On 02/22/2011 03:31 AM, Schmidt, Lindsey C (MU-Student) wrote:
What is plot.new? and how can i get it to work so i can load other data?
Hi Lindsey,
It looks to me as though spelling mistakes in your code are causing
failures to generate some of the objects that you then pass to "plot".
The pl
setClass created an S4 class, not an S3 class, and classes are not
objects (so don't try to use it in package.skeleton). [*]
Are you going to have "buzz" methods for S4 generics? If not, you
don't need the setClass line. If you do, you need to put the line
yourself in an R file in the packag
hi,
i am using* tm package* for text mining.
i have filtered two sets of terms whose frequency is greater than 5 in two
separate files.
now i want to find associations between every term of file1 with every term
of file2 in the form of
matrix of descending order in which rows represent terms of fi
> library(ROCR)
> library(e1071)
svmres.prob <- svm(traindx, traindy, probability=TRUE)
svmpred.prob <- predict(svmres.prob, testdx, probability = TRUE,
decision.values = TRUE, type="prob")
> print(length(attr(svmpred.prob, "probabilities")))
[1] 0
> print(attr(svmpred.prob, "probabilities"))
NULL
Pete
The original question "I would like to calculate the average of the mean
temperature for the summer months (Juli, August, September) for each of the
20 years"
Your codes gives mean for each of the 3 months, not for the 20 summers.
Here is a way to get the summer average for each of the 20
Hi List!
Suppose I have the following, please:
> setClass("buzz",representation(x="numeric"),S3methods=TRUE)
[1] "buzz"
> x <- rnorm(10)
> class(x) <- "buzz"
> plot.buzz <- function(x,y,...) {
+ plot.default(x,type="l",col="blue")
+ }
> f <- function(x) {
+ return(x^2)
+ }
> package.skeleton(list
Hi:
Here's one way:
# Manufacture some monthly data over 20 years
df <- data.frame(year = rep(1981:2000, each = 12),
month = factor(rep(month.abb, 20), levels = month.abb),
meantemp = rnorm(240,
mean = rep(c(30, 35, 40, 50, 60, 70, 75, 80
For 5 years
set.seed = 1
d=data.frame(year=rep(2007:2011,each=12), month=rep(1:12,5), meanTemp =
rnorm(60,10,5))
meanByMonth = ave(d$meanTemp, d$month, FUN = mean)[7:9]
HTH
Pete
--
View this message in context:
http://r.789695.n4.nabble.com/Calculate-a-mean-for-several-months-for-several
Hello everyone,
I have a dataset with 3 colums (Year, Month, MeanTemp). Now I would like to
calculate the average of the mean temperature for the summer months (Juli,
August, September) for each of the 20 years.
I'm sure it's somehow possible with a loop, but all I tried so far didn't
worked and
Hi,
I'd like to run regsubsets for model selection by exhaustive search. I have
a list with 20 potential explanatory variables, which represent the real and
the imaginary parts of 10 "kinds" of complex numbers:
x <- list(r1=r1, r2=r2, r3=r3, ..., r10=r10, i1=i1, i2=i2, i3=i3, ...,
i10=i10)
Is
I am working with panel data. I am using the plm package to do this.
I would like to do be able to adjust for autocorrelation, as one does with
glm models and correlation structures (eg corr=corARMA(q=4)) . In
particular, I want to employ MA(4) error structure.
Is there a way of doing thi
Hello,
I'm willing to plot a sequence of densities on a 3d graph, something like
-
x <- sapply(1:10, function(i)rnorm(1000))
f <- sapply(1:10, function(i)density(x[,i], from=-5,to=5)$y)
grid <- density(x[,1], from=-5,to=5)$x
win.gr
Hi all,
I wrote the following code for the problem I posted before. Can some one
recommend a better way to do this?
f1 <- function(lambda,z,p1){
lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8}
f2 <- function(p1,cl, cu){
0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)
On 2011-02-21 07:17, Francois Rousseu wrote:
I guess I will just get over my laziness and use the rep function instead! But
my goal was more to point out the weird names attribution behaviour.
Maybe not all that weird. Do str(x) between your names<-
assignments:
x <- 1:10
str(x)
names(x
Dear R users,
May I know is there a package that implements the Cochrane-Orcutt Prais Winsten
itterative for dealing with autocorrelation in a regression model?
