re committing that change.
Paul
On 1/03/19 8:13 AM, Lee Steven Kelvin wrote:
Hello all,
When producing a plot in R using the cairo_pdf device, the resultant plot does
not respect the plotting boundaries. Lines and shaded regions will spill over
the lower x-axis and the right-side y-axis (sid
ting-boundaries
Thank you in advance for any insights into this issue.
Sincerely,
Lee Kelvin
--
Dr Lee Kelvin
Department of Physics
UC Davis
One Shields Avenue
Davis, CA 95616
USA
Ph: +1 (530) 752-1500
Fax: +1 (530) 752-4717
[[alternative HTML version deleted]]
LS
> >> gas1 <- plsr(octane ~ NIR, ncomp = 10, data = gasTrain, validation =
> "LOO")
> >> where octane ~ NIR is the model that this example is fitting with.
> >> NIR is a collective of variables, i.e. NIR spectra consists of 401
> diffuse
&
NIR[0,i]*NIR[1,i] + ...
i.e. quadratic with interaction terms.
But I don't know how to formulate this.
May I have some help please?
Thanks,
Kelvin
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i.e. quadratic with interaction terms.
But I don't know how to formulate this.
May I have some help please?
Thanks,
Kelvin
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http
Dear R-User,
I have written a simple code to analyze some data using Bayesian logistic
regression via the R2WinBUGS package. The code when run in WinBUGS stops
WinBUGS from running it and using the package returns no results also.
I attach herewith, the code and a sample of the dataset.
Any
On Saturday, August 1, 2015 3:32 AM, David Winsemius dwinsem...@comcast.net
wrote:
On Jul 31, 2015, at 6:36 PM, Christopher Kelvin via R-help wrote:
Dear All,
I am performing some simulations for a new model. I run about 10,000
iterations with a sample of 50 datasets and this returns one set
Dear All,
I am performing some simulations for a new model. I run about 10,000 iterations
with a sample of 50 datasets and this returns one set of 50 simulated data.
Now, what I need to obtain is 10 sets of the 50 simulated data out of the
10,000 iterations and not just only 1 set. The model
-
(p[1])*sum(s)*log((exp(-(p[2])*sum(x
return(-log1)
}
}
start - c(1,1)
zz-optim(start,fn=z,data=q,hessian=T)
zz
m1-zz$par[2]
p-zz$par[1]
Thank you
Chris Kelvin
INSPEM. UPM
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)
-(scale)*sum(x) + (shape)*log((exp(-(scale)*sum(x-
(shape)*sum(s)*log((exp(-(scale)*sum(x
return(-log1)
}
start - c(1,1)
zz-optim(start,fn=z,data=q,hessian=T)
zz
Thank you
Chris Kelvin
- Original Message -
From: Berend Hasselman b...@xs4all.nl
To: Christopher
Thank you very much for everything. Your suggestions were very helpful.
Chris
- Original Message -
From: Berend Hasselman b...@xs4all.nl
To: Christopher Kelvin chris_kelvin2...@yahoo.com
Cc: r-help@r-project.org r-help@r-project.org
Sent: Thursday, September 20, 2012 10:06 PM
Subject
Hello,
I want to estimate the exponential parameter by using optim with the following
input, where t contains 40% of the data and q contains 60% of the data within
an interval. In implementing the code command for optim i want it to contain
both the t and q data so i can obtain the correct
((
for(i in 1:length(t)){
for(j in 1:length(v-1))
{ ifelse ((t[i]v[j] t v[j+1] ),{z[i]-v[j];s[i]-v[j+1]},NA)}}
return(cbind(z,s))}
y(t,v)
Chris Kelvin
Institute for Mathematical Research
UPM
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Dear all,
I want to determine the standard error or the mean squared error for the
parameter estimate for beta and eta base on the real data.
Any help on how to obtain these estimated errors.
library(survival)
d - data.frame(ob=c(149971, 70808, 133518, 145658, 175701, 50960, 126606,
82329),
Hello,
When i generate data with the code below there appear NA as part of the
generated data, i prefer to have zero (0) instead of NA on my data.
Is there a command i can issue to replace the NA with zero (0) even if it is
after generating the data?
Thank you
library(survival)
Hello,
May i know whether it is possible to generate data twice from Weibull
distribution and use one as the start time and the
other as the end time, below is my code.
Any suggestion on how to estimate the parameters of Weibull distribution with
interval data will be highly appreciated.
Thank
Hello,
I have tried obtaining the value of standard error from the code below but i
get different values when i compare it with the
standard error obtained from the hessian matrix. Can somebody help me out?
