For some simple reason, I am unable to see the mistake:
ggplot(filter(nlsw88, !(is.na(union))), aes(y = wage, x = union, fill = union))
+ geom_boxplot() + facet_wrap(~idblack)
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Advanced topics:
https://cran.r-project.org/web/packages/ipfr/vignettes/common_ipf_problems.html
Thank you for your time,
Kyle Ward
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x}')
}
@
Then you can put your subsection logic into subsection.Rnw, just make sure to
remove all of the default latex header stuff, just put in the blocks that you
want to process:
This comes from a subsection.
<>=
cat(”I’m in a subsection”)
@
This worked for me
Kyle
___
- SpatialPoints(PID208 [2:3])
mcp(xy, percent=95)
mcp(xy, percent=95)$area #to find actual area covered
plot(mcp(xy,percent=100))
Anybody know what I might be missing to successfully export the file for use in
GIS?
Thanks!
Kyle
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sking him to perform a post-mortem examination: he may be able to say
> what the experiment died of. ~ Sir Ronald Aylmer Fisher
> The plural of anecdote is not data. ~ Roger Brinner
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonabl
I am using LakeAnalyzer in Rstudio to produce heat maps and plots using
data from constant monitoring buoys. I have a prewritten script that is
functioning on a colleagues computer perfectly. I am using Rstudio 0.99.902
and R 3.3.1. I have added four packages to the project (lattice,
manipulate, ma
ed using GitHub, and more information as well as
the current development version can be found at
https://github.com/kylehamilton/lavaan.shiny
William Kyle Hamilton - Graduate Student
University of California, Merced - Psychological Sciences
psychology.ucmerced.edu - k
Good afternoon,
I recently received a ticket from a customer to upgrade from 3.1.1. to 3.2.1.
After the upgrade, when he tries to install a package he receives the error
below. Could you please advise as to what is wrong? Thank you.
Kyle
--- Please select a CRAN mirror for use in this
to
allow for easy data entry.
The package is developed using GitHub, and more information as well as
the current development version can be found at
https://github.com/kylehamilton/MAVIS
William Kyle Hamilton - Graduate Student
University of California, Merced - Psychological Sci
Are there any libraries for R that would enable the import of data stored
in star schema?
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PLEASE do read the posting guide htt
-0.08372245 0.02811895
> vcovHC(lm(y~x, data=test), type=c('HC'))[1,1]^0.5
[1] 0.5885252
> vcovHC(lm(y~x, data=test), type=c('HC'))[2,2]^0.5
[1] 0.1676871
(The HC3 method gives SEs which are consistent with those from lm().)
I don't understand how survreg() can do
> Survreg treats weights as case weights, and lm treats them as sampling
> weights.
> Here is a simple example. Data set test2 has two copies of every obs in data
> set test.
>
> > test <- data.frame(x=1:6, y=c(1,3,2,4,6,5))
> > test2 <- test[c(1:6, 1:6),]
>
> > summary(lm( y ~ x, data=test))$co
alog~uvlog-1, weights=1/(haerrlog^2))
surv_basic_weight = survreg(Surv(time=halog,
event=eventcode_det)~uvlog-1, dist='gaussian', init=c(3.33*1.8/9.97),
weights=1/(haerrlog^2), robust=T)
summary(basic_weight)$coef[2]
summary(surv_basic_weight)$table[1,2]
# if I leave off robust=T, surv
Hrm, thanks. The uncertainties are what they are, though (and the
model is what it is, too) -- is there an alternative to modifying
them? Maybe another type of analysis that handles upper limits?
Kyle
On Thu, Jun 13, 2013 at 5:46 AM, Andrews, Chris wrote:
> It seems a line through the ori
ta[2] is very close to 3*err[2] already
survreg(Surv(time = data, event = c(1,2,1))~model-1+cluster(id),
weights=1/(err^2), dist='gaussian', init=c(2.1))
Thanks,
Kyle
On Wed, Jun 12, 2013 at 6:51 AM, Terry Therneau wrote:
> I will assume that you are talking about uncertainty in the
for how to accomplish this?
Thanks,
Kyle
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and provide commented, minimal, self-contained, re
Oh! Great idea. I'll give that a try.
---Kyle.
On Fri, Apr 27, 2012 at 12:00 PM, William Dunlap wrote:
> You can use strwrap and paste to insert newlines into your labels. E.g.,
>
> > wrapped <- function(strings, width) vapply(strings,
> function(s)paste(collapse
Hello!
Does anyone know of a handy way to wrap the names.arg text in a barplot?
I'm creating a bar plot with rather long labels; I can adjust the margins,
but I'd also like to have the text wrap to about 4cm. Thanks!
