Hello all,
May be silly question, but what exactly is beta parameter in functions like
regmixEM from mixtools package?
I mean, how to determine this beta, if i have a set of metrics for each
case? Is there a function for that? I have try to put NULL at this
parameter, but function just do not
Newbie wrote:
Dear R-users
I need to calibrate kappa, rho, eta, theta, v0 in the following code, see
below. However when I run it, I get:
y - function(kappahat, rhohat, etahat, thetahat, v0hat)
{sum(difference(k, t, S0, X, r, implvol, q, kappahat, rhohat, etahat,
thetahat, v0hat)^2)}
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hello,
as far as I understood your problem, this function might do the trick:
CountNextBehavior - function (data.source, interest.behavior,
lev.ignore, interest.timeframe) {
## --
##
Dear Uwe,
Thanks for you suggestion. I should have spotted that (not thinking
about the order of commands properly). I still don't know why it failed
to work in RGui for me. It works now. I suspect it must be another
case of PEBCAK (Problem Exists Between Chair And Keyboard).
Best wishes,
Thanks Henrik, but I have 2 reasons for not using that approach:
A) If I don't map the drive until after R starts the UNC path is already
present in several places I know about and probably some I don't, leading to
the problems I started with.
So reason 'B' doesn't really matter to me, but as
Dear Mark,
Thank you very much for your kind advice.
Actually, I already performed penalized logistic regression by pentrace
and lrm in package rms.
The reason why I wanted to reduce dimensionality of those 9 variables
was that these variables were not so important according to the subject
Dear All,
is there a simple way to retain the class attribute of a column, if
merging two data.frames?
When merging the example data.frames form help(merge) I am unable to
keep the class attribute as set before merging (see below).
Two columns are assigned new classes before merge (myclass1,
Dear all,
I am forced to work in an environment without administrator rights.
When using R2.13.1 on Windows 7 (64-Bit), I found that I can´t install or
update any packages due to missing writing permissions.
I managed to get full access to a directory on my C:\ drive now - but how do I
You might try using outer to create a matrix that will help out:
Time - c(1000, 1050, 1100, 1500, 2500, 5000, 6500, 6600, 7000)
Time
[1] 1000 1050 1100 1500 2500 5000 6500 6600 7000
?outer
starting httpd help server ... done
x - outer(Time, Time, FUN = function(a, b){d - b-a; (d=0) (d =
Folkes, Michael:
I now realize I could write code to evaluate which of the first 7 days
in the year is a Monday and then I'd know the start of week 1 in each
year, and multiply from there.
But note that
library(surveillance) # ISO week
isoWeekYear(as.Date(2010-01-01))$ISOWeek
[1] 53
so
Dear Marc,
I would like to thank you for your answer.
Unfortunately still
setEPS()
postscript(file=exponcoverapprox.eps)
Tack on a day of the week (6 as the last day) for a point of reference
for the conversion:
dates - paste('6.', 0:53, '.2011', sep = '')
dates
[1] 6.0.2011 6.1.2011 6.2.2011 6.3.2011 6.4.2011
6.5.2011 6.6.2011
[8] 6.7.2011 6.8.2011 6.9.2011 6.10.2011 6.11.2011
6.12.2011 6.13.2011
[15]
On Aug 19, 2011, at 8:06 AM, Alaios wrote:
Dear Marc,
I would like to thank you for your answer.
Unfortunately still
setEPS()
postscript(file=exponcoverapprox.eps)
boxplot
(test
[30,1
:
500
],exponper
[90,1
:
500
],test
[150,1
:
500
],test
[210,1
:
500
],test
[270,1
:
500
Sometimes when I have a script that does not close out a graphics
device correctly (using PDF), I sometimes have problems opening up the
file. I use the following command to make sure all graphics devices
are closed before generating plots after a script has not terminated
correctly:
Dear al,
I would like to thank you for your replies.
I have tried with graphics.off() but did not help too.
I am also sorry that my example was not reproducible
So this one
setEPS()
postscript(file=mytest.eps)
On Aug 19, 2011, at 9:08 AM, Alaios wrote:
Dear al,
I would like to thank you for your replies.