Thank you
n=100;rr=1000
p1=1.2;b=1.5
sq11=sq21=0
for (i in 1:rr){
Hello,
I have tried obtaining the value of standard error from the code below but i
get different values when i compare it with the
standard error obtained from the hessian matrix. Can somebody help me out?
Thank you
n=100;rr=1000
p1=1.2;b=1.5
sq11=sq21=0
for (i in 1:rr){
Hello,
May i know whether it is possible to generate data twice from Weibull
distribution and use one as the start time and the
other as the end time, below is my code.
Any suggestion on how to estimate the parameters of Weibull distribution with
interval data will be highly appreciated.
Thank
Hello,
When i run the code below from Weibull distribution with 30% censoring by using
optim i get an error form R, which states that
Error in optim(start, fn = z, data = q, hessian = T) :
objective function in optim evaluates to length 25 not 1
can somebody help me remove this error. Is my
Hello,
I want to estimate weibull parameters with 30% censored data. I have below the
code for the censoring
but how it must be put into the likelihood equation to obtain the desire
estimate is where i have a problem with,
can some body help?
My likelihood equation is for a random type-I
Hello,
can i implement this as 10% censored data where t gives me failure and x
censored.
Thank you
p=2;b=120
n=50
set.seed(132);
r-sample(1:50,45)
t-rweibull(r,shape=p,scale=b)
t
set.seed(123);
cens - sample(1:50, 5)
x-runif(cens,shape=p,scale=b)
x
Chris Guure
Researcher,
Institute for
Hello,
I wish to censor 10% of my sample units of 50 from a Weibull distribution.
Below is the code for it.
I will need to know whether what i have done is correct and if not, can i have
any suggestion to improve it?
Thank you
p=2;b=120
n=50
r=45
t-rweibull(r,shape=p,scale=b)
Hello,
i need to simulate 100 times, n=40 ,
the distribution has 90% from X~N(0,1) + 10% from X~N(20,10)
Is my loop below correct?
Thank you
n=40
for(i in 1:100){
x-rnorm(40,0,1) # 90% of n
z-rnorm(40,20,10) # 10% of n
}
x+z
__
Hello!
I got something to ask..whether you can help me with the R program...i got this
for example 5x4 matrix..and i want to find:
i) mean for each row of the matrix
ii) median for each column of the matrix
and i need to do this using a loop function...below is my program..u try to
check it for
Dear All,
Can you help me, with the code below how do I obtain the fisher information
from it.
Is my q-replicate(1000,x) the right way to do simulation.
thank you.
x-rweibull(100,0.8,1.5)
q-replicate(1000,x)
z-function(p){
beta-p[1]
eta-p[2]
Hello,
i have used the code below to estimate the parameters of weibull distribution
and i want to obtain the fisher information
by providing the the next code but i receive errors anytime i try to, what do i
do?
by the way is my replication correct and is it placed at the right position for
Hello,
i have used the code below to estimate the parameters of weibull distribution
and i want to obtain the fisher information
by providing the the next code but i receive errors anytime i try to, what do i
do?
by the way is my replication correct and is it placed at the right position for
I need help,
the codes below estimates the weibull parameters with complete failure, my
question is how do i change the state to include
some censoring (may be right, type-I or type-II) to generate and estimate the
parameters.
thank you
x=rweibull(10,2,2)
library(survival)
I need help,
the codes below estimates the weibull parameters with complete failure, my
question is how do i change the state to include
some censoring (may be right, type-I or type-II) to generate and estimate the
parameters.
thank you
x=rweibull(10,2,2)
library(survival)
Hello,
If i write a function as below using log of weibull distribution i do not get
the required
results in estimating the parameters what do i do, please
a/b * (t/b)^a-1 * exp(-t/b)^a
n=500
x-rweibull(n,2,2)
z-function(p) {(-n*log(p[1])+n*log(p[2])-
Please, Help me,
How do I generate data from the weibull distribution if the data contain both
failure and interval censored,
For example, I want to generate n=100, shape=2 and scale =4 with 30% interval
censored.
What about right censoring
Thank you
[[alternative HTML version
please help;
I want to know how to generate an interval-censored data of about 20% and a
right censored data of about 30%
using the weibull distribution of say, x=rweibull(100,shape=1.2,scale=1.5)
[[alternative HTML version deleted]]
__
Can somebody help me,
How do I generate data from the weibull distribution if the data contain both
failure and interval censored,
For example, I want to generate n=100, shape=2 and scale =4 with 30% interval
censored.
