info.
Thanks,
Tim
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and provide commented, minimal, self-contained
s",
"Missed exercise"),
values = c(20, 4)
)
Note that this method then gets very close without the scale_shape_manual, too.
Hope that helps.
Tim
> Date: Mon, 30 Oct 2023 20:55:17 +
> From: Kevin Zembower
> To:
://bugs.r-project.org/show_bug.cgi?id=18379
If you are in this situation, the fix is to add a new User Environment Variable:
R_LIBCURL_SSL_REVOKE_BEST_EFFORT and set it to TRUE
Tim
>>>>>>
Date: Thu, 11 May 2023 09:39:25 +0300
From: Ivan Krylov
To: =?UTF-8?B?U2ViYXN0acOhbg==
If you only want the character strings, this seems a little simpler:
> strsplit("a bc,def, adef ,,gh", "[ ,]+", perl=T)
[[1]]
[1] "a""bc" "def" "adef" "gh"
If you need delime
is in R-4.2.1 Patched (I don't know if it has made it out to the full
distribution) and works in my 'corporate' environment. Perhaps it also applies
to your environment.
Tim
Date: Wed, 5 Oct 2022 10:34:02 +
From: PIKAL Petr
To: Ivan Krylov
Cc: r-help mailing list
Subject: Re: [R] R
Manipulating formulas within different models I notice the following:
m1 <- lm(formula = hp ~ cyl, data = mtcars)
m2 <- update(m1, formula. = hp ~ cyl)
all.equal(m1, m2)
#> [1] TRUE
identical(m1, m2)
#> [1] FALSE
waldo::compare(m1, m2)
#> `old$call[[2]]` is a call
#> `new$call[[2]]` is an S3
hs do not. For example, the content for "C:\Test\test.txt" can be
"Testing 123 测试". file.show("C:\\Test\\test.txt") would open this file and
display its content correctly without specifying the encoding parameter.
Thanks,
Tim
[[alternative HTML version deleted]
sed vector-based approach has been largely preserved but is no
longer recommended.
Examples for the usage of the new interface, the documentation of all
functions, as well as further changes can be found at
https://tim-tu.github.io/weibulltools/
If you notice a bug or have suggesti
Cheers Denes,
That's useful to know. I'll stick with the match.call version
instead. Interestingly I initially tried to post to R-devel (as I
thought it may involve something internal) but was asked to post here
instead.
Best
Tim
On Tue, 6 Oct 2020 at 08:22, Dénes Tóth wrote:
>
>
.
dots <- function (...) {
exprs <- substitute(list(...))
as.list(exprs[-1])
}
In the original, dots <- function(...) as.list(substitute(...())),
Does ...() get parsed in a special way?
Tim
On Tue, 6 Oct 2020 at 05:30, Bert Gunter wrote:
>
> You need to understand
Could someone explain what is happening with the ...() of the
following function:
dots <- function(...) as.list(substitute(...()))
I understand what I'm getting as a result but not why. ?dots and
?substitute leave me none the wiser.
regards
Hi All,
I am currently working on a project examining the health effects of physical
activity using a compositional data (CoDa) approach. I am using a
linear regression to measure the effect of physical activity. What I want to
do is predict an outcome, e.g. body mass index, based on the mean
You also should be able to reference locations from the 'root' of your project
using the here() package.
https://here.r-lib.org/
Best,
Tim Howard
> Date: Thu, 2 Apr 2020 10:21:47 +0100
> From: Rui Barradas
> To: Ivan Calandra
> Cc: "r-help@r-project.org"
>
In
y <- substr(x, i, 1)
your third integer needs to be the location not the number of digits, so change
it to
y <- substr(x, i, i)
and you should get what you want.
Cheers,
Tim
> Date: Sun, 21 Jan 2018 10:50:31 -0500
> From: Ek Esawi <esaw...@gmail.com>
&g
ct fit, but a p-value of 0.3112 doesn’t seem reasonable.
As a side note, the various r^2 values seem odd too.
