Hi R users,
I did do the research and work on for hours, but I still don't know how to
solve my silly problem. I try to creat a new variable in my dataset.
such as if diet==C vesl==P then trt=CP; if diet==C vesl==A
then trt=CA;. The following is my code (It does not work correctly).
On Wed, 18 Feb 2009, jjh21 wrote:
Hello,
I know that two possible approaches to dealing with clustered data would be
GEE or a robust cluster covariance matrix from a standard regression. What
are the differences between these two methods, or are they doing the same
thing? Thanks.
There are
Dear R user,
I am looking for SVM regression in R. It willl be
helpful for me if some one send me SVM regression code.
Thanks
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hi R users,
I want to build R-2.8.1 on AIX5.3, but I got following error message:
Error in dyn.load(file, DLLpath = DLLpath, ...) :
unable to load shared library
'/rnd/homes/jixu/tmp/R-2.8.1/library/stats/libs/stats.so':
rtld: 0712-001 Symbol d1mach was referenced
from module
Hi Chun,
I did do the research and work on for hours ... I try to creat a new
variable in my dataset.
Yes, looks like you did. Look at ?interaction, which gives you more
flexibility than ?:.
## Example
diet-sort(rep(x=c(C,T),4))
vesl-rep(x=c(A,P),4)
mydata-data.frame(diet,vesl)
mydata$trt
You seem not to have put the R bin directory in your path. That is
where Rterm, Rscript ... are installed (and the installer does not
change the PATH for you).
On Thu, 19 Feb 2009, mau...@alice.it wrote:
Sorry. This is still unclear to me.
I generated a file called Test.R that contains the
Hi Jin.
I found there is someone meet same problem when he build R-2.7.0 by searching
r-help, and he used a patch to resolve this issue. The patch¡¯s location is
http://prs.ism.ac.jp/~nakama/AIX/changefiles. But I can not find a patch
R-2.8.1 in this path.
Gnu patch is necessary.(AIX
Hello,
I am reading in a file called fit2.txt (Limma). fit2.txt has 38 columns but
when I dim(fit2) I only get 6 columns. The first column that it does not read
in is df.residual.
fit2-read.table(fit2,
file=fit2.txt,sep=\t,quote=,comment.char=,as.is=TRUE)
The first few lines of fit2.txt
I am writing a version of the subset function for
a new class. I don't understand the behavior of match.call
in this particular case, and I didn't seem to be able to
find much help in the language definition or the email archive.
Here follows a minimal example:
setClass(myClass,
there is svmpath package by Trevor Hastie
 Justin BEM
BP 1917 Yaoundé
Tél (237) 99597295
(237) 22040246
De : Alex Roy alexroy2...@gmail.com
à : r-help@r-project.org
Envoyé le : Jeudi, 19 Février 2009, 9h19mn 18s
Objet : [R] SVM regression code
Dear R
I am looking for a person who is able to translate a short R code to a C
code. I am not sure where to start searching.
It is only a short R code (the wrm.smooth code from the robfilter package).
Perhaps someone can give me an hint where to start.
Thanks in advance.
Sincerely,
Lars
--
View
Hello,
My problem is that I would like to merge multiple files with a common
column but merge accepts only two
data.frames to merge. In the real situation, I have 26 different
data.frames with a common column. I can of course use merge many times
(see below) but what would be more sophisticated
I want to construct a symmetric band matrix in the Matrix package from a matrix
where the first column contains data for the main diagonal, the second column
has data for the first subdiagonal/superdiagonal and so on.
Since the Matrix will be 10^5 x 10^5 or so, with perhaps 10-20 non-zero
Hi: Below is a TOTAL HACK and I don't recommend it but it does seem to
do what you want. I think that I remember Gabor saying that you can
merge multiple data frames using zoo but I don't know the specifics. I'm
sure he'll respond with the correct way. Below uses a global variable
to access
Hi all,
This could be naivety/stupidity on my part rather than a problem with model
output, but here goes
I have fitted a fairly simple model
m1-glm(count~siteall+yrs+yrs:district,family=quasipoisson,weights=weight,data=m[x[[i]],])
I want to know if yrs (a continuous variable) has a
Hi,
I think Reduce could help you.