I have tried with graphics.off() but did not help too.
I am also sorry that my example was not reproducible
It has never been reproducible because you have ignored the request 3
days ago to
Hello
Many thanks.
* is not a typo. The output is a description of a nonlinear system so
terms such as y(k-1)*y(k-2) are allowed. I wonder whether could be
ignored so that the outputs such as y_{01}(k-003)* would not show up.
Cheers
Ed
On Thu, Aug 18, 2011 at 3:40 PM, David Winsemius
On 08/17/2011 10:53 PM, Alex Ruiz Euler wrote:
Dear R community,
I have a 2 million by 2 matrix that looks like this:
x-sample(1:15,200, replace=T)
y-sample(1:10*1000, 200, replace=T)
x y
[1,] 10 4000
[2,] 3 1000
[3,] 3 4000
[4,] 8 6000
[5,] 2 9000
[6,] 3
On 08/18/2011 07:46 AM, Timothy Bates wrote:
This takes a few seconds to do 1 million lines, and remains explicit/for loop
form
numberofSalaryBands = 100 # 200
x= sample(1:15,numberofSalaryBands, replace=T)
y= sample((1:10)*1000, numberofSalaryBands, replace=T)
df
As I already stated in my reply to your earlier post:
resending the answer for the archives of the mailing list...
Hi Alex,
The other reply already gave you the R way of doing this while avoiding
the for loop. However, there is a more general reason why your for loop
is terribly inefficient. A
Hi,
My time zone in Montreal is Standard time zone:UTC/GMT -5 hours (see
http://www.timeanddate.com/worldclock/city.html?n=165).
Yet, in R (POSIXct objects) I must specify the opposite, i.e. UTC+5:
dateMontreal = as.POSIXct(2011-01-15 05:00:00, tz=EST)
dateMontreal2 =
On Aug 19, 2011, at 9:45 AM, Eduardo Mendes wrote:
Hello
Many thanks.
* is not a typo. The output is a description of a nonlinear system
so terms such as y(k-1)*y(k-2) are allowed. I wonder whether
could be ignored so that the outputs such as y_{01}(k-003)* would
not show up.
On Aug 19, 2011, at 10:21 AM, David Winsemius wrote:
On Aug 19, 2011, at 9:45 AM, Eduardo Mendes wrote:
Hello
Many thanks.
* is not a typo. The output is a description of a nonlinear
system so terms such as y(k-1)*y(k-2) are allowed. I wonder
whether could be ignored so that the
R Users:
Can anyone please help me with the following:
I'm unclear as to how to get format to do what I want.
I've tried the following and get unexpected results.
Input.
val-321.6
format(val, digits=1)
format(val, digits=2)
format(val, digits=3)
format(val, digits=4)
format(val,
On Aug 19, 2011, at 10:23 AM, Michael Karol wrote:
R Users:
Can anyone please help me with the following:
I'm unclear as to how to get format to do what I want.
I've tried the following and get unexpected results.
Input.
val-321.6
format(val, digits=1)
format(val, digits=2)
Hello,
I have a dataset with an Id columns like:
4/3003
55/333
66/22
I want to put leading zeros to get:
0004/3003
00055/333
66/22
How can I solve this?
Thanks
Vasco Cadavez
[[alternative HTML version deleted]]
__
On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:
Hello,
I have a dataset with an Id columns like:
4/3003
55/333
66/22
I want to put leading zeros to get:
0004/3003
00055/333
66/22
How can I solve this?
?sprintf
?formatC
--
David Winsemius, MD
West Hartford, CT
On Aug 19, 2011, at 11:17 AM, David Winsemius wrote:
On Aug 19, 2011, at 11:12 AM, Vasco Cadavez wrote:
Hello,
I have a dataset with an Id columns like:
4/3003
55/333
66/22
I want to put leading zeros to get:
0004/3003
00055/333
66/22
How can I solve this?
?sprintf
Dear R-Users,
I'm trying to setup a personal repository for a few packages I'm working on.
