fset and predicting the baseline survival
fit1_shrunk <-
flexsurvspline(Surv(rfstime,rfs)~offset(heuristic_lp_flex),k=2,data=rotterdam)
The error message is
Error in model.matrix.default(forms[[i]], m) :
model frame and formula mismatch in model.matrix()
Any suggestions gratefully received!!!
Kin
1 vs Session 4 (for
example)? How is that handled in the model? As of now, time (days since
baseline) is being treated as one unit, rather than four separate sessions.
Here is an example of my code: mdl.outcome <- lmer(outcome ~ time*Group +
(1 | PID), data = dta)
Thank you!!
laura
[[alt
;Error
in tcmat %*% linfct : requires numeric/complex matrix/vector arguments"
I see many tutorials online about adjusting in pairwise comparisons and
ANOVA but not in this situation. Any guidance you can provide would be much
appreciated.
Thank you!
Laura
[[alternative HTML version d
3 knots)", "FP1"),values=c("green","red","blue"))+theme_bw()
plot_part9 + labs(x="Number of positive nodes",y="Prediction",color="") +
theme(legend.position=c(0.8,0.8))
I have 2 concerns over this:
1. The plot
I am a beginner to R and I need to map some Atlantic puffin migration routes
onto a map of the Northern Hemisphere. I have a latitude and longitude point
per bird, per day. I would like to be able to plot the routes of all my
birds on one map and ideally so that I can see at which date they are at
almost the predicted value of 0.9968 of above. However, I need to get
the linear predictors to agree with the output first!
Many thanks for your help with this.
Kind regards,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.or
Hello,
I am a newbie in R and I have a problem.
I want this line:
newMyData <-MyData[!(is.na(col1)) | !(is.na(col2)) | !(is.na(col3)) | !(
is.na(col4)) | !(is.na(col5)),]
write in for loop. The number of col[number] may be different from 3 to 8
Is there any ideas?
Thank you
La
so is anyone aware of how to
extract the BIC in R?
Also, it is easy to obtain predictions from the model using code such as:
predict(fit_3k,type="hazard")
However, is there a way to extract the linear predictors for each individual?
Many thanks for your help.
Kind regards,
Laur
atively obtained from mod1$linear.predictor]
Many thanks for your assistance.
Kind regards,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
s a problem
when "sdata" is only 1 row, and ideally provide a solution?
Many thanks,
Laura
Dr Laura Bonnett
NIHR Post-Doctoral Fellow
Department of Biostatistics,
Waterhouse Building, Block F,
1-5 Brownlow Street,
University of Liverpool,
Liverpool,
L
Prediction Models book by Steyerberg.
Many thanks again for your assistance.
Kind regards,
Laura
From: harre...@gmail.com [mailto:harre...@gmail.com] On Behalf Of Frank Harrell
Sent: 14 September 2017 17:22
To: David Winsemius
Cc: Bonnett, Laura ; r-help@r-project.org
Subject: Re: [R] Help
Intercept -13.8515 0.9694 -14.29 <0.0001
x[1]0.0989 0.0103 9.58 <0.0001
x[2]0.9030 0.1510 5.98 <0.0001
x[3]1.3576 0.2570 5.28 <0.0001
x[4] 0.6884 0.2034 3.38 0.0007
x[5]0.6327 0.2003 3.16 0.0016
I was therefore hoping someone would
etc.
Can anyone suggest what might be causing these warning messages, and what steps
I can take to prevent them reoccurring as they are causing predictions for
every patient subgroup to have the same summary statistics.
Kind regards,
Laura
[[alternative HTML version deleted]]
f 170 estimates nested
within 26 studies, the coefficient is positive and as we would expect.
Does anyone have any suggestions for what might be going on or how I might
diagnose the problem with this model?
Thanks,
Laura
Laura Duncan, M.A.
Research Coordinator
Offord Centre for Child Studies
McMa
ndom effects
for each level specified correctly, does the rma.mv command automatically
adjust the weights in this way? Another way to ask this question is - what are
the default weights used by rma.mv when there are multiple estimates within
studies?
Thanks
Laura
Laura Duncan, M.A.