I understand that gls in nlme package does it properly, my question is
will this
two methods provide the same answer for linear mod
Here is the way I usually write my SQL:
> x <- c('
+ select comm, count(*) -- count the number of comms
+from myDF-- this is the dataframe
+group by comm -- grouping argument
+ ')
> x # as vector
[1] "\nselect comm, count(*) -- count the number of comms\n from
myDF-- this i
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius
> Sent: Monday, February 21, 2011 3:11 PM
> To: David Winsemius
> Cc: r-help@r-project.org; IgnacioQM
> Subject: Re: [R] How to delete rows with specific values o
On Feb 21, 2011, at 6:14 PM, Phil Spector wrote:
Mai -
sql=c("-- This is a comment line",
+ "select sysdate -- This is a comment Text" ,
+ " from dual ")
use = sub('--.*$','',sql)
use[use != '']
[1] "select sysdate " " from dual "
Although to get it to print the way yo
On Feb 21, 2011, at 6:05 PM, David Winsemius wrote:
On Feb 21, 2011, at 4:03 PM, IgnacioQM wrote:
I need to filter my data:
I think its easy but i'm stuck so i'll appreciate some help:
I have a data frame with 14 variables and 6 million rows. About
half of this
rows have a value of "0" in
Mai -
sql=c("-- This is a comment line",
+ "select sysdate -- This is a comment Text" ,
+ " from dual ")
use = sub('--.*$','',sql)
use[use != '']
[1] "select sysdate " " from dual "
Although to get it to print the way you listed, you need
to reduce the width of the line:
Hi,
I tried to remove the text starts by "--" to the end of the line as below
sql=c("-- This is a comment line",
"select sysdate -- This is a comment Text" ,
" from dual ")
>sql
[1] "-- This is a comment line"
[2] "select sysdate -- This is a comment Text"
[3] " from dual "
On Feb 21, 2011, at 4:03 PM, IgnacioQM wrote:
Hi,
I need to filter my data:
I think its easy but i'm stuck so i'll appreciate some help:
I have a data frame with 14 variables and 6 million rows. About half
of this
rows have a value of "0" in 12 variables (the other two variables
always
h
Hi,
I need to filter my data:
I think its easy but i'm stuck so i'll appreciate some help:
I have a data frame with 14 variables and 6 million rows. About half of this
rows have a value of "0" in 12 variables (the other two variables always
have values). How can I delete the rows in which all 12
[OOPS! The example I gave before does not correspond to the stated
seed. At the end is a revised example where I have checked that
it does correspond. Sorry.
]
On 21-Feb-11 21:38:11, Ted Harding wrote:
> [See at end]
>
> On 21-Feb-11 17:13:03, Larry Hotchkiss wrote:
>> Hi list:
>> Here is one
*Hi,
*Does anyone know how can I show an *ROC curve for R-SVM*? I understand in
R-SVM we are not optimizing over SVM cost parameter. Any example ROC for
R-SVM code or guidance can be really useful.
Thanks, Angel.
[[alternative HTML version deleted]]
_
Hi all,
I have the following two function f1 and f2.
f1 <- function(lambda,z,p1){
lambda*(p1*exp(-3*z-9/2)+(0.2-p1)*exp(4*z-8))-(1-lambda)*0.8}
f2 <- function(p1,cl, cu){
0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl+3)+(1-pnorm(cu+3)))+(0.2-p1)*(pnorm(cl-4)+(1-p
Don't worry about the sign. When predicting relative log hazard, high hazard
means short survival time so Dxy is negative. When predicting survival
probability (u specified), high prob. means long survival time so Dxy is
positive. You can just reverse the sign when u is not specified.
I did no
[See at end]
On 21-Feb-11 17:13:03, Larry Hotchkiss wrote:
> Hi list:
> Here is one approach that will generate two uniformly distributed
> variables with a correlation of 0.5 --
> [snip]
>> > -Urspr?ngliche Nachricht-
>> > Von:r-help-boun...@r-project.org
>> > [mailto:r-help-boun..