Tim Glover
Senior Scientist II (Geochemistry, Statistics), Americas - Environment &
Infrastructure, Amec Foster Wheeler
271 Mill Road, Chelmsford, Massachusetts, USA 01824-4105
T +01
Interesting, thanks for that. I came accross qcc but my quick scan of
the docs is that it only did xbars but maybe I need to re-read the
docs, I guess it does the individual plot versions (I-MR) too.
Tim
On 8 July 2017 at 20:53, Rui Barradas <ruipbarra...@sapo.pt> wrote:
> Hello,
&
Hi,
I've had a quick look through the package list, and unless I've missed
something, I can't seem to find anything that will do I-MR / Xbar-R /
Xbar-S control charts ?
Assuming there is something out there, can anyone point me in the
right direction ?
Thanks !
TIm
rpg is a package for working with postgresql: https://github.com/thk686/rpg
odeintr is a package for integrating differential equations:
https://github.com/thk686/odeintr
Cheers,
THK
http://www.keittlab.org/
[[alternative HTML version deleted]]
See error messages below. Thanks -- Tim
- When attempting to install, the print-out I get in the Console in RStudio
is any length of: > install.packages("swirl")
% Total% Received % Xferd Average Speed TimeTime Time
Current
Dload
;, "0.401", "0.358", "0.317", "0.39", "0", "0.371", "O", "0.479", "0.399", "0.374", "0.348", "0.354", "0.365", "0.371", "0"), .Dim = c(9L, 9L), .Dim
leshooting or finding out who the culprit is?
Thanks.
--
Tim Richter-Heitmann (M.Sc.)
PhD Candidate
International Max-Planck Research School for Marine Microbiology
University of Bremen
Microbial Ecophysiology Group (AG Friedrich)
FB02 - Biologie/Chemie
Leobener Straße (NW2 A2130)
D-28359 Bremen
Tel.:
sign NA
so i can proceed with
ggplot(final, aes(value.x,value.y, col=group)) + geom_point()
So, in the example, the pairs A1-A1, A1-A2, A2-A1, A2-A2 should be
identified as "both cats", hence should get the factor "cat".
Thank you very much!
Tim
Hi ,
I work for the NHS, and our IT service has been unable to download as its
anti-virus software says it contains an exploit.
Is this normal? Is there a way around this?
Kind regards,
Tim Kingston
Sent from my HTC
[[alternative HTML version deleted
correlation between two time series over nested time
periods?
Date: Thursday, May 14, 2015, 6:14 AM
On
2015-05-14 , at 02:11, Tim via R-help r-help@r-project.org
wrote:
Hello Tim,
Re:
I have two time series
Calculate and plot cross correlation
between two time series over
maximum lag), and plot A and B in a single plot?
Thanks and regards,
Tim
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!
Thanks, On 20.02.2015 20:36, David Winsemius wrote:
On Feb 20, 2015, at 9:33 AM, Tim Richter-Heitmann wrote:
Dear List,
Consider this example
df - data.frame(matrix(rnorm(9*9), ncol=9))
names(df) - c(c_1, d_1, e_1, a_p, b_p, c_p, 1_o1, 2_o1, 3_o1)
row.names(df) - names(df)
indx - gsub
rows x 3 columns). I could do
as.data.frame(t(df_x),
but maybe that is not elegant.
What would be the solution for splitting the dataframe by rows?
Thank you very much!
--
Tim Richter-Heitmann
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NA6
What am I missing?
Thanks,
Tim
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appreciated!
Tim
[[alternative HTML version deleted]]
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General linear constraints don't seem to work. I get an error message if I
have more constraint equations than variables. E.g. executing the following
code
print(R.version)
library('tmvtnorm')
cat('tmvtnorm version ')
print(packageVersion('tmvtnorm'))
## Let's constrain our sample to the dwarfed
]
})
unique_values - lapply(sub.matrices, function(x) x[upper.tri(x)])
names(unique_values) - unique(indx)
This code needs to be expanded to form sub.matrices for any combination
of unique indices in temp.
Thank you so much!