I am on R-Forge but I still need to have various versions of my package that
R-Forge does not build (for R 2.8.1 for example).
So I followed the instructions in this document:
Hhttp://
Dear R-community,
I have tried to estimate an accelerated failure time(AFT) and proportional
hazard (PH) parametric survival model with time-independent and
time-dependent covariates. For that purpose, I have used the eha package.
Please, consider this example:
weibullph -
Greetings,
I am having trouble getting the function reformulate_ATSP_as_TSP to work for
me. I have provided a simple example of some of the code I've been using.
In particular, I'm not sure why I'm getting the error
Error in dimnames(tsp) - list(lab, lab) :
length of 'dimnames' [1] not
Hi all,
I have a series of intra-day data. The variables exhibit a typical daily
pattern over the day. I need to diurnallly adjust the data. It takes the
follow form
1, regress y on a piecewise cubic spline of x with knots (a1,a2,a3,a4...). x
is the time of a day.
2, divide original series by
While R has library TSP to help solve traveling salesperson problems, does
anyone know if it has any libraries to help solve multiple traveling
salesperson problems? For instance, suppose one is planning school bus
routes and one has multiple buses. Thank you for your time.
--
View this message
On Aug 19, 2011 khosoda wrote:
I used x10.homals4$objscores[, 1] as a predictor for logistic regression
as in the same way as PC1 in PCA.
Am I going the right way?
Hi Kohkichi,
Yes, but maybe explore the sets= argument (set Response as the target
variable and the others as the predictor
Dear professor,
I am currently using Design package and the cph formula for assessing
multivariable analysis.
I am tryng to get the C-index for my survival model based on Dxy
coefficient.
I am confused since there is a negative value.
Do I need to used the absolute Dxy ?
Thanks to all of you for the suggestions and corrections.
Sharad
--
View this message in context:
http://r.789695.n4.nabble.com/splitting-sample-names-tp3753712p3755297.html
Sent from the R help mailing list archive at Nabble.com.
__
Indeed, as David pointed out, all the portion that used courier font (all the
good stuff) was absent from the email posting.
Thanks for your answers.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Dear R Users
Any idea if there exists any one dimensional Cox Process datasets in R?
'Spatstat' is very comprehensive but doesn't seem to have any examples of 1D
(time series) Doubly Stochastic Poisson Process data. (I am aware it can be
simulated)
Thank you,
Copying list one what was sent in reply. Anybody have a better solution?
On Aug 19, 2011, at 11:57 AM, Vasco Cadavez wrote:
Thanks,
A solution can be by substring to remove the /
then numeric will be ok! What you think?
How can I remove the /
with sub or gsub:
sprintf(%010.0f,
On Fri, Aug 19, 2011 at 9:19 AM, David Winsemius dwinsem...@comcast.net wrote:
Copying list one what was sent in reply. Anybody have a better solution?
No sure my solution is better, but it avoids the integer conversion
and retains the /.
I wrote a function that padds entries of input
Hello, all users here:
Recently i am doing a project of survival analysis. I collect the
characteristics of patients and have got some factors which are related with
the cancer.
When i come to overcome relations between genotypes (Snps) and
survival time via stratified analysis, i
Hi everybody,
I used randomForest to regress invertebrates abundances in least impaired
river reaches from some environmental parameters. Then I used these models
to predict invertebrates abundances in impaired reaches.
Now I would like to model the deviation (observation - prediction) with a
I have a plot created with strat.plot() from package rioja. When
the plot is created with scale.percent=FALSE, each x axes is labeled at
0 and its maximum. However, when scale.percent=TRUE, the x axes are not
labeled. I need to use scale.percent=TRUE and I need labels for the x axes.
I installed some downloaded packages in R. I always do
$sudo R CMD INSTALL anRpackage.tar.gz
By default it is storing these packages into my directory
/home/mary/R/x86_64-pc-linux-gnu-library/2.13/.
However I want them to be systemwide into /usr/local/lib/R/site-library/
folder.