Rese
Dear all,
I am using an unbalanced panel dataframe to run my analysis. The df is
based on a national survey and I know the reason why the df is unbalanced.
I tried to understand if plm package corrects for unbalanced panel, so that
I can use the plm package without problem.
I create a plm.data u
the glm() function.
>
> Jean
>
> On Tue, Sep 20, 2016 at 5:04 AM, laura roncaglia <
> roncaglia.la...@gmail.com> wrote:
>
>> I am a beginner user of R. I am using a national survey to test what
>> variables influence the partecipation in complementary pen
I am a beginner user of R. I am using a national survey to test what
variables influence the partecipation in complementary pensions (the
partecipation in complementary pension is voluntary in my country).
Since the dependent variable is a dummy (1 if the person partecipate and 0
otherwise) I want
I am a beginner user of R.
I am writing the master thesis using a data frame from a national survey.
The data frame contains several variables, one of which contains the survey
weights.
I need to apply the survey weights to the data frame, in order to use the
data frame with the plm package (I ne
---The las post rejected two files I had attached, so I modified
it.---
Hi. I am trying to make a nls fit for a little bit complicated expression that
includes two integrals with two of the fit parameters in their upper limits.
I got the error "Error in nlsModel(formula, mf, star
Hi. I am trying to make a nls fit for a little bit complicated expression that
includes two integrals (please find enclosed the equations).
I got the error "Error in nlsModel(formula, mf, start, wts) :
singular gradient
matrix at initial parameter estimates". First of all, I have searched
already
j,input,method='mpl',start=2),silent=TRUE);
resmatpar[j,5] <- summary@estimate;
resmatllk[j,5] <- summary@loglik;
d <-
c(resmatllk[j,1],resmatllk[j,2],resmatllk[j,3],resmatllk[j,4],resmatllk[j,5]);
copchoice[j] <- which(d==max(d));
param[j] <- resmatpar
be straightforward but has proven the opposite
Many thanks
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project
Hola,
Estoy todavía dando mis primeros pasos en R y una de las cosas que tengo que
hacer es trabajar con un csv de 16 GB. Consta de 10 columnas, 7 númericas
He probado varias cosas entre ellas 'colbycol' pero nada, mi ordenador de
queda frito.
Mi ordenador tiene 8GB de RAM y Windows 8
¿Debo tr
Dear all,
I am running a gls and I would like to check the vif of my model. It seems that
the vif function in the car package and the vif.mer function available online
do not work for gls. Would you know of a method to measure variance inflation
factors for GLS?
Thank you
Laura
Thanks! That worked. However, what error distribution is assumed? I was hoping
to use a negative binomial.
-Original Message-
From: Milan Bouchet-Valat [mailto:nalimi...@club.fr]
Sent: Friday, April 11, 2014 11:37 AM
To: Lee, Laura
Cc: r-help@r-project.org
Subject: Re: [R] weights error
gn(id=~1, strata=~STRATA, weights = ~weight, data=data)
SNB3 <- svyglm(NB3, design=dstrat)
The error that is returned is:
Error in model.frame.glm(formula = list(coefficients = c(-6.96858807641624, :
object '.survey.prob.weights' not found
I would appreciate any assistance in solving
0)”
Any help with this would be much appreciated.
Many thanks,
Laura
structure(list(Participant = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L
problem?
Many thanks,
Laura
structure(list(condition = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 2L
)), .Names = c("subject", "conditionNo",
> "state", "latency"), row.names = 3:8, class = "data.frame")
>
> Thanks again,
>
> Laura
__
R-help@r-project.org mailing list
https://stat.ethz.ch/m
Hi everyone,
I have a large dataset with missing values. I tried using glmnet, but it seems
that it cannot handle NA values in the design matrix. I also tried lars, but I
get an error too. Does anyone know of any package for computing the lasso
solution which handles NA values?