--
Tim Richter-Heitmann (M.Sc.)
PhD
Another source of large datasets is the Public Data Sets on AWS
http://aws.amazon.com/public-data-sets/
Tim Hoolihan
@thoolihan
http://linkedin.com/in/timhoolihan
On Oct 28, 2014, at 7:00 AM, r-help-requ...@r-project.org wrote:
--
Message: 2
Date: Mon, 27 Oct
How can I create an improved version of a method in R, and have it be used?
Short version:
I think plot.histogram has a bug, and I'd like to try a version with a fix.
But when I call hist(), my fixed version doesn't get used.
Long version:
hist() calls plot() which calls plot.histogram() which
, as the data sets are
pretty large.
--
Tim Richter-Heitmann (M.Sc.)
PhD Candidate
International Max-Planck Research School for Marine Microbiology
University of Bremen
Microbial Ecophysiology Group (AG Friedrich)
FB02 - Biologie/Chemie
Leobener Stra�e (NW2 A2130)
D-28359 Bremen
Tel.: 0049(0
:10), graphics)
hist(1:10)
Haley
On Thu, Oct 9, 2014 at 1:14 AM, Tim Hesterberg timhesterb...@gmail.com
wrote:
How can I create an improved version of a method in R, and have it be
used?
Short version:
I think plot.histogram has a bug, and I'd like to try a version with a
fix
this sequential ordering to key.xtickfun? May i
ask for an example code?
Thank you very much!
--
Tim Richter-Heitmann (M.Sc.)
PhD Candidate
International Max-Planck Research School for Marine Microbiology
University of Bremen
Microbial Ecophysiology Group (AG Friedrich)
FB02 - Biologie/Chemie
created by envfit to rows meeting a
criterion in $pval (via unlist and which, i suppose). However, i
have difficulties to work out the correct code.
Any help is much appreciated!
--
Tim Richter-Heitmann (M.Sc.)
PhD Candidate
International Max-Planck Research School for Marine Microbiology
Hello,
I am using R 3.1.1 on a (four year old) MacBook, running OSX 10.9.4.
I just tried making and labeling a plot as follows:
x-rnorm(10)
y-rnorm(10)
plot(x,y)
title(main=random points)
which produces a scatter plot of the random points, but without the title
and without any numbers
file
seems fine with me on surface (i dont know how the numbers are stored
internally). I just see correct numbers, also the View command
yields the correct content.
Anyone knows help? Its pretty annoying.
Thank you!
--
Tim Richter-Heitmann
[[alternative HTML version deleted
with lines.
Also of interest would be annotating the whispers with their sample ID
(because the whiskers basically represent the values for y1,2 (11,12;
21,22)).
Any help is welcome! I am new to R, so please bear with me. Thank you!
--
Tim Richter-Heitmann (M.Sc.)
PhD Candidate
International
are not too stupid.
--
Tim Richter-Heitmann (M.Sc.)
PhD Candidate
International Max-Planck Research School for Marine Microbiology
University of Bremen
Microbial Ecophysiology Group (AG Friedrich)
FB02 - Biologie/Chemie
Leobener Straße (NW2 A2130)
D-28359 Bremen
Tel.: 0049(0)421 218-63062
Fax: 0049(0
scale
gg - ggplot(pred, aes(x=Age, y=fit)) +
geom_line(size = 2) + theme_bw() +
geom_ribbon(aes(ymin = fit - 1.96 * se.fit,
ymax = fit + 1.96 * se.fit,), alpha = 0.2, color =
transparent) +
labs(x = Age, y = Log odds (Better))
gg
-Tim
On Wed, Apr 16, 2014 at 7:03 PM, Michael
Sorry I jumped the gun. That does not provide you with the same plot as gg2
that you are aiming for.
-T
On Wed, Apr 16, 2014 at 7:37 PM, Tim Marcella timmarce...@gmail.com wrote:
I think all you have to do is add type=response to your call for the
predictions.
Does this work for you
Hi,
I cannot figure out how or if I even can plot the results from a nested
multinomial logit model. I am using the mlogit package.