I tried
$sudo
Hi, friends.
I keep coming to you because I'm so new to R and can't seem to figure out
some simple things. Sorry.
Consider the following code. I want to load a table and write out the
structure to a PDF document. I just can't seem to manage writing
non-graphic output to PDF. Any help? I've
Hi all,
I have two data frames, two columns each, 1000s of rows. Each row
represents a segment of the genome where a deletion has occurred.
First column is start position of the deletion in genomic distance,
second is end position.
So, e.g., first 3 rows of data frame A is:
1003 1023
5932 6120
On 11-08-16 9:50 PM, Eric Heupel wrote:
I have what is probably a noob question, but
I am trying to create a 3d plot to illustrate the range of values for the
following simple function:
A = B*(C/D)
B, C, and D are independent variables whose range are equal (e.g. 1 to 3
inclusive)
I
Take a look at:
R CMD INSTALL --help
and you will realize that you need to specify the library path, e.g. R
CMD INSTALL anRpackage --library=/usr/local/...
or take a look at ?install.packages and use the second argument, e.g.
install.packages('anRpackage', lib = '/usr/local/...')
Regards,
At 18:10 19/08/2011, Mary Kindall wrote:
I installed some downloaded packages in R. I always do
$sudo R CMD INSTALL anRpackage.tar.gz
By default it is storing these packages into my directory
/home/mary/R/x86_64-pc-linux-gnu-library/2.13/.
However I want them to be systemwide into
Many thanks.
The still untested worked.
Cheers
Ed
On Fri, Aug 19, 2011 at 11:23 AM, David Winsemius dwinsem...@comcast.netwrote:
On Aug 19, 2011, at 10:21 AM, David Winsemius wrote:
On Aug 19, 2011, at 9:45 AM, Eduardo Mendes wrote:
Hello
Many thanks.
* is not a typo. The
Dear all,
I have what is a bit of a confusing question, so I hope that I can explain
clearly. Thank you for your help in advance.
I would like to do a replacement procedure on several strings, but the way
that I am currently going about it is not working.
I have defined len, which is a series
r-help-boun...@r-project.org wrote on 08/19/2011 12:15:39 PM:
[image removed]
[R] how to merge distance data based on location
Matthew Keller
to:
r help
08/19/2011 12:18 PM
Sent by:
r-help-boun...@r-project.org
Hi all,
I have two data frames, two columns each, 1000s
-- Forwarded message --
From: Ole Peter Smith ole@gmail.com
Date: Fri, Aug 19, 2011 at 1:40 PM
Subject: Re: [R] Leading zeros
To: David Winsemius dwinsem...@comcast.net
I'm all new to R, assisting the last days of topics from the sideline.
I am, however, a longterm
I'm using chisq.test() on a matrix of categorical data, and I see that the
residuals attribute of the returned object will give me the Pearson residuals.
That's cool. However, what I'd really like is the standardized (adjusted)
Pearson residuals, which have a N(0,1) distribution. Is there a way to
Hi everybody.
I'm trying to use R with Sweave but I have a problem perhaps with the
directory path of sweave in R.
The windows path is this:
C:\Program Files (x86)\R\R-2.9.2\share\texmf\Sweave
When I run the latex file with R, the program works well, without any
errors, but when I create the
There is a strsplit() function (syntax is (stringToBeSplit, splitAt) ) that
may be of use. I haven't followed the thread so I don't know how well it
handles the original problem.
Michael Weylandt
On Fri, Aug 19, 2011 at 12:41 PM, Ole Peter Smith ole@gmail.com wrote:
-- Forwarded
B is the specification for time-varying covariates. Otherwise, your model
will think that each row is one independent observation that either had an
event or was censored at time or total_time.
HTH,
Daniel
javier palacios wrote:
Dear R-community,
which of the following two formats is
On Aug 19, 2011, at 1:28 PM, Stephen Davies wrote:
I'm using chisq.test() on a matrix of categorical data, and I see
that the
residuals attribute of the returned object will give me the
Pearson residuals.