I have the same problem with my macbook air (OS X 10.9)
--
View this message in context:
http://r.789695.n4.nabble.com/RCmdr-issues-tp4680537p4680804.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
h
o
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Laura Cisneros
> Sent: Montag, 15. Juli 2013 03:09
> To: r-help@r-project.org
> Subject: [R] Replacing values of a matrix with values from corresponding
>
y replace 12 and 1 (from the randomized matrix) with 4
and 1 (from the original matrix). Then do this for each row.
Any suggestions on how I may be able to accomplish this would be greatly
appreciated.
Laura
--
Laura Cisneros, Doctoral Candidate
Dept. of Ecology and Evolutionary Biology
U
is a bit
about my R abilities at the moment, I am quite happy to run the code a few
times to get the information that I would need.
Thanks a lot for any help,
Laura
structure(list(subject = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L
for any help,
Laura
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
mp1"];
"SMP" -> "item4" [label="smp2"];
"SMP" -> "item6" [label="smp3"];
"SMP" -> "item8" [label="smp4"];
"SMP" -> "item10" [label="smp5"];
"SMP"
g on other columns - whereas I would want to impute data between
zero and the censored value.
Any guidance/advice would be very much appreciated.
Laura
Dr Laura MacCalman Msci MSc PhD Gradstat
Senior Statistician
Institute of Occupational Medicine
Research Avenue North
Riccarton
Edinburgh
EH1
Thanks,
Laura
Laura Thomas
PhD Student- Sport and Exercise Psychology
Department of Sport and Exercise
Penglais Campus
Aberystywth University
Aberystwyth
01970621947
l...@aber.ac.uk
www.aber.ac.uk/en/sport-exercise/
__
R-help@r-project.org mailing
appreciated.
Many Thanks,
Laura
Laura Thomas
PhD Student- Sport and Exercise Psychology
Department of Sport and Exercise
Penglais Campus
Aberystywth University
Aberystwyth
01970621947
l...@aber.ac.uk
www.aber.ac.uk/en/sport-exercise/
__
R-help@r
Where can I mark the option to not receive emails from users?
--
Laura de Mergelina
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
Hi Gavin,
Many thanks for your help and interest.
Regards,
Laura
> Subject: RE: [R] Error in plot.envfit(ef, p.max = 0.1) :(subscript)
> logical subscript too long
> From: gavin.simp...@ucl.ac.uk
> To: lauram...@hotmail.com
> CC: r-help@r-project.org
> Date: Mon, 31 Dec 2
this? I can only see things involving plot.predict.crr which
doesn't seem to be what I need?
Many thanks,
Laura
Usung Windows 7 and R 2.14.1
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mai
a.rm=TRUE)efplot(species,
dis="sites")plot(ef,p.max=0.1)
Error in plot.envfit(ef, p.max = 0.1) : (subscript) logical subscript too long
Many thanks!
Laura
[[alternative HTML version deleted]]
__
R-hel
\dev\R-2.15.2\bin\i386
Also when I type where gcc I get:
E:\dev\Rtools\gcc-4.6.3\bin\gcc.exe
C:\strawberry\c\bin\gcc.exe
I cannot understand what I'm doing wrong. I'm sorry if this question has
been asked before but I couldn't find anything useful.
I would appreciate any help. Tha
-05
[[2]]$upper
[1] 0.0002945504
[[2]]$se
[1] 5.296472e-05
However, the predicted value is now 1.7x10^-4. The standard error value makes
sense for this predicted value, but I'm not sure why the predicted value has
changed. I would appreciate any assistance.
Cheers,
know
why it is happening. Does anyone of you know what can be happening?
Thanks in advance,
Laura Echeverri
--
View this message in context:
http://r.789695.n4.nabble.com/Simulating-a-VEC-tp4650463.html
Sent from the R help mailing list archive at Nabbl
ient results
so I'd rather take advantage of the full data.
Many thanks for your advice on this.
Kind regards,
Laura Schwirz
On 17 November 2012 09:56, yrosseel wrote:
> Dear Laura,
>
> John is correct. The error is produced by the sem() function in the lavaan
> package. Th
etcor, N=1861, meanstructure=F,fixed.x=F)
Warning message:
In if (orthogonal) { :
the condition has length > 1 and only the first element will be used
> sem.cdu= sem(cdu, hetcor, meanstructure=F,fixed.x=F)
Warning message:
In if (orthogonal) { :
the condition has length > 1 and only the first
!
