On Thu, 12 Apr 2012, jpm miao wrote:
Example:
Will ts objects be obsolete or modified?
No, I don't think so.
a[,1]
1983 Q1 2.747365190
1983 Q2 2.791594762
1983 Q3 -0.009953715
1983 Q4 -0.015059485
1984 Q1 -1.190061246
1984 Q2 -0.553031799
1984 Q3 0.686874720
1984 Q4
On Thu, 12 Apr 2012, jpm miao wrote:
It seems to only works for zoo or ts objects?
It was tested for zoo, ts, and plain numeric vectors.
I tried to run it for xts objects, and error message occurs.
I had a quick look if this can be avoided. But xts' merge() method is not
fully compatible
Hi
The function loess takes very long time if the dataset is very huge
I have around 100 records
and used only one independent variable. still it takes very long time
Any suggestion to reduce the time
-
Thanks in Advance
Arun
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Dear all,
There have been some recent major updates to the TractoR and RNiftyReg
packages, which are available for medical image analysis applications.
If you use R for these purposes, I hope you will find the updates
useful.
RNiftyReg is an image registration package, which provides a clean
ok this is these are the final results: by relation free vs sliced vs scale
component function
relation free
http://r.789695.n4.nabble.com/file/n4551068/relation_free.png
relation sliced
http://r.789695.n4.nabble.com/file/n4551068/relation_sliced.png
scale component function
Hi
see inline
Hello,
I've got a small dataset on box turtle shell measurements that I would
like to perform a detrended correspondence analysis on. I thought that
it
would be interesting to examine the morphometrics for each species in
the
area of overlap and in areas where neither
Hello everybody,
I know this is pretty basic stuff, but could anyone explain me how to
recode a single value of a variable
into a missing value?
I used to do it like this:
myData[myData$var1==5;var1]-NA # recode value 5 into NA
But the column var1 already contains NAs, which
I also had the same problem.
Being on Linux, I prefer Walmes' command line method but I found that putting
\usepackage[utf8x]{inputenc}
as the first instruction in the master .Rnw file also does the trick. No need
to change anything after that in the file, at least not for me!
Rgds,
Rainer
On
Hi David,
You bring up a good question. I am not sure what is the right way to
solve it. But here is a simple solution I put together:
x = c(1:10,5)
y = x
x[c(2,3)] - NA
# reproducing the problem:
y[x==5]
na2F - function(x) {
x2 - x
x2[is.na(x)] - F
x2
}
na2F(x==5)
# solved
Le jeudi 12 avril 2012 à 12:29 +0300, Tal Galili a écrit :
Hi David,
You bring up a good question. I am not sure what is the right way to
solve it. But here is a simple solution I put together:
x = c(1:10,5)
y = x
x[c(2,3)] - NA
# reproducing the problem:
y[x==5]
na2F -
Le jeudi 12 avril 2012 à 11:08 +0200, David Studer a écrit :
Hello everybody,
I know this is pretty basic stuff, but could anyone explain me how to
recode a single value of a variable
into a missing value?
I used to do it like this:
myData[myData$var1==5;var1]-NA # recode
On 12.04.2012 03:51, winie wrote:
I am trying to schedule my R script using cron, but it is not working. It
seems R can not find packages in cron. Anyone can help me? Thanks.
The following is my bash script
# source my porfile
. /home/winie/.profile
# script.R will load packages
R CMD BATCH
On 12.04.2012 05:49, arunkumar wrote:
Hi
The function loess takes very long time if the dataset is very huge
I have around 100 records
and used only one independent variable. still it takes very long time
Any suggestion to reduce the time
Use another method that is computationally
Looking for r-software resource persons for the Sensitization and Capacity
Building Workshop on Free Software and Knowledge Initiatives in
Agricultural Research for Development to be held in the month of July 2012
at Hyderabad. https://sites.google.com/site/fskiard/
1. Some data structured the way you are using would have been helpful.
I used Tal Galil's play data and set up a dataframe with the variable names you
are using:
structure(list(var1 = c(1, NA, NA, 4, 5, 6, 7, 8, 9, 10, 5),
var2 = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 5)), .Names = c(var1,
var2),
Dear all, I am fitting a LOGIT model on this Data...
Data - structure(c(1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1,
0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1,
0, 1, 1, 0, 1, 0, 47, 58, 82, 100, 222, 164, 161, 70, 219, 81,
209, 182, 185, 104, 126, 192, 95, 245,
Dear people,
I have got a question concerning the underlying numerical codes when
reading an SPSS file into R.