That's cool. However, what I'd really like is the standardized
(adjusted)
Pearson
Dear all
?Hmisc::rcorr states that it takes as main argument a numeric
matrix. But is it normal that it fails in such an ugly way on a data
frame? (See below.) If the function didn't attempt any conversion to a
matrix, I would have expected it to state that in the error message
that it didn't
Dear all
Is there an easy way to display only one half (top-right or
bottom-left) of a correlation matrix?
require(Hmisc)
rcorr(as.matrix(mtcars[ , 1:4]))
mpg cyl disphp
mpg 1.00 -0.85 -0.85 -0.78
cyl -0.85 1.00 0.90 0.83
disp -0.85 0.90 1.00 0.79
hp -0.78 0.83 0.79
On Fri, Aug 19, 2011 at 11:11 AM, Rebecca Gray atlas...@gmail.com wrote:
Dear all,
I have what is a bit of a confusing question, so I hope that I can explain
clearly. Thank you for your help in advance.
I would like to do a replacement procedure on several strings, but the way
that I am
r-help-boun...@r-project.org wrote on 08/19/2011 01:50:48 PM:
[image removed]
[R] display only the top-right half of a correlation matrix?
Liviu Andronic
to:
r-help@r-project.org Help
08/19/2011 01:55 PM
Sent by:
r-help-boun...@r-project.org
Dear all
Is there an easy
On Fri, Aug 19, 2011 at 11:50 AM, Liviu Andronic landronim...@gmail.com wrote:
Dear all
Is there an easy way to display only one half (top-right or
bottom-left) of a correlation matrix?
require(Hmisc)
rcorr(as.matrix(mtcars[ , 1:4]))
mpg cyl disp hp
mpg 1.00 -0.85 -0.85 -0.78
I want to employ a parsimonious model to draw nomograms, as the full
model is too complex to draw nomograms readily (several interactions of
continuous variables). However, one interesting variable stays or
leaves based on whether I choose p value or AIC options to
fastbw(). My question
Thanks for the code corrections. I see how for loops, append and
naively populating a NULL vector can be so resource consuming. I tried
the codes with 20 million observations in the following machine:
processor : 7
cpu family : 6
model name : Intel(R) Core(TM) i7 CPU Q 720
On Fri, Aug 19, 2011 at 2:23 PM, danielepippo dan...@hotmail.it wrote:
Hi everybody.
I'm trying to use R with Sweave but I have a problem perhaps with the
directory path of sweave in R.
The windows path is this:
C:\Program Files (x86)\R\R-2.9.2\share\texmf\Sweave
When I run the latex file
Here is yet another way of prepending leading zeros on a string:
x - c('123/1234', '234/12', '21342342134/34', '99')
n - 10 # upto 10 leading zeros (max length of string)
leading - paste(rep('0', n), collapse = '')
# add up to 10 zeros and then truncate to max length, but at least keep
On Fri, Aug 19, 2011 at 9:02 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
Use as.dist: here's an example.
Seems promising, but for one issue: I would like to keep the diagonal
and thus specify 'diag=T', but then as.dist() replaces the diagonal
values with zero. (See below.) Is there a
On Fri, Aug 19, 2011 at 12:32 PM, Liviu Andronic landronim...@gmail.com wrote:
On Fri, Aug 19, 2011 at 9:02 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
Use as.dist: here's an example.
Seems promising, but for one issue: I would like to keep the diagonal
and thus specify 'diag=T',
On 19.08.2011 13:32, Christoph Scherber wrote:
Dear all,
I am forced to work in an environment without administrator rights.
When using R2.13.1 on Windows 7 (64-Bit), I found that I can´t install or
update any packages due to missing writing permissions.
I managed to get full access to a
On 19.08.2011 15:50, Paul Hiemstra wrote:
On 08/17/2011 10:53 PM, Alex Ruiz Euler wrote:
Dear R community,
I have a 2 million by 2 matrix that looks like this:
x-sample(1:15,200, replace=T)
y-sample(1:10*1000, 200, replace=T)
x y
[1,] 10 4000
[2,] 3 1000
[3,] 3
On 19.08.2011 21:15, Ista Zahn wrote:
On Fri, Aug 19, 2011 at 2:23 PM, danielepippodan...@hotmail.it wrote:
Hi everybody.