Kyle H. Ambert
Doctoral Candidate, Bioinformatics
Oregon Health
You could try the sink function! I use that from time to time:
?sink
Kyle H. Ambert
Fellow, National Library of Medicine
Department of Medical Informatics & Clinical Epidemiology
Oregon Health & Science University
ambe...@gmail.com
__
y way around this, some setting in devSVG that I can turn on that
will capture these things and make them syntactically correct for SVG or is
this a bug in devSVG (in that case I'd like to report the bug). Or is this
actually valid .svg in which case GIMP and Qt both have bugs.
Thanks
Kyle
NOV23 10 group2
Any help would be appreciated.
Thanks,
Kyle
The information of this email and in any file transmitted with it is strictly
confidential and may be legally privileged.
It is intended solely for the addressee. If you are not the intended recipient,
any copying
ip file right afterwards - so it
would seem as though I have the correct permissions. (A little Googling
also showed that some people have success when simply retrying to do the
install - no luck there, either.)
Any ideas on how to work around this? I really need that upda
ld be very helpful.
Technical Notes:
OS: win7 32bit
Compiler: mingw32
R: 2.11.1
Thanks
Kyle
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g C so this may be a very basic question. If anyone
could point me to a good reference on this it would be very helpful.
Technical Notes:
OS: win7 32bit
Compiler: mingw32
R: 2.11.1
Thanks
Kyle
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If it's specifically drawing the grid that's the trouble, I think the
simplest approach is to use the grid() function in the base graphics
package.
Kyle H. Ambert
Fellow, National Library of Medicine
Department of Medical Informatics & Clinical Epidemiology
Oregon Health & Scie
r blackboost methods, which do not return scalar
coefficients that are readily extractable.)
On Sun, Feb 7, 2010 at 6:31 PM, David Winsemius wrote:
>
> On Feb 7, 2010, at 5:03 PM, Kyle Werner wrote:
>
>> I'm running R 2.10.1 with mboost 2.0 in order to build predictive
>
I'm running R 2.10.1 with mboost 2.0 in order to build predictive
models . I am performing prediction on a binomial outcome, using a
linear function (glmboost). However, I am running into some confusion
regarding centering. (I am not aware of an mboost-specific mailing
list, so if the main R list i
lues in the vector assigned
to the "times" argument). It might be helpful to know a bit more about the
specific problem you're trying to solve though.
Kyle H. Ambert
Fellow, National Library of Medicine
Department of Medical Informatics & Clinical Epidemiology
Oregon Health
d use something like
points(c(1:10), col="red")
Or, if I wanted to generate a barplot using a shading color other than gray,
barplot(c(1:10), col="steelblue")
Does that answer your question?
Kyle H. Ambert
Fellow, National Library of Medicine
Department of Medical Informati
I am using the gbm package for generalized boosted regression models,
and would like to be able to extract the coefficients produced for
storage in a database.
I am already using R to automatically generate formulas that I can
export to a database and store. For example, I have been using Dr.
Harr
Dear David,
Thank you for the reference to Frank Harrell's excellent text. I will
read up to correct my statistical deficiencies offline.
Thank you.
On Sun, Oct 25, 2009 at 1:24 PM, David Winsemius wrote:
>
> On Oct 25, 2009, at 12:55 PM, Kyle Werner wrote:
>
>> David,
likelihood ratio) + 2k,
with the best model (assuming the same observations) having the lowest
AIC. I hope that my understanding of this fundamental formula is
correct, but please let me know if not.
Thanks.
On Sun, Oct 25, 2009 at 10:51 AM, David Winsemius
wrote:
>
> On Oct 25, 2009, a
I am trying to obtain the AICc after performing logistic regression
using the Design package. For simplicity, I'll talk about the AIC. I
tried building a model with lrm, and then calculating the AIC as
follows:
likelihood.ratio <-
unname(lrm(succeeded~var1+var2,data=scenario,x=T,y=T)$stats["Model
ype=\"fitted.ind\")
>From there, I could cobble together a dataframe of the actual results
in the new dataset with the predicted probabilities based on the
model, and regress from there. This allowed me to generate my
statistic of interest (the C-index).
Again, thank you,
Kyle
On Thu,
cients to the variables). It doesn't simply apply the new data
to the model from logit.lrm that I generated before.
So, can someone point me in the right direction for evaluating the
model that I built with trainingData, but getting the C-statistic
against my validationData?
Thanks so
different coefficients
to the variables). It doesn't simply apply the new data to the model from *
logit.lrm* that I generated before.
So, can someone point me in the right direction for evaluating the model
that I built with trainingData, but getting the C-statistic against my
validationData?
}, \Sexpr{covmat[3,1]})^T$
% end code
but, of course, this is poor way of going about it. Any suggestions?
Thanks in advance.
Regards,
Kyle
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, is there another approach to this
problem someone can suggest?
Thanks,
Kyle
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Thanks, Barry. I'll use that in the future.
---Kyle.