DF1 - data.frame(var1 = letters[1:5], a = rnorm(5))
DF2 - data.frame(var1 = letters[3:7], b = rnorm(5))
DF3 - data.frame(var1 = letters[6:10], c = rnorm(5))
DF4 - data.frame(var1 = letters[8:12], d = rnorm(5))
g - merge(DF1, DF2, by.x=var1, by.y=var1, all=T)
Sorry, that was a typo in the email, not the model. So I still have the
problem.
Cheers, Simon.
- Original Message -
From: Ted Harding ted.hard...@manchester.ac.uk
To: Simon Pickett simon.pick...@bto.org; r-help@r-project.org
Sent: Thursday, February 19, 2009 10:56 AM
Subject:
Thanks, both solutions work fine. I tried these solutions to my real
data, and I got an error
Error in match.names(clabs, names(xi)) :
names do not match previous names
I refined this example data to look more like my real data, this also
produces the same error. Any ideas how to prevent this
I'd suggest to make the data available on the web. Then we can take a
closer look. You or some mail tool in between removed the tabs from the
message, hence we cannot reproduce in any way.
Best,
Uwe Ligges
Sally wrote:
Hello,
I am reading in a file called fit2.txt (Limma). fit2.txt has 38
On 19-Feb-09 10:38:50, Simon Pickett wrote:
Hi all,
This could be naivety/stupidity on my part rather than a problem
with model output, but here goes
I have fitted a fairly simple model
m1-glm(count~siteall+yrs+yrs:district,family=quasipoisson,
Cheers Mark,
I did originally think too, i.e. that not including the main effect was the
problem. However, the same thing happens when I include main effects
test1-glm(count~siteall+yrs*district,family=quasipoisson,weights=weight,data=m[x[[i]],])
I have the initial matrice:
*data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste(Q,1:4,
sep=),2), Boolean = rep(c(Y,N),4))*
Subject Quarter Boolean
1100 Q1 Y
2100 Q2 N
3100 Q3 Y
4100 Q4 N
5101 Q1 Y
6101
Dear R users,
thanks to Samuel for making the package knnFinder available to the public. I was
wondering if there is an easy way to only build and store the kdd tree
in a first step and perform NN queries from then on ?
It seems that nn() does both simultaneously.
Thanks!
Markus
I have the initial matrice:
*data.frame(Subject=rep(100:101, each=4), Quarter=rep(paste(Q,1:4,
sep=),2), Boolean = rep(c(Y,N),4))*
Subject Quarter Boolean
1100 Q1 Y
2100 Q2 N
3100 Q3 Y
4100 Q4 N
5101 Q1 Y
6101
Hello
I have a dataset named b2 with 1521 rows, in that dataset i have 64 rows
containing specific information.
the rownumbers with specific info are:
+ i
[1] 22 53 104 127 151 196 235 238 249 250 263 335 344 353
362 370 389 422 458 459 473 492 502 530 561 624 647
Hi,
Look, is simple with reshape:
x - data frame(...)
reshape( x, idvar = Subject, direction = wide, timevar = Quarter)
Regards,
Patricia
Date: Thu, 19 Feb 2009 11:02:58 +0100
From: pcando...@gmail.com
To: r-help@r-project.org
Subject: [R] table with 3 varialbes
I have the initial
Wacek Kusnierczyk wrote:
to contribute my few cents, here's a simple benchmarking routine,
inspired by the perl module Benchmark. it allows one to benchmark an
arbitrary number of expressions with an arbitrary number of
replications, and provides a summary matrix with selected timings.
Another option using Recall,
merge.rec - function(.list, ...){
if(length(.list)==1) return(.list[[1]])
Recall(c(list(merge(.list[[1]], .list[[2]], ...)), .list[-(1:2)]), ...)
}
my.list - list(DF1, DF2, DF3, DF4)
test2 - merge.rec(my.list, by.x=var1, by.y=var1, all=T)
If you don't mind I've added this example to the R wiki,
http://wiki.r-project.org/rwiki/doku.php?id=tips:data-frames:merge
It would be very nice if a R guru could check that the information I
put is not complete fantasy. Feel free to remove as appropriate.
Best wishes,
baptiste
On 19
Hi Simon: In below , test1 spelled out is count ~ siteall + yrs +
district + yrs:district so this is fine.
but in test2 , you have years interacting with district but not the main
effect for years. this is against the rules of marginality so I still
think there's a problem. I would wait for
Hi,
I think you should take a look to ?reshape.