Thank you
[[alternative HTML version deleted]]
R-help mailing list submissions
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide
above?
A B C D ...
0 4 2 0
2 2 0 1
0 1 4 1
2 2 0 0
...
Thanks for help!
Kelvin
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PLEASE do read the posting guide http
0.3 0.1 0.1
1 0.1 0.1 0.2 0.1
2 0.1 0.2 0.2 0.2
3 0.2 0.1 0.1 0
...
Thanks your help!
Kelvin
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this library before, and if so how?
Thankyou in advance,
Lee Kelvin
--
School of Physics Astronomy
University of St Andrews
North Haugh
St Andrews KY16 9SS
United Kingdom
Phone (+44) [0]1334461668
Email l...@st-andrews.ac.uk
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R-help@r-project.org
Thanks Uwe. The wordnet is working now. My previous configuration that
causes the error was:
R - version 2.9.1
wordnet - version 0.1-4
rJava 0.6-2
After updating rJava to version 0.7-0 everything works now.
Uwe Ligges-3 wrote:
Looks like nobody answered so far:
Kelvin Lam wrote
failed for 'wordnet'
My guess is I miss something from the namespace but unfortunately I can't
get to the internet. Does anyone know what I'm missing (from the
namespace?). Thank you very much!
Kelvin
--
View this message in context:
http://www.nabble.com/Problems-with-loading-%27wordnet%27
\)
[2] c(\aid\, \assistance\, \help\)
How can I get a one character string at the end that looks like this:
[1] aid assist assistance help aid assist help
Thanks for the help!
Kelvin
--
View this message in context:
http://www.nabble.com/extracting-text-from-wordnet-using
...@gmail.com
wrote:
Try this:
Str - c(c(\aid\, \assist\, \assistance\, \help\),
c(\aid\, \assistance\, \help\))
unlist(sapply(Str, function(x)dget(textConnection(x)), USE.NAMES =
FALSE))
On Sat, Aug 15, 2009 at 6:33 PM, Kelvin Lam lamk...@gmail.com wrote:
Dear group
Use odbcConnectExcel() in the RODBC package.
rajclinasia wrote:
Hi Every one,
I have a problem with Reading Excel file into R 2.9.0 version. In older
versions it is working with xlsReadWrite package. But in 2.9.0 version
there is no package like that. so help me out in this aspect.
Hi all,
I wonder how you can replace all words that need to be changed using
replacePatterns(). The following is my code. I want to replace both abc
and def to Yes . However, I can only replace the first occurrence in
sample[[1]].
sample[[1]]
[1] abc def ghi
change - c(abc,def)
R function. Ergo, possibility of confusion.
Bert Gunter
Genentech Nonclinical Biostatisics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Kelvin Lam
Sent: Tuesday, August 11, 2009 9:43 AM
To: r-help@r-project.org
Hi all,
I wonder if there's any way to reshuffle the text collection by the document
meta values. For instance, if I have 5 documents that correspond to the
following meta data:
MetaID Sex Age
0 M38
0 M46
0 F 24
0 F 49
0 F 33
Can I
I should elaborate the situation a bit more. We store our data in UNIX and
have been using UNIX SAS for our work. My Biostat dept has 40 SAS users
from which at most 10 also use R. The Epi/Grad Students/Investigators
combine for another 30-40 not-so-frequent SAS users let alone R. So we are
but so far haven't managed
to find it. I am extremely grateful if someone can point me how to go about it.
Thanks.
Kelvin
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PLEASE do read the posting guide http://www.R
To: kelvin lau kelvin...@yahoo.com
Cc: r-help@r-project.org
Date: Wednesday, July 15, 2009, 6:25 PM
one way is the following:
dat - read.table(textConnection(
0 0 1 0 0 1 0 1
0 0 0 0 0 0 0 0
1 0 0 1 1 0 1 0
0 0 1 1 0 0 0 0
1 1 0 0 0 0 1 1
0 0 0 0 0 0 0 0
0 1 0 1 1 0 1 0
1 1 1 1 1 1 1 1
1 1
. I
wonder if any of you can give me some tips on how to modify it or any
experience modelling other survival models besides Coxph. Thank you
very much!
Regards
Kelvin Lam, MSc.
Analyst, Programming Biostatistics
Institute for Clinical Evaluative Sciences (ICES)
G179 - 2075 Bayview Avenue
Hi group,
Can someone kindly inform which package deals with Bivariate Survival
Analysis. I did a search but nothing came up. Thanks!
Regards,
Kelvin
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