Thanks for the attempt and required output. How about this?
firststep = DT[,cbind(expand.grid(B,B),v=1/length(B)),by=C][Var1!=Var2]
setkey(firststep,Var1,Var2,C)
firststep = firststep[,transform(.SD,cv=cumsum(v)),by=list(Var1,Var2)]
setkey(firststep,Var1,Var2,C)
DT[, {x=data.table(expand.grid(B,B
Hi:
Perhaps another possibility is ?capture.output
Dennis
On Mon, Feb 21, 2011 at 10:26 AM, Tatyana Deryugina wrote:
> Thanks, Ista! I looked at sink() and it looks like it might work.
> However, it seems as though you need to use it with the "print"
> command. I have a lot of regression output
If I read the help file correctly, another alternative is as
follows:
sink(file="my_R_log.txt", split=TRUE)
"split = logical: if TRUE, output will be sent to the new sink
and to the current output stream"
Hope this helps,
Spencer
On 2/21/2011 11:45 AM
Hi:
Here's one way:
f <- function(x) sin(x) + pi/4
g <- function(x) {
sin(x) * (x >= 0 & x <= pi) + sin(x) * (x >= -2 * pi & x <= -pi) -
(pi/4) * (x > -pi & x < 0) - pi/4 * (x > pi & x < 2 * pi)
}
x <- seq(-2 * pi, 2 * pi, length = 200)
plot(x, f(x), type = 'l', col = 'blue', yli
Please start R and enter the following commands:
library(splancs)
area = 6*4
lambda = 1.5
N = rpois(1,lambda*area)
u = runif(N,-2,4)
v = runif(N,0,4)
h = chull(u,v)
h = c(h,h[1])
plot(u[h],v[h],"l",asp=1)
points(u,v)
If you do not get the plot you're expecting, then
repost
1) adding an informa
Hi:
Try the vegan and biodiversityR packages; both contain functions for
producing species accumulation curves.
On Mon, Feb 21, 2011 at 9:38 AM, Vanessa Francisco wrote:
> Hello! I'm a PhD student working with coral reef fish diversity in Mexico.
> I want to do species accumulation curves but I
Rohit-
If I understand you correctly, and your list's name is mylist,
then
mapply('*',mylist,as.numeric(names(mylist)))
will do what you want. In the future, please provide a reproducible
example.
- Phil Spector
On Feb 21, 2011, at 12:38 PM, Vanessa Francisco wrote:
Hello! I'm a PhD student working with coral reef fish diversity in
Mexico.
Are you hiring?
I want to do species accumulation curves but I have differential
sampling
effort for each "sample".
Do you know or have developed an R script
Hi Tatyana,
Well, most things are implicitly printed. For the most part you should
be able to just
sink(file="my_R_log.txt")
and write your r-code as you normally would. Anything that would
usually be printed to the screen is instead sent to my_R_log.txt
instead, including things that are implici
Hi:
Assuming dd is the name of your data frame,
> dd$diff <- with(dd, q2 - q1)
> dd
TIME ID q1 q2 diff
11 1187 3 2 -1
21 1706 3 30
31 1741 2 42
42 1187 3 2 -1
52 1706 3 30
62 1741 2 42
is one way to do it.
HTH,
Dennis
On Mon, Feb 21,
Windows is perfectly capable of handling UTF-8, but its native
encoding is UTF-16LE. Applications on Windows are meant to work with
text data in the UTF-16LE encoding. If it needs to be converted to or
from another encoding then there are services that do this (which
work). There are countless pro
Hi R-helpers,
I am using nnet package for Neural Network. I have understood almost all
the basic parameters and the way they are used. I would like to know how to
give the followinf parameters as input.
1.weights
2.wts
3.mask
4.contrasts
5.subset
6.For which kind of datasets we can set entropy
Dear forumities,
It's seem that there is no straight forward way to calculate R2 of a cluster
solution in R. So, I would like to know if I'm right when calculating a R2-like
statistic for a given clustering solution. In fact, I have different cluster
solution for a given set of data. I would l
If I understand you correctly - you can add a column to the output table for
your color variable?.
Otherwise, if you need to write the *text* in a different color, that's
probably something you want to do outside of R...
Rob Tirrell
On Mon, Feb 21, 2011 at 10:14, Yan Jiao wrote:
> Dear brainy
Thanks, Ista! I looked at sink() and it looks like it might work.