Regards
Patricia
Date: Thu, 19 Feb 2009 11:02:58 +0100
From: pcando...@gmail.com
To: r-help@r-project.org
Subject: [R] table with 3 varialbes
I have the initial matrice:
*data.frame(Subject=rep(100:101, each=4),
I think you are looking for something like:
ifelse(1:nrow(b2) %in% i, 1, 0)
Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)
joe1985 wrote:
Hello
I have a dataset named b2 with 1521 rows, in that
That's perfectly fine. I figured out how to to this with my second example
DF1 - data.frame(var1 = letters[1:5], a = rnorm(5), b = rnorm(5), c = rnorm(5))
DF2 - data.frame(var1 = letters[3:7], a = rnorm(5), b = rnorm(5), c = rnorm(5))
DF3 - data.frame(var1 = letters[6:10], a = rnorm(5), b =
good point! Provide your own set of x,y,z co-ords, mine are pretty big
but you can use any.
library(akima)
fr3d = data.frame(x,y,z)
xtrp - interp(fr3d$x,fr3d$y,fr3d$z,linear=FALSE,extrap=TRUE,duplicate=
strip)
op - par(ann=FALSE, mai=c(0,0,0,0))
filled.contour(xtrp$x, xtrp$y, xtrp$z, asp =
Yes, even better
DF1 - data.frame(var1 = letters[1:5], a = rnorm(5), b = rnorm(5), c = rnorm(5))
DF2 - data.frame(var1 = letters[3:7], a = rnorm(5), b = rnorm(5), c = rnorm(5))
DF3 - data.frame(var1 = letters[6:10], a = rnorm(5), b = rnorm(5), c
= rnorm(5))
DF4 - data.frame(var1 = letters[8:12],
Hi Simon,
I want to know if yrs (a continuous variable) has a significant unique
effect in the model,
so I fit a simplified model with the main effect ommitted...
[A different approach...] This is not really a sensible question until you
have established that there is no significant
Maybe reshape will help you, but I'm in doubt that your posted desired
result fits your given data - e.g shouldn't subject 101 Q3 give Y?
xx-data.frame(Subject=rep(100:101, each=4),
Quarter=rep(paste(Q,1:4,sep=),2), Boolean = rep(c(Y,N),4))
why use ifelse? Shouldn't
b2$totalvac-rep(0,1521)
b2$totalvac[i]-1
do the trick?
joe1985 schrieb:
Hello
I have a dataset named b2 with 1521 rows, in that dataset i have 64 rows
containing specific information.
the rownumbers with specific info are:
+ i
[1] 22 53 104 127 151 196
Dear list,
is it possible to change the background color of dotplot's points? I tried
in many ways but unsuccessfully
Thanks in advance
Gianandrea
require(lattice)
dotplot(variety ~ yield | site, data = barley, groups = year, pch=21)
dotplot(variety ~ yield | site, data = barley, groups =
Hi all,
I got one problem with comparing strings like if any string is like
*RIGHT, EPICARDIUM: FOCUS, GRAY-WHITE, SINGLE, APPROX 0.6 CM IN DIAMETER*.
and i have to compare *GRAY-WHITE*with the above string
or otherwise i have to compare*TUMOR BENIGN* this string
How about this:
dat-c(0.00377467,0.00377467,0.00377467,0.00380083,0.00380083,0.00380083,0.00380959,
+ 0.00380959,0.00380959,0.00380083,0.00380083,0.00380083)
dat[seq(1, by=3, to=length(dat))] - 0
dat
[1] 0. 0.00377467 0.00377467 0. 0.00380083 0.00380083
0. 0.00380959
'grep' will tell you if there is a match in the string;
x - c(*RIGHT, EPICARDIUM: FOCUS, GRAY-WHITE, SINGLE, APPROX 0.6 CM IN
DIAMETER*.,*MEDULLRY TUMOR BENIGN,TYP PHEOCHROMOCYTOMA*)
grep(GRAY-WHITE, x)
[1] 1
grep(TUMOR BENIGN, x)
[1] 2
On Thu, Feb 19, 2009 at 7:39 AM, venkata kirankumar
Hi
r-help-boun...@r-project.org napsal dne 19.02.2009 13:39:42:
Hi all,
I got one problem with comparing strings like if any string is like
*RIGHT, EPICARDIUM: FOCUS, GRAY-WHITE, SINGLE, APPROX 0.6 CM IN
DIAMETER*.
and i have to compare *GRAY-WHITE*with the above string
Hi Simon,
[On my response] ...not really a sensible question until...