I used the package foreign and when I look at a variable I get the verbal
codes.
I would like to know how it is possible to get the underlying numerical
codes as output, which are the
winie winieb...@gmail.com
on Wed, 11 Apr 2012 18:51:28 -0700 writes:
I am trying to schedule my R script using cron, but it is not
working. It seems R can not find packages in cron. Anyone can help me?
Thanks.
The following is my bash script
# source my porfile
Christofer Bogaso bogaso.christofer at gmail.com writes:
Dear all, I am fitting a LOGIT model on this Data...
snip ---
glm(Data[,1] ~ Data[,-1], binomial(link = logit))
Call: glm(formula = Data[, 1] ~ Data[, -1], family = binomial(link = logit))
Coefficients:
(Intercept)
Thanks - I checked through and it looks as if all of the geneids are
formatted similarily so I don't know which one would be causing an error.
Interestingly, your sapply method works on the same data. So I'm happy
although still confused, because the strsplit method worked the other
day with a
I was working with some excel files with a lot of data.
And by hand it is impossible to handle them.
So they are now converted to .csv.
With headers above the columns, like:
Data1, Data2, Data3
But now I needed to calculate the log2 value of Data1 and place the result
under Data2.
I can't
Hi,
You can use write.table with Append = TRUE to keep on writing in the same
csv file, but I don't know how to write in an specific cell of your csv
file.
http://stat.ethz.ch/R-manual/R-devel/library/utils/html/write.table.html
write.table (data, file = example.csv, sep = ;, col.names = NA,
I have a general question how to understand the message with non-zero exit
status when new package is installed. Based on my experience, this message
implies a package dependence. Am I correct to understand this message? Are
there anyone who can provide some reference or explain details about the
myData[myData$var1==5;var1]-NA # recode value 5 into NA
try
myData[!is.na(myData$var1) myData$var1==5;var1]-NA
or, more simply,
myData$var1[myData$var1==5]-NA
***
This email and any attachments are confidential. Any
Hi
I have dataset with n variables, say n =3 x y z
the formula should be form = x+y+z+x*x +x*y+x*z +y*y + y*z+z*z
my code has to for loop
Is there any way to reduce two loops
for (i in 1:n)
{
for (j in i:n)
{
dat= cbind(dat,dat[,i]*dat[,j])
}
}
Hello,
I'm the maintainer of a syntax highlighting tool highlight.js[1].
Recently the Kaggle project has announced they wanted to sponsor the
development of the R highlighting definition for it[2]. I wanted to drop
a line about it here since I suspect this list has much more R
programmers on
Dear R users,
I still did not receive an answer to my question and went through the
archive with no luck.
So what does tiff(filename =Rplot%03d.tif) mean ? Why the following code
does produce two files?
tiff(filename =Rplot%03d.tif,width=24,height=20,units=cm,res=300,
pointsize=10,
Thanks Ken for your reply. No doubt your english is quite tough!! I
understand something is not normal with the 5th explanatory variable
(se:2872.17069!) However could not understand what you mean by You
seem to be getting complete separation on X5 ?
Can you please be more elaborate?
Thanks,
On
Hi Marion,
Did you look at the help file? Did you try use.value.labels = FALSE?
Best,
Ista
On Thu, Apr 12, 2012 at 6:28 AM, Marion Wenty marion.we...@gmail.com wrote:
Dear people,
I have got a question concerning the underlying numerical codes when
reading an SPSS file into R.
I used the
Dear all, can somebody please help me how to calculate McFadden
R-square for a LOGIT model? Corresponding definition can be found
here:
http://publib.boulder.ibm.com/infocenter/spssstat/v20r0m0/index.jsp?topic=%2Fcom.ibm.spss.statistics.help%2Falg_plum_statistics_rsq_mcfadden.htm
Here is my
On 12-04-2012, at 13:07, John S wrote:
Dear R users,
I still did not receive an answer to my question and went through the
archive with no luck.
So what does tiff(filename =Rplot%03d.tif) mean ?
Just read the documentation for tiff: ?tiff
Why the following code
does produce two
You should look up the Hauck-Donne phenomenon, which shows that
with binomial GLMs, the standard error can grow faster than
the effect size. Complete separation results, for example,
when one predictor (or a combination of several predictors)
perfectly predicts the response. Something like this
Bert and Gabor,
Thanks for clarifying. Much appreciated.