However, it seems as though you need to use it with the "print"
command. I have a lot of regression output that I would like to store
in a log file. How do I use sink with that?
Best,
Tatyana
Hi all,
I'm having a problem with the auto.key function in xyplot.
This is simplified version of my data:
q Weeks Dq
-10 1 2.1122
1 1 1.9904
10 1 1.739
-10 3 2.5942
1 3 1.9714
10 3 1.8745
-10 5 2.5743
1 5
Hi R community,
I have a question I'm sure is very simple for most of you.
I have a list, with each element being a matrix and the names of the
elements are numbers (like 1,3,...). I can extract the matrices and the
names individually. Now, I want to multiply each of the names to the
individual l
?InternalMethods
?S3groupGeneric
?S4groupGeneric
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Erin Hodgess
Sent: Friday, February 18, 2011 10:45 PM
To: R help
Subject: [R] question about generics
Dear R People:
Is there a way
To the more experienced R users,
I am not a professional programmer so please excuse me if my questions seem
naive.
I have only begun using R to solve some problems and I'm already regretting
it
1.
I am trying to plot different functions in the same figure. The first
function
f(x)= sin(x)+(pi/
What is plot.new? and how can i get it to work so i can load other data?
> library(splancs)
> area = 6*4
> lambda = 1.5
> N = rpois(1,lambda*area)
> u = runif(N,-2,4)
> v = runif(N,0,4)
> plot(u,v,asp=1)
> h = chull(u,v)
> h = c(h,h[1])
> plot(u[h],v[h],"1",asp=1)
Error in plot.xy(xy, type, ...) :
On 2011-02-20 20:02, Karmatose wrote:
>> I'm trying to include multiple variables in a non-parametric analysis
>> (hah!). So far what I've managed to
>> figure out is that the NPMC package from CRAN MIGHT be able to do what I
>> need...
Also look at packages nparcomp and coin (+ multcomp). Both
Hello! I'm a PhD student working with coral reef fish diversity in Mexico.
I want to do species accumulation curves but I have differential sampling
effort for each "sample".
Do you know or have developed an R script that consider the differential
effort of each sample?
PRIMER and other programs us
Hello
I have a question about how to regress panel data in R. What are the
appropriate commands and preparations that need to be made when regressing
panel data?
/Andreas Otterstrom
[[alternative HTML version deleted]]
_
Hi R-users,
I've found the error
In these rows:
# Model (a)
testtemp <- lm.bp(doctorco~sex+age+income, prescrib~sex+age+income,
data=bootdata)
betafound <- c(testtemp$beta,testtemp$beta3)
results[i,] <- betafound
betafound must to be equal to:
betafound <- c(testtemp$beta1,testtemp$beta
I am still struggling (I'm an R novice). Basically I just want to sum the
values per group if the year condition is met. I have the feeling that using
a loop would work, but I am not really familiar with loops. Something like
this?
for(DF$C in 1:length(DF$C))
{
DF<-which(DF$year
Thank you very much Prof Harrell!
Sorry that I am new to this forum, and so ain't familiar with how to post
message appropriately.
I repeated the same procedure using a dataset from the {survival} package.
This time I used the {rms} package, and 100 bootstrap samples:
> library(rms)
> library(s
Dear all,
I want to perform paired Wilcoxon signed ranks test on my data.
I have pairs defined by ID and TIME variables.
How can I calculate difference in variables q1, q2 in each pair?
TIME ID q1 q2
1 1187 3 2
1 1706 3 3
1 1741 2 4
2 1187 3 2
2 1706 3 3
2 1741 2 4
Please, any clue!
:)
--
**
Dieter (et. al.)
I am weak and therefore yield to temptation... This is OT for R, so
stop reading and discard now if you're looking for real R Help.
(see also one inline coment below)
Mount soapbox; begin rant http://ift.confex.com/ift/2005/techprogram/paper_27139.htm
The importance of this ide
Dear brainy R users,
I need to output a matrix, with two colors , meaning some elements using
different color
I normally use write.table(table.m, file="table file name.csv",
sep=","), how could I pass the index for the color ?