On reading through this...what I mean is that yours seems not to be a
sensible approach, the question itself may be reasonable. What you want to
be doing is testing whether the interaction term (yrs:district) gets
dropped.
Perhaps you can try this,
d - c(0.00377467, 0.00377467, 0.00377467, 0.00380083,
0.00380083, 0.00380083,
0.00380959, 0.00380959, 0.00380959, 0.00380083, 0.00380083,
0.00380083)
c( t( cbind(matrix(d, ncol=3, byrow=T), 0)))
I don't know how to avoid the transpose operation that
Dear all,
I tried a simple naive Bayes classification on an artificial dataset, but I
have troubles getting the predict function to work with the type=class
specification. With type= raw, it works perfectly, but with type=class I
get following error :
Error in as.vector(x, mode) : invalid 'mode'
On Thu, Feb 19, 2009 at 1:19 AM, jdeisenberg catc...@catcode.com wrote:
Nicole Hackman wrote:
Hello, I have a very simple data set i imported from excel including 96
averages in a column along with 96 standard errors associated with those
averages (calculated in excel). I plotted the 95
to avoid the transposition you can use
rbind(matrix(d, nrow=3), 0)
baptiste auguie schrieb:
Perhaps you can try this,
d - c(0.00377467, 0.00377467, 0.00377467, 0.00380083,
0.00380083, 0.00380083,
0.00380959, 0.00380959, 0.00380959, 0.00380083, 0.00380083,
0.00380083)
c( t(
Doran, Harold wrote:
lm(y ~ x-1)
solve(crossprod(x), t(x))%*%y# probably this can be done more
efficiently
You could do
crossprod(x,y) instead of t(x))%*%y
that certainly looks more readable (and less error prone) to an R newbie
like myself :-)
2009/2/19 baptiste auguie ba...@exeter.ac.uk
Perhaps you can try this,
d - c(0.00377467, 0.00377467, 0.00377467, 0.00380083, 0.00380083,
0.00380083,
0.00380959, 0.00380959, 0.00380959, 0.00380083, 0.00380083,
0.00380083)
c( t( cbind(matrix(d, ncol=3, byrow=T), 0)))
I don't
Hi Kenn,
Thanks for the suggestions, I'll have to see if I can figure out how to
convert the relatively simple call to lm with an equation and the data file
to the functions you mention (or if that's even feasible).
Not an expert in statistics myself, I am mostly concentrating on the
Maja,
The need to interpret parameters in log-linear models (and therefore,
the need to understand how the model is parameterized) often vanishes
if you visualize the fitted model or the residuals in a mosaic display.
e.g., ucb1 asserts Admit is jointly independent of Gender and Dept ---
fits
And I would like to group the Subject by Quarter using as a result in the
table the value of the third variable (Boolean). The final result would
give:
Subjet Q1 Q2 Q3 Q4
1100 Y Y Y Y
2101 N N N N
Are you sure that this is the final result you want ?
You could use the reshape
Sorry, you don't need 'melt' :
cast(truc, Subject ~ Quarter)
2009/2/19 David Hajage dhajag...@gmail.com
And I would like to group the Subject by Quarter using as a result in the
table the value of the third variable (Boolean). The final result would
give:
Subjet Q1 Q2 Q3 Q4
1100 Y
actually
c(rbind(0,matrix(d, nrow=3)))
which has the bonus of giving the desired result ;)
baptiste auguie schrieb:
Perhaps you can try this,
d - c(0.00377467, 0.00377467, 0.00377467, 0.00380083,
0.00380083, 0.00380083,
0.00380959, 0.00380959, 0.00380959, 0.00380083, 0.00380083,
Hi all,
can any one suggest how to convert one string into binary formate to store
in another file
and to use for farther searching
with using of binary files only that search can process for that i need to
convert string into binary files
thanks in advance
[[alternative HTML version
2009/2/19 Thomas Lumley tlum...@u.washington.edu:
On Wed, 18 Feb 2009, Uwe Ligges wrote:
dobomode wrote:
Hello R-help,
I am trying to import a large dataset from SPSS into R. The SPSS file
is in .SAV format and is about 1GB in size. I use read.spss to import
the file and get an error
Gabor Grothendieck wrote:
On Wed, Feb 18, 2009 at 7:27 AM, Esmail Bonakdarian esmail...@gmail.com wrote:
Gabor Grothendieck wrote:
See ?Rprof for profiling your R code.