One more question. Do either of you know the location of the code
that does the lookup of the default arguments?
I would like to be able to capture the implicit dependency of the
function on the default arguments.
Hence, I would write
Thanks Berend for your answer!
I read the documentation but I dont understand filename =Rplot%03d.tif ?
Do you mind sending me what you tried?
On Thu, Apr 12, 2012 at 7:34 AM, Berend Hasselman b...@xs4all.nl wrote:
On 12-04-2012, at 13:07, John S wrote:
Dear R users,
I still did
So what does tiff(filename =Rplot%03d.tif) mean ?
Just read the documentation for tiff: ?tiff
A look at ?sprintf might also be useful if you're unfamiliar with the C format
specification. %03d is the format spec for a 3-digit decimal number with
leading zeroes.
On 12.04.2012 13:49, John S wrote:
Thanks Berend for your answer!
I read the documentation but I don’t understand filename =Rplot%03d.tif ?
Do you mind sending me what you tried?
This is a bug in R. I'll take a look.
Best,
Uwe Ligges
On Thu, Apr 12, 2012 at 7:34 AM, Berend
Hi Ista,
thank you for your reply!
Yes, I looked at the help sheet but didn't use this command because I would
still like the categorial variables to be converted into factors and have
the value labels from SPSS imported into R as levels.
With the
use.value.labels = FALSE?
command I do get
On 12.04.2012 12:05, Chel Hee Lee wrote:
I have a general question how to understand the message with non-zero exit
status when new package is installed. Based on my experience, this message
implies a package dependence. Am I correct to understand this message? Are
there anyone who can provide
Dear R communitiy,
I am trying to use multiple functions for aggregation within a function
call for dcast. However this seems to result in an error. Also I have not
managed to make dcast() work with fun.aggregate=sd. Please find attached
some example code using the ChickWeight data.
Many thanks
Please ignore this mail. I got a solution by using 'pscl' package!
On Thu, Apr 12, 2012 at 4:56 PM, Christofer Bogaso
bogaso.christo...@gmail.com wrote:
Dear all, can somebody please help me how to calculate McFadden
R-square for a LOGIT model? Corresponding definition can be found
here:
Christofer Bogaso bogaso.christofer at gmail.com writes:
Dear all, can somebody please help me how to calculate McFadden
R-square for a LOGIT model?
[snip]
library(sos)
findFn(McFadden)
brings you to:
http://finzi.psych.upenn.edu/R/library/pscl/html/pR2.html
i.e. install the sos
On Apr 12, 2012, at 5:18 AM, Yellow wrote:
I was working with some excel files with a lot of data.
And by hand it is impossible to handle them.
So they are now converted to .csv.
With headers above the columns, like:
Data1, Data2, Data3
But now I needed to calculate the log2 value of Data1
Hi!
I'm doing a PCA on some testdata (only 2 features) currently to see what
happens. I read a lot of times one should center ones data, so i left the
option on.
But now i am scratching my head on how to correctly transform new data to be
comparable to the data used in the pca.
pca$x are the
Hi,
I'm trying to do an exploratory factor analysis for a uni assignment
but I keep getting this error message coming up
[1] One of your variables is a constant. Constants are disallowed as
part of a scale.
when I turn the data into a matrix and look at the stats. I've been
told that
I have dataset with n variables, say n =3 x y z
the formula should be form = x+y+z+x*x +x*y+x*z +y*y + y*z+z*z
my code has to for loop
Is there any way to reduce two loops
for (i in 1:n)
{
for (j in i:n)
{
dat= cbind(dat,dat[,i]*dat[,j])
}
}
Yes; use outer
But
Hello.
I am running a multivariate multilevel mixed effects model, and am trying
to understand what the interaction term tells me.
A very simplified version of the model looks like this:
model - lmer (phq ~ -1 + as.factor(index_phq) * Neuro + ( -1 +
as.factor(index_phq)|UserID), data=data)
The
Hi, I've been looking at ways to make pyramid plots in R. I like the
pyramid.plot method in plotrix as it seems the simplest to use and building
them in ggplot looks a bit more code intensive than I'd like, being as I'm new
to R. This package does pretty much what I need it to do, however I
On 12-04-12 12:13 AM, Duncan Mackay wrote:
At 12:03 12/04/2012, you wrote:
I had the same problem! So, as I'm a linux user,
I prefer use linux terminal. On terminal I type this to compile
R CMD Sweave --encoding=utf-8 myfile.Rnw
and the compilation is successful. Try to set the encoding
Ah.. forget it, something must have gone wrong while i was testing (probably
messed up some apply()). Seems you can just subtract the returned means - i
hope at least.