Many thanks
yan
**
This is asking FAR too much under Windows, which has no UTF-8 locales.
In particular, cat() (on which write() is based) will convert to the
native locale, even if you manage to input the string as an R UTF-8
string.
And conversion is a OS service, so you are getting the conversion
Windows see
As I partially showed, I guess that the problem of assigning the value
99 to any object in the workspace can be dealt with a more precise
regular expression.
Maybe that would do:
ls(pattern="^a[1-9]$")
or
ls(pattern="^a[0-9]+$")
In any case, I agree that alternatives are better
Ivan
Le 2/21
If instead of having a1, a2, etc. as global variables you put them into a list
then this becomes simple.
The general rule is that if you ever want to do the same (or similar) think to
a set of variable, then they should not have been separate variables, but part
of a bigger one. Lists work wel
Hi list:
Here is one approach that will generate two uniformly distributed
variables with a correlation of 0.5 --
> N <- 1000
> x <- sample(1:9,N,replace=TRUE)
> uni <- runif(N)
> y <- (uni<0.5)*x+(uni>0.5)*sort(x)
> (y.freq <- table(y) )
y
1 2 3 4 5 6 7 8 9
Hi,
This works for me:
pat <- ls(pattern="^a") ## I would anchor "a" to the beginning with "^"
for safety!
for (i in seq_along(pat))assign(pat[i], value=99)
Or this with lapply:
lapply(pat, FUN=function(x) assign(x, value=99, envir=.GlobalEnv))
See ?assign
HTH,
Ivan
Le 2/21/2011 17:22, Nuno
On Feb 21, 2011, at 9:53 AM, Erich Neuwirth wrote:
We want to generate a distribution on the unit square with the
following
properties
* It is concentrated on a "reasonable" subset of the square,
and the restricted distribution is uniform on this subset.
* Both marginal distributions are uni
Hi Nuno,
Yes, you can do
for(i in ls(pattern="a"))
{
assign(i, 99)
}
but honestly this is a bad idea. It will try to assign the value of 99
to any object in your workspace that contains an a, which sounds
really scary to me.
Better I think to use a list:
ab.list <- list(a1=1, a2=2, a3=3, a4=4
Thomas,
I wasn't able to reproduce your finding. The last two characters in my
'out.txt' file were just as expected. But, I'm in an UTF-8 locale. Your
locale affects the encoding of characters on your platform. If you're
not in a UTF-8 locale, then characters are converted from your native
encodi
Hi Patrick,
You didn't say how exactly you tried to "Discretize" your variable,
but have you tried
?factor
?as.character
?
Best,
Ista
On Mon, Feb 21, 2011 at 3:34 PM, Patrick Skorupka
wrote:
> Hi everyone,
>
> I am new to field of data mining as well as particularly using R respectively
> RWe
Hi Tatyana,
I think you are looking for ?sink
Best,
Ista
On Mon, Feb 21, 2011 at 2:11 PM, Tatyana Deryugina wrote:
> Hi everyone,
>
> Is there a way to make R save the workspace output (just the results,
> not the objects themselves) as you go? I'm running analysis that takes
> a long time to ru
On Mon, Feb 21, 2011 at 12:13:09PM +0100, Sylvia Tippmann wrote:
> Hello,
>
> I'd like to fit a logit function to my data.
> The data is distributed like a logit (like in this plot on wikipedia
> http://en.wikipedia.org/wiki/File:Logit.png)
> but the values on the x-axis are not between 0 and 1.
Dear R colleagues,
This seems pretty straight forward but I have been banging my head on this for
some time and can't seem to find a solution
suppose I have something like
a1<-1; a2<-2; a3<-3; a4<-4; b1<-3; b2<-4
I would like to quickly assign to objects with a certain pattern, e.g., those in
Thanks!
On Mon, Feb 21, 2011 at 11:40 AM, wrote:
> Well, you can lose B by just adding to X in the first for-loop, can't you?
> For (...) X <- X + A[...]
>
> But if you want elegance, you could try:
>
> X = Reduce("+",lapply(1:(p+1), function(i) A[i:(n-p-1+i),i:(n-p-1+i)]))
>
> I imagine someone
Hi everyone,
I am new to field of data mining as well as particularly using R respectively
RWeka for writing my master thesis.