I'm trying to use R with Sweave but I have a problem perhaps with the
directory path of sweave in R.
The windows path is this:
C:\Program Files
Dear All,
Is there a simulator that can generate observations from a generalized poisson
distribution?
Thanks and regards,
Chee
[[alternative HTML version deleted]]
__
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On Fri, Aug 19, 2011 at 9:38 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
if as.dist doesn't work, use brute force:
x = matrix(rnorm(5*100), 100, 5)
mat = signif(cor(x), 2);
mat[lower.tri(mat)] =
data.frame(mat)
Yes, brute force works. This isn't quite how I wanted to do this,
On Fri, Aug 19, 2011 at 4:25 PM, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
On 19.08.2011 21:15, Ista Zahn wrote:
On Fri, Aug 19, 2011 at 2:23 PM, danielepippodan...@hotmail.it wrote:
Hi everybody.
I'm trying to use R with Sweave but I have a problem perhaps with the
directory
Dear R-users
I am slowly migrating my mex files (MATLAB - Fortran and C) to R. To get my
own functions available on R section I have decided to learn how to build a
R package. I choose a simple example with a few Fortran and R functions
(wrapper).
The fortran sources are located at src and the
On 19.08.2011 16:45, Benoit Bruneau wrote:
Dear R-Users,
I'm trying to setup a personal repository for a few packages I'm working on.
I am on R-Forge but I still need to have various versions of my package that
R-Forge does not build (for R 2.8.1 for example).
So I followed the instructions
On 19.08.2011 22:53, Eduardo Mendes wrote:
Dear R-users
I am slowly migrating my mex files (MATLAB - Fortran and C) to R. To get my
own functions available on R section I have decided to learn how to build a
R package. I choose a simple example with a few Fortran and R functions
(wrapper).
On Fri, Aug 19, 2011 at 2:55 PM, javier palacios xpfen...@gmail.com wrote:
Dear R-community,
I have tried to estimate an accelerated failure time(AFT) and proportional
hazard (PH) parametric survival model with time-independent and
time-dependent covariates. For that purpose, I have used the
I was unable to find an answer to my problem. I would like to label
the y axis of a plot with a rate and would like to use a dot (•)
rather than a multiplication sign (x).
ylab = quote(Speed~(cmxsec^2))
Thanks in advance.
keith
--
M. Keith Cox, Ph.D.
Alaska NOAA Fisheries, National Marine
Hi
I have modified the path to
dyn.load(paste(Sys.getenv(R_LIBS_USER),/fortran/src/fortran.so,sep=))
and the package could installed, loaded and the lines with dyn.load worked.
It does not look like a pretty solution but works on my linux (I am not sure
if it works on my mac or windows).
I am
Hi,
I would like to draw horizontal lines above a bar graph, in order to display
the p-values of a Fisher test. Here is an
examplehttp://thejns.org/action/showPopup?citid=citart1id=f3-1060501doi=10.3171%2Fped.2007.106.6.501of
the type of display I would like to have. Is there a way to draw the
Hello,
I'm having trouble figuring out how to calculate a p-value for a 1-tailed
test of beta_1 in a linear model fit using command lm. My model has only 1
continuous, predictor variable. I want to test the null hypothesis beta_1
is = 0. I can calculate the p-value for a 2-tailed test using
Thanks,
Javier
--
View this message in context:
http://r.789695.n4.nabble.com/COXPH-TIME-DEPENDENT-tp3754837p3756128.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
On Fri, Aug 19, 2011 at 04:43:35PM -0400, Chee Chen wrote:
Dear All,
Is there a simulator that can generate observations from a generalized
poisson distribution?
Thanks and regards,
Chee
You can use rzigp() in the ZIGP package. rzigp(n,mu,phi,omega=0) for
generalized poisson.