Does anyone have a lead on any tutorials? Both of the vignettes are lacking
plotting instructions.
Thanks, Tim
--
Tim Marcella
[[alternative HTML
CSHR.shore.fly - coxph(Surv(entry, exit, to == 1) ~ shore.cat, data
glba.mod)
My variable shore.cat is violating the proportional hazards assumption so I
am trying to add in an interaction with time. Do I interact exit? entry? or
the range of the two?
Thanks, Tim
--
Tim Marcella
508.498.2989
) or not, and then once a reaction has been made, which one is chosen
and how these decisions relate to perpendicular distance to the ship's path.
Thanks, Tim
--
Tim Marcella
508.498.2989
timmarce...@gmail.com
[[alternative HTML version deleted]]
__
R-help@r
Hi,
I was trying to gather/combine the results returned from the mclapply function.
This is how I do it if only one data object is returned:
#= This works fine ===
library(multicore)
frandom1 - function(iter,var1 = 3,var2 =2){
mat - matrix(rnorm(var1*var2),nrow=var1,ncol=var2)
to worry about this and can just report the original std.errors
and associated p values from the segemented object in the pub.
Question # 2: If the count model uses the std.errors, how can I reformulate
this equation to generate the original std.errors.
Thanks, Tim
[[alternative HTML version
this
be accounted for in the hurdle() function?
Thanks, Tim
[[alternative HTML version deleted]]
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Hi,
My data is characterized by many zeros (82%) and overdispersion. I have
chosen to model with hurdle regression (pscl package) with a negative
binomial distribution for the count data. In an effort to validate the
model I would like to calculate the RMSE of the predicted vs. the observed
Hi,
Sorry for the newbie question! My code:
x - 'a'
ifelse(x == 'a',y - 1, y - 2)
print(y)
Shouldn't this assign a value of 1? When I execute this I get:
x - 'a'
ifelse(x == 'a',y - 1, y - 2)
[1] 1
print(y)
[1] 2
Am I doing something really daft???
thanks!
sessionInfo()
R version
, a few
users having this problem:
Whenever i am starting R, i am welcomed with:
Warning message:
In normalizePath(path.expand(path), winslash, mustWork) :
path[1]=//x/home$/tim/Documents/R/win-library/3.0: Access is denied
And i am not allowed to install packages or manipulate objects in my
I could use a little help writing a panel function to append text to each
panel of a lattice::barchart(). Below is a modified version of the barley
dataset to illustrate.
data(barley)
# add a new variable called samp.size
barley$samp.size-round(runif(n=nrow(barley), min=0, max=50),0)
# Below is
looking for. I'll look into trellis.focus to see if the light bulb comes
on in my head. I'd still appreciate a more efficient way to do this given
the number of plots and panels I need.
Thanks,
Tim
On Tue, Nov 26, 2013 at 2:08 PM, Duncan Mackay dulca...@bigpond.com wrote:
Hi
Try
barchart(yield
.
Tim
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and provide commented, minimal, self-contained, reproducible code.
was worried it was
detecting and treating polygons as classes, somehow (ecoregions in example
below).
I had already reached out to Kincaid and Olsen but had not received an answer
yet so I moved on to R-help. I'll go back to them.
Thanks again.
Best,
Tim
Law, Jason jason
), it looks like this and works
perfectly:
oil2 - data.frame (
names = c('YEAR', 'TX', 'CA', 'AL', 'ND'),
oil_2011 = c(2011, 2, 4, 2, 21000),
oil_2012 = c(2012, 3, 25000, 21000, 6)
)
attach(oil2)
oil2[c(oil_2012 == max(oil_2012)),]
Any help is much appreciated.
Thanks, Tim
= y, repos = NULL, lib =
C:/pathToLocationWhereWritesAreAllowed)
#now manually copy the folders to the R installation and overwrite the existing
ones.