I intend to create some specific J48 classification trees with the
RWeka_classifiers_tree function. When I run the source code it says “cannot
handle numeric class”. I
Hi all,
I'm having a problem with the auto.key function in xyplot.
I changed the symbols an lines styles using these commands
trellis.par.set(superpose.symbol=list(pch=c(0,1,2,3,4,5,6,8,15,16)))
trellis.par.set(superpose.symbol=list(col=c(rep("black",11
trellis.par.set(superpose.symbol=list(
Here is a solution using contour:
x <- y <- seq(-1.5,1.5,length=100)
z <- outer(x, y, function(x,y) x^2+y^2)
contour(x,y,z, levels=1)
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help
one simple idea is to generate correlated normals (vector multivariate normal),
and then use the cumulative distribution function F_i of component i such:
F_i(X_i), which is uniform.
Kjetil
(this will not preserve tha value of the correlation coefficient, so
you must experiment)
On Mon, Feb 21,
I guess I will just get over my laziness and use the rep function instead! But
my goal was more to point out the weird names attribution behaviour.
I'm using named vectors of dates that can be associated to 2 stages (incubation
and rearing in nesting birds) to overlay a vector of observed d
> Date: Mon, 21 Feb 2011 15:53:26 +0100
> From: erich.neuwi...@univie.ac.at
> To: marchy...@hotmail.com
> CC: soren.fau...@biology.au.dk; r-help@r-project.org
> Subject: Re: [R] Generating uniformly distributed correlated data.
>
> We want to generate
Hi,
Sure.
Check fitdistr from MASS or fitdist from fitdistrplus package.
Best,
Gergely
On Mon, Feb 21, 2011 at 3:29 PM, Jim Silverton wrote:
> Is there any R package that can fit a beta distribution in R?
>
> --
> Thanks,
> Jim.
>
>[[alternative HTML version deleted]]
>
> __
It is indeed an interesting behavior and I have no idea what you could
do except what you did, though I would use:
names(x) <- rep("A", length(x))
But I don't really understand why you want to give the same name to all
elements? There might be another way around depending on your goal
Ivan
L
Well, you can lose B by just adding to X in the first for-loop, can't you?
For (...) X <- X + A[...]
But if you want elegance, you could try:
X = Reduce("+",lapply(1:(p+1), function(i) A[i:(n-p-1+i),i:(n-p-1+i)]))
I imagine someone can be even more eleganter than this.
rad
-Original Messag
We want to generate a distribution on the unit square with the following
properties
* It is concentrated on a "reasonable" subset of the square,
and the restricted distribution is uniform on this subset.
* Both marginal distributions are uniform on the unit interval.
* All horizontal and all vert
Try this:
names(x) <- rep("A", length(x))
On Mon, Feb 21, 2011 at 11:44 AM, Francois Rousseu <
francoisrous...@hotmail.com> wrote:
>
> Hello R users
>
> I was trying to find a less annoying way of naming vectors than:
>
> x<-1:10
> names(x)[1:length(x)]<-"A"
>
> So I tried:
>
> x<-1:10
> names(
It's nice to see all those solutions, but I'm wondering how it would be
helpful to have the display like this.
I'm a bit curious because for me the R output formatting is not very
important.
Ivan
Le 2/21/2011 15:09, (Ted Harding) a écrit :
On 21-Feb-11 13:55:24, Peter Ehlers wrote:
On 2011-0
Hello R users
I was trying to find a less annoying way of naming vectors than:
x<-1:10
names(x)[1:length(x)]<-"A"
So I tried:
x<-1:10
names(x)<-"A" #but this gave only the first element named (as described in
the help files)
and
x<-1:10
names(x)[]<-"A" #but this gave all element
Hi everyone,
Is there a way to make R save the workspace output (just the results,
not the objects themselves) as you go? I'm running analysis that takes
a long time to run and I want to be able to interrupt it without
losing all the output to date. Is there an alternative to putting
"save.image()
Is there any R package that can fit a beta distribution in R?
--
Thanks,
Jim.
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Hi Dennis,
That's very helpful. The plot appears appears to be of the data and not the
fitted linear trajectories, however, as the lines are not linear. Is it
possible to plot the fitted linear trajectories (according to the
fixed/random intercepts and slopes of the lme model)?