If lm is the culprit, rewriting your lm calls using lm.fit might help.
Yes, based on my informal benchmarking, lm is the
James,
as I previously told you in my broken English, probably the function
you're looking for is not filled.contour but image and contour
The following code makes exactly what you ask for
data(akima)
akima
akima.smooth -
with(akima, interp(x, y, z, xo=seq(0,25, length=500),
You can find the complete list at:
http://cran.r-project.org/web/views/MachineLearning.html
Max
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
I am using DLM to fit a state space model. The covariance matrix of states (W)
is given by:
a 0 a 0
0 0 0 0
a 0 a 0
0 0 0 0
where a is a parameter to be estimated. Even though the matrix is positive
semidefinite, sometimes DLM gives me an error that W is not a valid variance
matrix. As far
thanks all for the correction, funny how it's often the complicated
solution that comes to mind first.
baptiste
On 19 Feb 2009, at 13:41, Eik Vettorazzi wrote:
actually
c(rbind(0,matrix(d, nrow=3)))
which has the bonus of giving the desired result ;)
baptiste auguie schrieb:
Perhaps
The zoo package has a multi-way merge for zoo objects. Its
just do.call(merge, z) where z is a list of zoo objects.
In detail:
set.seed(1)
DF1 - data.frame(var1 = letters[1:5], a = rnorm(5), b = rnorm(5), c = rnorm(5))
DF2 - data.frame(var1 = letters[3:7], a = rnorm(5), b = rnorm(5), c =
Dear List,
I have this column/vector:
vec - c(function, missing, string)
and want to compute a second column/vector:
- value if the pattern unc is found: 1
- value if the pattern iss is found: 2
- value if none of the patterns is found: 0
This should be the result:
vec2
[1] 1 2 0
Any help?
On 19/02/2009 9:26 AM, Stefan Uhmann wrote:
Dear List,
I have this column/vector:
vec - c(function, missing, string)
and want to compute a second column/vector:
- value if the pattern unc is found: 1
- value if the pattern iss is found: 2
- value if none of the patterns is found: 0
This
- Original Message -
From: baptiste auguie ba...@exeter.ac.uk
To: Gundala Viswanath gunda...@gmail.com
Cc: r-h...@stat.math.ethz.ch R-help r-h...@stat.math.ethz.ch
Sent: Thursday, February 19, 2009 7:12:23 AM GMT -06:00 US/Canada Central
Subject: Re: [R] Insert value in a Vector
A Todos
Todavía hay sitios para los próximos cursos. Para mas información y el
formulario de inscripción pónganse en contacto con
train...@mango-solutions.com, o visiten nuestro sitio Web
www.mango-solutions.com http://www.mango-solutions.com/ .
The R Language
Introducción al idioma R
You asked about survival curves with age scale versus follow-up scale.
fit1 - coxph(Surv(time/365.25, status) ~ t5 + id + age, data=stanford2)
surv1- survfit(fit1)
surv1
n events median 0.95LCL 0.95UCL
157.000 102.000 1.999 0.898 3.608
summary(surv1, times=3)
time n.risk
Dear Chun-Hao,
How about this?