--
Laura Catalina Echeverri Guzmán
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide
tes/vars.pdf>
The example is on pages 24-25
Best,
Laura
--
View this message in context:
http://r.789695.n4.nabble.com/Interper-output-from-cajorls-and-VECM-tp4639963p4647314.html
Sent from the R help mailing list archive at Nabble.com.
__
R-
lp me with this issue.
Thank you!
--
Laura Catalina Echeverri Guzmán
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-projec
e sum as in:
Prediction<-exp(predict(ZIP,newdata=effort,type="response"))
sum(Prediction)
Thanks again in advance!
Cheers,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/ma
total number of turtles predicted for the
new effort data. I did also log-transform the effort in the new effort data
before applying predict (correct?). I will try the manual method as you
recommended to help me understand the code.
Thanks,
Laura
From: Alain Zuur [via R] [mailto:ml-node+s78969
led effort), would
I use sum(predict(ZIP,newdata=effort,type="response"))? I want to be certain
I am understanding the coding--this is my first time using the predict
function.
Thanks,
Laura
-
Laura M. Lee
Senior Stock Assessment Scientist
North Carolina Division of Marine Fis
Alain-
Thanks again for the response. I guess my question is more related to R, which
I'm learning as I go along. Could you provide guidance as to how I would code
this in R?
Thanks,
Laura
From: Alain Zuur [via R] [mailto:ml-node+s789695n4635920...@n4.nabble.com]
Sent: Monday, July 09, 2
Thanks for your reply. I do have a copy of "Zero Inflated Models and
Generalized Linear Mixed Models with R" and have been using that as a guide.
I applied the predict function (type="count") to the dataset for which I
built the model to compare the predicted bycatch numbers to the observed to
ensu
advance!
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal
message:
In sqrt(diag(solve(object$infomat))) : NaNs produced
Can anyone explain why this is the case?
Very many thanks,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-he
Thank you!
On Tue, Mar 20, 2012 at 8:39 PM, David Winsemius wrote:
>
> On Mar 20, 2012, at 7:52 PM, Laura Rodriguez Murillo wrote:
>
> Dear list,
>> I have a data frame where one of the columns are p values with scientific
>> notation mixed with regular numbers with
Dear list,
I have a data frame where one of the columns are p values with scientific
notation mixed with regular numbers with decimals.
>a=data frame
>a
P OR N
0.50 0.7500 237
0.047 1.1030 237
0.124 0.7742 237
0.124 0.7742 237
0.0080 1.1590 237
0.50 0.7500 237
4.5e-07 1.2 237
5.6e-04 0.
Hi Jim,
thanks for your reply, I will try using this instead!
best,
Laura
On Feb 24, 2012, at 3:09 AM, Jim Lemon wrote:
> On 02/24/2012 09:30 AM, Laura Matrajt wrote:
>>
>> Hi,
>> I found some weird behavior of the function multhist in the plotrix package
>>
Hi,
I found some weird behavior of the function multhist in the plotrix package
and I would like to ask you about it:
so, it works well to put two histograms next to each other, but I don't
understand the breaks
so for example,
library(plotrix)
mh <- list(rnorm(200, mean=100, sd=10), rnorm(20
not appropriate.
Are you able to provide any suggestions as to how I might achieve the graph of
actual vs. Predicted survival times?
Many thanks,
Laura
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: 07 February 2012 16:59
To: Bonnett, Laura
Cc: 'r-h
$tsecond,pch=4,col="red")
This leads to the attached plot. It doesn't seem correct though as the
predicted times all start over 500 days.
Any suggestions would be very welcome.
Many thanks,
Laura
Actual vs. Survival LJB.pdf
Description: Actual vs. Survival LJB.pdf
__
Hello. I am using the zeroinfl package to fit a zero-inflated negative
binomial. The explanatory variables are Year and Depth x STemp (interaction). I
am in need of guidance for extracting the year effect and the associated
precision.
Thank you for your time.
Cheers,
Laura
ependently of calling the output?