Am 12.04.2012 um 14:30 schrieb Jessica Streicher:
Hi!
I'm doing a PCA on some testdata (only 2 features) currently to see
Hi Duncan
Thank you for the tips
I tried
\usepackage[latin1]{inputenc}
but it still bailed up.
tried showNonASCII on the file with the ° typed
as Alt-248 (used to doing as DOS value)
becomes Alt-176 in ASCII
% 1° line
resulted in
483: % 1b0 line
The showNonASCII is a nifty function
The same data, with the proper citations to Michelson(1882) and
Stigler(1977) are contained in the HistData package as data(Michelson)
See
library(HistData)
example(Michelson)
On 4/11/2012 7:42 AM, Duncan Murdoch wrote:
On 12-04-11 12:43 AM, Křištof Želechovski wrote:
URL:
Hi Rainer
Thanks for an alternative. For the record I tried
your latex solution on my Windows 7
\usepackage[utf8x]{inputenc}
but it failed on the Alt-248
Regards
Duncan
At 19:13 12/04/2012, you wrote:
I also had the same problem.
Being on Linux, I prefer Walmes' command line method but
On 12.04.2012 13:53, Uwe Ligges wrote:
On 12.04.2012 13:49, John S wrote:
Thanks Berend for your answer!
I read the documentation but I don’t understand filename
=Rplot%03d.tif ?
Do you mind sending me what you tried?
This is a bug in R. I'll take a look.
I think I have a fix, will
--- begin included message ---
Hello R users,
I am analizing survival data (mostly uncensored) and want to extract the
most out of it.
Since I have more than one factor, I?ve read that the survival regression
can help to test the interactions between factors, and then decide how to do
the
Hello,
Is it possible to selectively display labels on a histogram?
Thanks
Carol
[[alternative HTML version deleted]]
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PLEASE do read the posting guide
Any help ?
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Hello,
I am trying to do contrasts after applying a binomial mixed effect model
with the function lmer. I have to extract the fix effect values, but as I
write fixef(model), I get this error message:
Error in UseMethod(fixef) :
no method for 'fixef' with objects of class mer
Has anybody some
Dear all,
This is probably more related to statistics than to [R] but I hope someone
can give me an idea how to solve it nevertheless:
Assume I have a variable y that is a function of x: y=f(x). I know the
average value of y for different intervals of x. For example, I know that
in the
Hi
Any help ?
Call 999
Regards
Petr
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Hi
Hello,
Is it possible to selectively display labels on a histogram?
What labels?
Like that?
x-rnorm(1)
hist(x)
hist(x, axes=F, xlab=bla, ylab=ble, main=bleble)
axis(1, at=c(-4, -1, 1, 4))
Regards
Petr
Thanks
Carol
[[alternative HTML version deleted]]
Hi,
The code I have used is
summary(prefdata)
prefdata[prefdata5]=NA
summary(prefdata)
prefdata2=as.matrix(prefdata[3:22])
stats=paf(prefdata2)
[1] One of your variables is a constant. Constants are disallowed as
part of
could you provide some data / sample data you used for prefdata and
swertie v_coudrain at voila.fr writes:
I am trying to do contrasts after applying a binomial mixed effect model
with the function lmer. I have to extract the fix effect values, but as I
write fixef(model), I get this error message:
Error in UseMethod(fixef) :
no method for 'fixef' with
There are numerous JS libraries for syntax highlighting. I have never
tried highlight.js yet, but I wrote something for SyntaxHighlighter
which might be easily translated to highlight.js:
http://yihui.name/en/2010/09/syntaxhighlighter-brush-for-the-r-language/
Regards,
Yihui
--
Yihui Xie
This is the R-help mailig list, not Nabble, hence quote the original
message!
Anyway, the answer is: See ?options and its argument nwarnings.
Uwe ligges
On 12.04.2012 14:49, helin_susam wrote:
Any help ?
--
View this message in context:
Hi,
h - htmlpCatsupa/sup/ppDog/p/html
sub(sup.*sup,,h)
see http://en.wikibooks.org/wiki/R_Programming/Text_Processing for more
information.
Regards!