diet-sort(rep(x=c(C,T),4))
vesl-rep(x=c(A,P),4)
mydata-data.frame(diet,vesl)
mydata$trt-with(mydata,paste(diet,vesl,sep=))
mydata
HTH,
Jorge
On Thu, Feb 19, 2009 at 2:53 AM, Chun-Hao Tu tc...@hotmail.com wrote:
Hi R users,
I did do the research and work on
On Thu, Feb 19, 2009 at 9:50 AM, Jorge Ivan Velez
jorgeivanve...@gmail.com wrote:
mydata$trt-with(mydata,paste(diet,vesl,sep=))
Besides the above (good!) solution, you might want to understand why
your original solution didn't work:
mydata$trt-ifelse(mydata$diet==C mydata$vesl==A, CA,
+
I am using glm.nb, a ~b*c ( b is categorical and c is continuous). when I
run this model I get the warning message:
Warning messages:
1: In theta.ml(Y, mu, sum(w), w, limit = control$maxit, trace =
control$trace :
iteration limit reached
2: In theta.ml(Y, mu, sum(w), w, limit = control$maxit,
Dear Mark and Simon,
I assume from the variable names that siteall and district are factors and
that yrs is numeric. If that's the case, then the second model formula, ~
siteall + district + yrs:district, nests yrs within district, that is, will
fit a separate slope for years within each level of
Hi,
I'm using read.table in a loop, to read in multiple files. The problem
is that when a file is missing there is an error message and the loop is
broken; what I'd like to do is to test for the error and simply do
next instead of breaking the loop. Anybody knows how to do that?
Example:
To All,
I'm using R 2.8.1. I have attached two images, one shows what I get and
the other shows what I want. All I want is the plotting region. Surely
there must be a way of plotting individual regions/components in R?
Patrizio, I prefer filled.contour() for my data. I still have the same
Just thought I'd let you guys know about this site I stumbled across:
http://riki.wikidot.com/ http://riki.wikidot.com/
It is obviously in its early stages (as it does not have any content yet)
but is looking like a good place to build a simple knowledge base for the R
software.
Anyway, if any
I don't want to be mean, I really like wikidot, but isn't it a better
solution to use the R wiki instead?
http://wiki.r-project.org/rwiki/doku.php
G.
On Thu, Feb 19, 2009 at 4:44 PM, thefurryblur wrcst...@gmail.com wrote:
Just thought I'd let you guys know about this site I stumbled across:
Hi Stephane,
see ?try
hth.
Stephane Bourgeois schrieb:
Hi,
I'm using read.table in a loop, to read in multiple files. The problem
is that when a file is missing there is an error message and the loop is
broken; what I'd like to do is to test for the error and simply do
next instead of
The idea of the biglm function is to only have part of the data in memory at a
time. You read in part of the data and run biglm on that section of the data,
then delete it from memory, load in the next part of the data and use update to
include the new data in the analysis, delete that, read
There's got to be a better way to use order() on a matrix than this:
y
2L-035-3 2L-081-23 2L-143-18 2L-189-1 2R-008-5 2R-068-15 3L-113-4
3L-173-2
3981 1 221 12
2
8571 1 221 22
2
I'm trying to create a colored map that would show the number of students
per state.
My data frame consists of two columns - state and count.
I'm using the following code
library(maps)
map(usa)
library(plotrix)
state.col-color.scale(gre$count,0,0,c(0,1))
map(state,fill=TRUE,col=state.col)
I'm
Dear All,
I have a query : what is the command to count number of repeated words in a
column.
for ex:
a =
oranges
oranges
apples
apples
grape
oranges
apple
pine
the result should be
oranges 3
apples 3
grape 1
pine 1
is there an easy way for this.
Thanks,
Nataraju
GM R D
Bangalore
--
How about this:
x - matrix(sample(0:1,100,TRUE),10)
# create a list of all the columns to sort
col.list - lapply(seq(ncol(x)), function(a) x[,a])
# now sort the matrix
x[do.call(order, col.list),]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]001001
On Thu, Feb 19, 2009 at 5:40 PM, Aaron Mackey ajmac...@gmail.com wrote:
There's got to be a better way to use order() on a matrix than this:
y
2L-035-3 2L-081-23 2L-143-18 2L-189-1 2R-008-5 2R-068-15 3L-113-4
3L-173-2
3981 1 221 12
?table
On Thu, Feb 19, 2009 at 11:48 AM, Nattu natar...@gmail.com wrote:
Dear All,
I have a query : what is the command to count number of repeated words in a
column.
for ex:
a =
oranges
oranges
apples
apples
grape
oranges
apple
pine
the result should be
oranges 3
apples 3
Dear Greg and Dobo,
The vif() in the car package computes VIFs (and generalized VIFs) from the
covariance matrix of the coefficients; I'm not sure whether it will work
directly on objects produced by biglm() but if not it should be easily
adapted to do so.