On Fri, Dec 5, 2008 at 11:01 AM, Barry Rowlingson <
[EMAIL PROTECTED]> wrote:
> 2008/12/5 Chris Poliquin <[EMAIL PROTECTED]>:
> > Hi,
> >
> > I have about 900 files that I need to run the same R sc
Have you thought about using one of the Python/R interface modules?
http://www.omegahat.org/RSPython/
http://rpy.sourceforge.net/
Admittedly, I have not had much success in getting these to work on my
machine, but I know others who have.
Kyle H. Ambert
Graduate Student
Department of Medical
ot;)
dat <- tis(runif(9*12 - 1), start=strt) # ends in 11/2008
par(mfrow=c(2,1))
tisPlot(dat, xTickFreq="monthly", xTickSkip=6) # graph ends in 11/2008,
but looks to start some time in late 1999
tisPlot(dat, xTickFreq="monthly", xTickSkip=12) # graph looks to end
around
TeX as it is and the problem has been solved to my
satisfaction I have chosen to put off exploration of this package for
another day.
Best,
Kyle
On Sun, Nov 30, 2008 at 8:42 AM, Vincent Goulet <
[EMAIL PROTECTED]> wrote:
> Kyle,
>
> In addition to "listings" alr
athbf{X}^T\mathbf{X})^{-1}) \times p \times F_{p,
n-p, .95}}
Thanks, and sorry for such a dumb question. Either I am not searching for
the right thing or this hasn't already been addressed in the lists (perhaps
because it is so easy).
Kyle
[[alternative HTML
be 'plot(runif(1000))', perhaps manually
enclosed in a verbatim environment
\end{document}
I am running R-2.6.2 on Ubuntu Hardy Heron.
Thanks for all your help with Sweave. I think it is a fantastic tool.
Kyle
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_
doing some tricky
stuff from within R such that when their code is preprocessed it is compiled
as LaTeX source, but this seems like an ugly hack for something so simple.
Please forgive me if this has been addressed elsewhere; I have found
documentation on Sweave to be rather sparse.
Bes
if(i==df$ID[j]){
tmp<-data.frame(df[j,])
}
}
}
tmp
}
but this seems inefficient. As I understand it, the subset function won't
really solve my problem, but it seems like there must be something out there
that will that I must be forgetting. Does anyone know of a way
if(i==df$ID[j]){
tmp<-data.frame(df[j,])
}
}
}
tmp
}
but this seems inefficient. As I understand it, the subset function won't
really solve my problem, but it seems like there must be something out there
that will that I must be forgetting. Does anyone know of a way
oblem, but my new job likes to include them in, say, MS word documents for
which I cannot easily use pdf.
TIA,
Kyle
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PLEASE do
Peter,
Ah, I was under the impression that I had to 'dev.off()' after every plot.
Naturally I had wondered about that strange argument that did not seem to do
anything...
Problem solved, thanks everyone.
Many thanks,
Kyle
On Mon, Sep 29, 2008 at 9:43 AM, Peter Dalgaard <[EM
R? I have only ever
previously used devices besides x11 by analogy with printing things to x11.
I am running R-2.7.2 on Red Hat Linux.
TIA,
Kyle
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this:
new<-model.matrix(lm.out)
new[,1]<-lm.out$model[,1]
This seems to do what I need.
*
Dr. J. Kyle Roberts
Department of Literacy, Language, and Learning
School of Education and Human Development
Southern Methodist University
P.O. Box 7
Dear All,
I want to extract the original dataset from a lm output. I know that I
can get most of it from
model.matrix(lm.out)
but I need the dependent variable to be in the first column. Any ideas?
Thanks,
Kyle Roberts
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Dear All,
How come par(mfrow=c(1,2)) works with boxplot, but not with bwplot?
This works
par(mfrow=c(1,2))
boxplot(dv~index, depend)
boxplot(dv~index, depend2)
This does not work
par(mfrow=c(1,2))
bwplot(index~dv, depend)
bwplot(index~dv, depend2)
Thanks,
Kyle
then using pseudorandom variables is the
incorrect way of going about it. Your suggestion of sample(c(rep
(0,545),rep(1,75))) seems to me to be the best way of going about it since
conceptually this is what you are doing: taking permutation of a fixed set
of numbers.
Best,
Kyle
_
urn(CAIC)
}
where K.corr is the finite-sample correction for K, for ML model fits.
I am posting this so that 1) This code can be of use to any other
souls in the statistical wilderness trying to do model selection with
mixed models, and 2) So that wiser minds can point out any errors in
Does anyone know how to graph a line according to a specified equation? I'd
like to plot the following hyperbola:
Y=139.35/(1+(0.174*X))
I know there's a way to do this, but I'm having a ridiculous time trying to
remember how.
Thanks!
Kyle H. Ambert
Department of Behavior
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