Thanks again!
-
Dear all,
[Note: This was originally posted over at Stack Overflow, but it was
suggested that the problem may be an error in R on Windows, so I have
cross posted over here too. ref: http://goo.gl/pd54D)
The following, when copied and pasted directly into R works fine:
> character_test <- functio
> Date: Mon, 21 Feb 2011 10:45:17 +0530
> From: reynoldspravin...@gmail.com
> To: r-help@r-project.org
> Subject: [R] Regarding Soft Independent Modeling Computational Analysis
>
> Hi
> I'm a B.E student pursuing my Project in the CEERI unit of the Counci
Hello
I have a data set with outlier and it is not normally distributed. I would
instead like to use a more robust distribution like t-distribution.
My question is if the coefficients of the regression are different from zero,
but assuming a t-distribution.
Could someone hint me what package t
Hello,
I'd like to fit a logit function to my data.
The data is distributed like a logit (like in this plot on wikipedia
http://en.wikipedia.org/wiki/File:Logit.png)
but the values on the x-axis are not between 0 and 1.
I don't think using a glm is the solution because I simply want to
infer th
Hi,
I am trying to determine the MLE of the following function:
http://r.789695.n4.nabble.com/file/n3317341/untitled.bmp
I have defined both parts of the equation as separate functions and looped
over the t and G values to get summations of each part.
The lamda function has 3 unknowns which
Dear Sue,
I am also having problems with this. As far as I can gather the predict
function will work with geeglm to give you predicted values from the model
but it does not produce the standard errors automatically.
I have posted a similar question and have had no answers. However, I know
that
On 21-Feb-11 13:55:24, Peter Ehlers wrote:
> On 2011-02-21 04:21, Antje Niederlein wrote:
>> Thanks for every helpful answer :-) !
>> I thought it was something "easier" but as long as there is a solution
>> it's fine for me.
>>
>> Ciao,
>> Antje
>
> Here's one more that I use:
>
> cat( 1:10, sep
Juergen Rose wrote:
> What are the differences between blas-reference, blas-atlas and
> blas-goto.
They are different. The reference BLAS is (relatively) slow; the other ones
are fast. If you’re happy with the performance of the reference BLAS, or
don’t use much matrix algebra-dependent code, t
On 2011-02-21 04:21, Antje Niederlein wrote:
Thanks for every helpful answer :-) !
I thought it was something "easier" but as long as there is a solution
it's fine for me.
Ciao,
Antje
Here's one more that I use:
cat( 1:10, sep="\n" )
But this won't give you the row numbers.
[I keep a functio
On 2011-02-20 20:02, Karmatose wrote:
Hi folks, sorry if this has been answered before, I searched long and hard
before deciding to make a thread.
I'm trying to include multiple variables in a non-parametric analysis
(hah!). So far what I've managed to figure out is that the NPMC package from
C
Hi!
21.02.2011 08:50, Schmidt, Lindsey C (MU-Student) wrote:
> What is plot.new? How can I fix this data or add plot.new so it works?
?plot.new
?plot
plot(u[h],v[h],type="l",asp=1) seems to work for me...
HTH,
Kimmo
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Am Montag, den 21.02.2011, 11:25 +0100 schrieb Karl Ove Hufthammer:
> Juergen Rose wrote:
>
> >> eigen(D)
> >
> > *** caught segfault ***
> > address (nil), cause 'unknown'
> >
> > Traceback:
> > 1: .Call("La_rs", x, only.values, PACKAGE = "base")
> > 2: eigen(D)
> >
> > All systems are Gentoo
By careful programming practices you should be able to avoid the
problem. There is no way to prevent it since the program is only
doing what you ask it to do and if you make a mistake and ask it do to
something, it is not the program's fault. One of the nice things (&
bad things) about R is that
> Date: Mon, 21 Feb 2011 13:03:53 +0100
> From: erich.neuwi...@univie.ac.at
> To: soren.fau...@biology.au.dk; r-help@r-project.org
> Subject: Re: [R] Generating uniformly distributed correlated data.
>
> hw<-function(r){
> (3-sqrt(1+8*r))/4
> }
>
>
>
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