I hope this helps,
John
On Wed, 18 Feb 2009, maiya wrote:
I realise that in the case of loglin the parameters are clacluated post
festum from the cell frequencies,
however other programmes that use Newton-Raphson as opposed to IPF work the
other way round, right?
In which case one would expect the output of
James Nicolson wrote:
To All,
I'm using R 2.8.1. I have attached two images, one shows what I get and
the other shows what I want. All I want is the plotting region. Surely
there must be a way of plotting individual regions/components in R?
Patrizio, I prefer filled.contour() for my data. I
Thanks to all, do.call(order, as.data.frame(y)) was the idiom I was
missing!
-Aaron
On Thu, Feb 19, 2009 at 11:52 AM, Gustaf Rydevik
gustaf.ryde...@gmail.comwrote:
On Thu, Feb 19, 2009 at 5:40 PM, Aaron Mackey ajmac...@gmail.com wrote:
There's got to be a better way to use order() on a
You are computing the hat matrix to predict stack.loss, but stack.loss is a
column in the A matrix, so you predictions are all perfect (given stack.loss,
what is stack.loss, fairly simple answer, all errors are 0). I think you want
to redo this using only the 3 columns other than stack.loss in
Here's a bit of a hacky way to do it:
#get the names of each state
state=map('state',plot=F)$names
#set up some random state-color data
cols =
as.data.frame(cbind(state=states,x=sample(1:10,length(states),replace=T)))
#do the plot
map('usa')
for(this_state in state){
The %in% operator returns a vector of logicals the same length as the vector to
the left. The if program flow operator expects a single logical value, not a
vector, since you are giving it a vector it looks at just the 1st element,
ignores the rest and gives the warning. This warning should be
I do not know about the ubuntu instructions, they would not help me on
Suse. The wine version is 1.1.9.
I thought that was the latest,but when I checked latest is 1.1.15, which
does indeed throw the blackbox error.
So, now it does not work for me either
Sorry I gave bad advise
kees
On Thu, 19
Hi,
Where can I find the source code for nlm()? I dowloaded the R2.8.1.tar.gz
file and looked at all the .c and .f files, but couldn't find either nlm.c
or nlm.f
There is an nlm.r file, but that is not useful.
Thanks for any help,
Ravi.
On Thu, Feb 19, 2009 at 8:30 AM, Esmail Bonakdarian esmail...@gmail.com wrote:
Hi Kenn,
Thanks for the suggestions, I'll have to see if I can figure out how to
convert the relatively simple call to lm with an equation and the data file
to the functions you mention (or if that's even
Hello Stephane,
here is something you could try,
filelist - c(file1.txt, file2.txt, file3.txt)
for (i in 1:3) {
tmpList-try(read.table(filelist[[i]]), silent=TRUE)
if(inherits(tmpList, try-error))
{print(paste(error opening file , filelist[[i]]))
} else {
Hello R users.
There is a paper from Ruey Tsay with the title: Testing and Modelling
Threshold Autoregressive Processes, published in 1989 in the Journal of the
American Statistical Association (March, Vol. 84, No. 405).
Mr. Tsay describes a very interesting way of identifying and modelling
Hi,
I have a basic and simple question on how to code pairwise (multiple) mean
compariosn between levels of a factor using one of the Duncan, Tukey or LSD.
Thanks in advance,
Saeed
--
View this message in context:
Hi 2 questions-
1. Is there a package that will allow me to run R scripts (entirely) from
Java?
2. If so, is there a way to capture the output of those scripts, (including
images) and embed them in my SWT java app?
My challenge is I have a java app that does some statistical chores- it
would be
On 2/19/2009 11:51 AM, Saeed Ahmadi wrote:
Hi,
I have a basic and simple question on how to code pairwise (multiple) mean
compariosn between levels of a factor using one of the Duncan, Tukey or LSD.
Here is one approach:
library(multcomp)
summary(glht(lm(Petal.Width ~ Species, data =
It seems to be in optimize.c
Rgonzui has a very nice search facility for source of R or CRAN packages
(however it is against R 2.8.0 source):
http://rgonzui.nakama.ne.jp/R/markup/R-2.8.0/src/main/optimize.c?fm=cq=nlm#
l378
-Christos
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