public spanking from Ben Bolker on the unnecessary
use of "which" in this instance.
> MM <- matrix(c(1:10,sample(-10:10,10)),nrow=10)
> MM
[,1] [,2]
[1,]1 -1
[2,]25
[3,]3 -2
[4,]4 -3
[5,]50
[6,]67
[7,]7 -9
[8,]
At 07:58 09/02/2009, Thomas Lumley wrote:
On Sun, 8 Feb 2009, Tom Backer Johnsen wrote:
Dirk Eddelbuettel wrote:
On 8 February 2009 at 20:36, Tom Backer Johnsen wrote:
| Dear me. Is the installation of R under Ubuntu really that
complex? I | have a dual boot machine (Linux / Windows, where
On Mon, Feb 9, 2009 at 8:03 AM, clion wrote:
>
> this is good, but it doesn't solve my main problem (which I unfortunately
> din't post - very sorry )
> I would like to filter may data , for example by:
>
> dat.sub<-dat[dat$age>10 & dat$NoCaps>2,]
>
> So I need a column where the number of Captur
Not sure what "a" really is. It's not a vector or a list according to
the R interpreter.
> a <- structure(1:3,x=3)
> mode(a)
[1] "numeric"
> is.vector(a)
[1] FALSE
> is.list(a)
[1] FALSE
Experimentation shows that simple indexing provides the functionality
you request while append does not.
If your matrix is called mat, how about
mat[which(mat[,2] > 0), ]
mat[which(mat[,2] < 0), ]
-Ian
mentor_ wrote:
>
> Hi,
>
> I have a matrix with negative and positiv values.
> How can I get either the negative or positive values from the matrix?
>
> Matrix:
> [,1] [,2]
> [1,]1
In SAS, for a two-way (or 3-way, stratified) table, the CMH option in
SAS PROC FREQ gives
3 tests that take ordinality of the factors into account, for both
variables, just the column variable
or neither. Is there an equivalent in R?
The mantelhaen.test in stats gives something quite different
Megh,
The problem is due to jump discontinuity in your function at x=0. It is
always good practice to plot the function over the range of interest.
x <- seq(-20, 20, by=0.01)
plot(x, th.price(x), type="l")
This will reveal the problem. The function value jumps from -384.4 to 36.29
at x=0.
I
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