Many thanks,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guid
Thank you!
-Original Message-
From: Milan Bouchet-Valat [mailto:nalimi...@club.fr]
Sent: 04 January 2012 09:23
To: Bonnett, Laura
Cc: 'r-help@r-project.org'
Subject: Re: [R] R 2.14.0 Design library
Le mercredi 04 janvier 2012 à 09:04 +, Bonnett, Laura a écrit :
> Dea
Dear all,
I have recently upgraded to R 2.14.0 with Windows 7. I wish to use the command
'cph' but the Design library is no longer on the list of installable packages.
How can I install Design so that I may use the 'cph' function?
Many thanks,
Laura
[[alte
7;qnorm' may
be an option but I'm lacking the necessary parameters within the framework
above.
Many thanks,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
P
Thanks for these various tips.
Sarah, this is not a howework, but a simplified dataset speecificly for
this question.
Laura
.
2011/11/14 Dennis Murphy
> Groupwise data summarization is a very common task, and it is worth
> learning the various ways to do it in R. Josh showed you one
p me with this?
Thanks
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented
while(data[i,2]<5)
data[i,]<-rmnorm(n=1,mean=c(mean(sanad[,1]),mean(sanad[,2]),mean(sanad[,3])),varcov=covmat)
while(data[i,2]>86)
data[i,]<-rmnorm(n=1,mean=c(mean(sanad[,1]),mean(sanad[,2]),mean(sanad[,3])),varcov=covmat)
}
r
Dear all,
I am a little bit confused because of the returned p-value by summary.lm and
also summary.rq
I thought if the pvalue is <= 0.05 the difference is significant. But the R
help says it is a two-sided pvalue. So does that mean the pvalue has to be
<= 0.025 and >= 0.975?
Be
Hello,
I used the TTR package in R to calculate moving averages. I have a monthly
time series and I would like to calculate the moving average over 10 years
with an offset of 1 year.
It should be something like sma.365 <- SMA(data, n=120)
Does anyone know how to include in offset?
Thanks a
replacement length"
Given I'm using the code attached can anyone offer a solution?
Many thanks,
Laura
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bonnett, Laura
Sent: 16 September 2011 08:22
To: 'Ken'
Cc: r-help@r
model
cox1.
Thanks,
Laura
-Original Message-
From: Ken [mailto:vicvoncas...@gmail.com]
Sent: 15 September 2011 21:43
To: Bonnett, Laura
Cc: Steve Lianoglou; r-help@r-project.org
Subject: Re: [R] Where to put tryCatch or similar in a very big for loop
What type of singularity exactly, if
Hi Steve,
Thanks for your response. The slight issue is that I need to use a different
starting seed for each simulation. If I use 'lapply' then I end up using the
same seed each time. (By contrast, I need to be able to specify which starting
seed I am using).
Thanks,
Laura
---
ried putting try around the whole for loop
without any success.
Many thanks for any suggestions you can offer.
Laura
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.o
gendat2). I was considering using the
'tost' function within the equivalence package but that doesn't seem
appropriate either.
Many thanks,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://
do not match those of SAS.
Additionally, can you use survfit with crr? If not, any suggestions as to how
I may obtain the necessary estimates in a competing risks setting?
Thank you for any help you can give.
Laura
[[alternative HTML version d
","10","LTG"))
# 1 year results
estimate1y <- c(NA,dattabrem1$HR)
lowerd1y <- c(NA,dattabrem1$CIlower)
upperd1y <- c(NA,dattabrem1$CIupper)
# 2 year results
estimate2y <- c(NA,dattabrem2$HR)
lowerd2y <- c(NA,dattabrem2$CIlower)
upperd2y <- c(NA,dattabrem2
Hello
I have a problem with the following function
(http://www.oga-lab.net/RGM2/func.php?rd_id=ismev:gev.fit):
gev.fit(xdat, ydat = NULL, mul = NULL, sigl = NULL, shl = NULL,
mulink = identity, siglink = identity, shlink = identity,
muinit = NULL, siginit = NULL, shinit = NULL,
show
ling
these without Dates.
R News 4/1 doesnt provide solution to this neither.