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PLEASE do read the
No, data labels on the histogram bars. labels = T in hist displays all data
labels.
thanks
From: Petr PIKAL petr.pi...@precheza.cz
Cc: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch
Sent: Thursday, April 12, 2012 4:03 PM
Subject: Re: [R] selective labels
Dear users,
I'm quite a new french R-user, and I have a problem about doing a
correlation matrix.
I have temperature data for each weather station of my study area and for
each year (for example, a data file for the weather station N°1 for the year
2009, a data file for the N°2 for the year
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Yellow
Sent: Thursday, April 12, 2012 2:18 AM
To: r-help@r-project.org
Subject: [R] How insert data to a column in existing csv file?
I was working with some excel files with
I would suggest using the XLConnect package so that you can read/write
the Excel files directly. You can read in the files to a dataframe,
make your transformations and then write the result back out to the
Excel file.
On Thu, Apr 12, 2012 at 11:29 AM, Nordlund, Dan (DSHS/RDA)
Hi
No, data labels on the histogram bars. labels = T in hist displays all
data labels.
You could find it probably quicker in documentation.
Plotting command usually creates (invisibly) the object which can be saved
and changed.
h-hist(x)
h is list which you can use or modify
for
Dear List,
I don't get scatter3d to color the sheres according to the '|' argument.
library(car)
scatter3d(prestige ~ income + education|type, data=Prestige)
The spheres on my screen are all colored the same and they are not
conditional on Prestige$type.
On the other hand: Fit3d and Ellipse3d
Hi
the solution with linear programming works very well!
Even if I take more 'realistic' matrices with dimensions of about 25x60,
the calculation of the basis goes well (if I increase the number of
iterations to 1000).
Meanwhile I also found an algorithm for an exact calculation of the extreme
Gavin,
Thank you for this example. I would use the as.pyramidLikert function in
the HH package.
The resulting plot is a standard lattice plot, so any fine tuning uses that
standard lattice features.
Rich
## install.packages(HH) ## if necessary
library(HH)
mydata - cbind(mpop,fpop)[10:1,]
Is this what you mean?
x - rnorm(12)
histogram(x)
histogram(x, scales=list(x=list(at=c(-1, 1), labels=c(minus one, one
On Thu, Apr 12, 2012 at 9:28 AM, carol white wht_...@yahoo.com wrote:
Hello,
Is it possible to selectively display labels on a histogram?
Thanks
Carol
I do something similar my script is structured as follows: perhaps it will
help
library(car)
library(rgl)
library(mgcv)
scatter3d(MADep2009 ~ MADwet2009 + MedAD2009 | as.factor(VegClass), data =
SFCN,
sphere.size=1.5, surface=FALSE, parallel=FALSE, elliposiod=TRUE,
surface.col=c(green, red,
Alison,
You've got two geneids with two periods (instead of just one period).
gene.list - strsplit(as.character(Rumino_Reps_agreeWalign$geneid),\\.)
Rumino_Reps_agreeWalign[sapply(gene.list, length)!=2, ]
geneid count_Conser count_NonCons count_ConsSubst
count_NCSubst
7
Hi Marion,
I don't have access to SPSS, so it would be hard for me to figure this
out for you. I would think that string variables would be imported as
strings or factors even when use.value.labels is true, but can't
verify that.
Good luck,
Ista
On Thu, Apr 12, 2012 at 7:59 AM, Marion Wenty
Dear R-list,
I am in the process of translating a long function written in Matlab
into R (mainly because I am a big of fan of R, and folks will not
have to pay to use it :). In the translation of this function
I got stack because they use spdiags, which, as far as I can tell
it is not available
Hi Jokel,
On Thu, Apr 12, 2012 at 7:57 AM, Jokel Meyer jokel.me...@googlemail.com wrote:
Dear R communitiy,
I am trying to use multiple functions for aggregation within a function
call for dcast. However this seems to result in an error. Also I have not
managed to make dcast() work with
Dear René,
I've confirmed that the spheres aren't coloured correctly on my Ubuntu system
(the first colour is used for all of the spheres), and I know that this works
right on Windows, as you mentioned. I'm curious to try it on my Mac, but don't
have that handy at the moment.
I also looked at
Hi everyone,
I read your messages and I tried some things out.
The Data2 is filled with NA.
So what I did was:
NameFile.csv$Data2[is.na(NameFile.csv$Data2)] = log2Values
And now it worked. :)
Seems like I needed to replace it, and not only write it.
Thanks.
--
View this message in
Hi all,
I'm just getting started in R. My problem is the following:
I have a data frame (v1) with lots of production data measurements.