Michael
Dear R-help members. I am an 'R-learner' (about 6 hours so far) using the
lattice library to create a ranked dotplot and am colour coding the dots by
a variable called Commodity. However when i use autokey to make a legend
the size (cex) and symbol (pch) do not match what is on the dotplot.
With Dotplot, I'm trying to make a figure that will ultimately have the same
x-axis (which will be my response variable and the error bars), but the
y-axis will consist of a different label for every point. Here's my code:
Dotplot(fTaxonGrouped ~ Cbind(normSlope,normLwr,normUpr)|fGroup,
Hi,
Try this,
library(gridExtra)
example(grid.table)
or addtable2plot() in plotrix, or textplot() in gplots, or Hmisc using
latex, or Sweave, ...
HTH,
baptiste
PS: please read the posting guide
On 20 August 2011 05:14, Ed Heaton heat...@comcast.net wrote:
Hi, friends.
I keep coming to
It is most likely due to your ordering of y values. You need to write
key manually to reflect the change. Without providing reproduciable
data, you may not get specific help.
Weidong Gu
On Fri, Aug 19, 2011 at 6:23 PM, markm0705 markm0...@gmail.com wrote:
Dear R-help members. I am an
On Aug 19, 2011, at 6:36 PM, Marlin Keith Cox wrote:
I was unable to find an answer to my problem. I would like to label
the y axis of a plot with a rate and would like to use a dot (•)
rather than a multiplication sign (x).
ylab = quote(Speed~(cmxsec^2))
?plotmath # seemed like the
x - rep(00,2)
y - c(23/45,67/8)
substr(x,1+nchar(x)-nchar(y), nchar(x)) - y
x
---
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Hello All,
I have a question about glm in R. I would like to fit a model with glm
function, I have a vector y (size n) which is my response variable and I have
matrix X which is by size (n*f) where f is the number of features or columns. I
have about 80 features, and when I fit a model using
convert your matrix to a data frame:
df - as.data.frame(mymatrix)
then you can simplify your formula and specify where the data is coming from:
glm.fit - glm(y~., data=df)
the . in the formula means all columns in your dataframe (except
y, if it is in df)
On Sat, Aug 20, 2011 at 10:43 AM,
On Aug 19, 2011, at 5:07 PM, sw1 wrote:
With Dotplot,
Are you sure that you are using lattice? Maybe you out to look more
closely at:
?Dotplot
I'm trying to make a figure that will ultimately have the same
x-axis (which will be my response variable and the error bars), but
the
Sorry about the nabble problem. At any rate, do require(Hmisc) then ?label
to see how to associate a vector of labels with all the variables in a data
frame at once.
Frank
do999 wrote:
Indeed, as David pointed out, all the portion that used courier font (all
the good stuff) was absent from
I don't see anything wrong with using as.matrix. The documentation doesn't
say it will support a data frame.
Frank
Liviu Andronic wrote:
Dear all
?Hmisc::rcorr states that it takes as main argument a numeric
matrix. But is it normal that it fails in such an ugly way on a data
frame? (See
Replace Design with rms (for general reasons not related to your question;
Design is about to be obsolete).
Negate Dxy. The linear predictor for the Cox model is relative log
hazard. Higher hazard means shorter survival time. For other survival
models the model is stated in terms of
On Aug 19, 2011, at 8:43 PM, Andra Isan wrote:
Hello All,
I have a question about glm in R. I would like to fit a model with
glm function, I have a vector y (size n) which is my response
variable and I have matrix X which is by size (n*f) where f is the
number of features or columns. I
On 20/08/11 10:20, Andrew Campomizzi wrote:
Hello,
I'm having trouble figuring out how to calculate a p-value for a 1-tailed
test of beta_1 in a linear model fit using command lm. My model has only 1
continuous, predictor variable. I want to test the null hypothesis beta_1
is= 0. I can
Dear Prof. Broström,
I have searched in the reference manual inside the package eha, updated
recently. I did not find any description on how to enter id in the aftreg
function except the description of the argument. Can you refer to a specific
part of the manual? Do you mean another
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