Have read and struggled with this stuff for 3 days.
Anyone else who could help on this?
Regards,Laura.
2011/7/12 Gabor Grothendieck
> On Tue, Jul 12, 2011 at 6:58 AM, B Laura wrote:
> > Hello all,
> >
CEST" "2000-04-29 CEST" "2000-05-06 CEST" "2004-10-27 CEST"
> start <- format(as.Date(start,"%Y-%m-%d"),"%Y-%m-%d")
> start
[1] "2002-09-03" "2000-07-26" "2003-01-03" "2001-06-28" "
nes, I
would be happy to remove them if possible.
I tried already the trunc() function but without succes. The result doesn't
change.
> Dataset$days2End2.in.R.TRUNC <- with(Dataset, trunc(End2)- trunc(Start))
> Dataset$days2End2.in.R.TRUNC
Time differences in days
[1] 730.0
trunc(End2)- trunc(Start))
> Dataset$days2End2.in.R.TRUNC
Time differences in days
[1] 730. 699. 671. 640.0417 610.0417 579.0417 549.0417 518.0417
487.0417 457.0417 426. 396.
attr(,"tzone")
[1] ""
I would be happy if someone could light up this thing.
Many tha
issue.
> as.numeric(as.Date("2000-1-1")-as.Date(subDataset$transplant.date))
[1] 1 1 0 0 0 0 -2 -2 -2 -2 -2 -3 -3 -3 -3
Truncation doesn't help either
>
trunc(as.Date("2000-1-1"),"days")-trunc(as.Date(subData
more concordant than
another over two survival functions hence I fitted two rcorrcens functions.
E.g. if I have a development data set with a variable for age and a
validation data set for age then I want to know if the concordance is the
same over the development and validation data sets.
Thank you,
Laur
.481 -0.038 0.038 0.011 3.44 6e-04 1500
Many thanks,
Laura
P.S. I'm using Windows XP, R 2.9.2
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posti
Dear R users,
I am in trouble using the function
solve(matrix)
in a loop.
When it happens that I have a singular matrix, I get an error message saying
so and the loop stops.
My question is if there in any way to keep on going in the loop.
Thank you in advance,
Laura
' is missing
>
I don't care what particular value it is, I just want to know if they are
all equal.
What am I doing wrong, please?
Thanks,
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.
Hi!
I was wondering if PERT or CPM was implemented in R.
I looked in the search engines but didn't find anything.
Since there are so many packages, I thought I'd double check via the
discussion.
Thanks,
Laura.
[[alternative HTML versi
Hello:
Here are some basic class items:
> x <- 1:10
> class(x)
[1] "integer"
> x.mat <- matrix(1:6,nrow=2)
> class(x.mat)
[1] "matrix"
> class(x>3)
[1] "logical"
> test <- function() { plot(1:10) }
> class(test)
[1] "function"
>
Is there something that says "vector", please? Or does it go to nu
Hi
I'm trying to learn about S4 methods, classes, etc.
Is it better to use initialize or use a construction function, please?
Thank you.
Laura Smith
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hi,
I have a table in R with data I needed and need to create a contingency table
out of it. The table I have so far looks like this:
Binger
r
DietType No Yes
Dangerous 15 12
Healthy52 9
None 134 24
Unhealthy 72 23
These are the error message
e and as a bargraph as I have not been
able to do it myself. If somebody could help me with doing this and send me an
overview in a document on how it should look, I would really appreciate it.
Thank you.
Laura
Diet: Binger-yes: Binger-No: Total:
None
Hi,
I'm having difficulty with getting a table to show with
multiple rows and columns. Below is the commands that I've typed in and
errors that I am getting. Thank you.
Laura
Table trying to enter:
Diet: Binger-yes: Binger-No: Total:
N
Hi,
I apologize for sending this again but realized that the data I had originally
sent in my first email did not show correctly and wanted to send it again to
give a better idea of how I am trying to show my data within R. Thanks!
Laura
Diet: Binger-yes
y
23
72
95
Dangerous
12
15
27
Laura
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.o
1 - 100 of 200 matches
Mail list logo