Each row contains a single measurement ('ARI_MIT') with a timestamp. I
want to lump the data by months with their mean and standard
deviation.
I have already
The formula is created sepearately and data is created for the formula and
then use the modelling lm
-
Thanks in Advance
Arun
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Hi,
I want to calculate pointwise mutual information between label 2-gram, and
words in my corpus 1-gram. Any suggestions as to how to go about it?
l =label
w = word
C = reference collection
I want to calculate following:
p(w,l| C)
p(w| C)
p(l | C)
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Hi All.
I have problem with generating series of plots. In detail:
I have file of data which I insert into dataframe:
data-read.table(file.txt, header=TRUE, sep= )
Data in this file are prepared in such a way, that the header of each
column has a value of one of examined parameter - something
Thank you very much. I started R again and loaded less libraries and it
worked! I didn't know about this problem, the next time, I'll check for
this before looking a whole afternoon for alternative methods ;).
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This is my wrong, I am so sorry for this. I just wanted to ask something
which I did not know. Thanks for answer.
Regards
Helin.
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Hi Gang,
I realize this post is not directly related to programing issues with R,
however, it appears this may be the best place to ask my question.
I am putting forward a request that R be considered approved software in
my organization. Never an easy task, this is made much more difficult
Hi every one. I have a exponential function (3 fitting parameters) that I
would like to use to produce data (6 series) without having to use a loop.
Here
wl = seq(300,500,1)
k1 = c(1.2e-6, 4.9e-6, 9.6e-6, 2.7e-10, 6.7e-8, 7.44e-6)
k2 = c(726, 352, 128, 5232, 1538, 128)
k3 = c(-176, -224, -257,
Thanks to all for your help! That works!
Orange
On Sat, Apr 7, 2012 at 7:25 PM, Gabor Grothendieck
ggrothendi...@gmail.comwrote:
Yes, I had fixed that on my end but didn't notice I had copied the old
version. I think its obvious enough that anyone will fix it
themselves. Regards.
On
Sorry, I did not understand. What does it mean call 999 ?
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Dear Terry
 here is the survreg line from which I understand that gender is significant
survreg(formula = Surv(dias, status) ~ trat * sexo * rep, dist = weibull)
                  Value    Std. Error     Â
z       p
sexom           -0.2187 Â
Dear all,
I have been searching far and wide for a solution to the problem of
negative intracluster correlation in the case of a binomial response
variable (also known as under-dispersion). Prentice (1986)
(http://www.jstor.org/stable/2289219) developed an extension of the
beta-binomial model
Hey there,
I want to plot 5 parabola functions, which happen to be
f(x) = 0.25x² + 6,47x -32.6
g(x)=0.99x² -6x -195
j(x)= 0.77x² +14x -495
k(x)=0.001x² + 65x -785
l(x) = 0.9x² -2x -636
in the same graph. Sadly I even do not really understand how to plot just
one
Welcome to R and the list.
Others may suggest books ( Nutshell was my first ) but first there are
some things that will help you
both in programming and getting help on the list.
You should post executable code in your question. So, build a toy example
of the data.frame you have
and show
Hi,
At the moment I am studying R at school.
But I found this site very useful to explain the plot functions to me:
http://www.harding.edu/fmccown/r/
Maybe that can help you too?
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Just a joke -- since you post through Nabble instead of being a list
subscriber, the vast majority of us didn't see your original post and
instead only saw your cry for help
Michael
On Thu, Apr 12, 2012 at 10:08 AM, helin_susam helin.su...@gmail.com wrote:
Sorry, I did not understand. What does
Slightly on topic, just yesterday, the CFPB announced they'll be using
R in their work: http://blog.revolutionanalytics.com/applications/
Depending on the scale of your organization, it may not be the
open-source nature that's quite the problem of the IT Gang but
rather the support (or lack
Perhaps ?outer -- well, not outer directly, but a multivariate outer
-- I keep this one around for personal use:
`mouter` - function(..., FUN = *){
dotArgs - list(...)
FUN - match.fun(FUN)
if(length(dotArgs) == 1L)
return(unlist(dotArgs))
if (length(dotArgs) == 2L)
Very interesting book!
However, it doesn't cover multivariate models (I have 9 moderately
correlated, categorical dependent variables).
Again, I'm trying to find out whether 5 time-varying variables
(dichotomous; five different life events yes/no; subjects can have
several life events at the same
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