Tim
Date: Wed, 02 Oct 2013 11:17:29 -0400
From: Tim Howard tghow...@gw.dec.state.ny.us
To: r-help@r-project.org
Subject: [R] update.packages
=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] tools_3.0.2
Thanks for any help
Tim
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Hi,
I was trying to install the GGally package, but was getting errors. Here is
what I get:
install.packages(GGally)
Installing package(s) into â/Users/ts2w/Library/R/2.15/libraryâ
(as âlibâ is unspecified)
Warning in install.packages :
 package âGGallyâ is not available (for R
#question I have the following data set:
Date-c(9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/8/2010)
EstimatedQuantity-c(3535,2772,3279,3411,3484,3274,3305)
ScowNo-c(4001,3002,4002,BR 8,4002,BR 8,4001)
dataset- data.frame(EstimatedQuantity,Date,ScowNo)
#I'm trying to convert the
Yes! This does this trick. Thank You!
Tim
peter dalgaard pda...@gmail.com 1/26/2013 11:49 AM
On Jan 26, 2013, at 16:32 , Tim Howard wrote:
Duncan,
Good point - I guess I am expecting too much. I'll work on a global replace
before import or chopping with strsplit while importing.
FYI
=showCaseDownloadForm
(see off-airport AEA 2005 for a csv with issues.)
Thank you for the help. I really do appreciate the suggested solutions.
Also, thanks to John Kane for the link to csvEdit.
Tim
Duncan Murdoch murdoch.dun...@gmail.com 01/25/13 6:37 PM
On 13-01-25 4:37 PM, Tim Howard wrote:
David,
Thank
for a while!
Thank you in advance,
Tim
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[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
Tim Howard 1/25/2013 1:42 PM
All,
I have some csv files I am trying to import. I am finding that quotes inside
strings are escaped in a way R
with the escape character within
a single string.
Tim
David Winsemius dwinsem...@comcast.net 1/25/2013 2:27 PM
On Jan 25, 2013, at 10:42 AM, Tim Howard wrote:
All,
I have some csv files I am trying to import. I am finding that quotes inside
strings are escaped in a way R doesn't expect
there would be a solution
within read.csv() ... or perhaps scan()?, I just can't figure it out.
best,
Tim
David Winsemius dwinsem...@comcast.net 1/25/2013 4:16 PM
On Jan 25, 2013, at 11:35 AM, Tim Howard wrote:
Great point, your fix (quote=) works for the example I gave. Unfortunately
)
width-c(slim,wide)
l - vector(list,length(height))
names(l) - height
for(i in 1:length(l)){
l[[i]] - vector(list,length(width))
names(l[[i]]) - width
}
Best
Tim
Date: Fri, 21 Dec 2012 11:34:06 +0100
From: Jessica Streicher j.streic...@micromata.de
To: David Winsemius dwinsem
of interest on the new table.
}
When f is called with 1:n, the table it creates should be the same
as the original table. When called with a bootstrap sample of
values from 1:n, it should create a table corresponding to the
bootstrap sample.
Tim Hesterberg
http://www.timhesterberg.net
(resampling
Le mercredi 12 septembre 2012 à 07:08 -0700, Tim Hesterberg a écrit :
One approach is to bootstrap the vector 1:n, where n is the number
of individuals, with a function that does:
f - function(vectorOfIndices, theTable) {
(1) create a new table with the same dimensions, but with the counts
bootstrap and jackknife methods won't work right.
Tim Hesterberg
http://www.timhesterberg.net
New: Mathematical Statistics with Resampling and R, Chihara Hesterberg
On Fri, Aug 31, 2012 at 12:15 PM, David L Carlson dcarl...@tamu.edu wrote:
Using a data.frame x with columns bins and counts:
x
I am trying to learn how to reshape my data set. I am new to R, so please
bear with me. Basically, I have the following data set:
site-c(A,A,B,B)
bug-c(spider,grasshopper,ladybug,stinkbug)
count-c(2,4,6,8)
myf - data.frame(site, bug, count)
myf
site bug count
1A spider 2
Hi,
Here is the corrected code:
library(ggplot2)
ids - paste('id_',1:3,sep='')
before - sample(9)
after - sample(1:10,9)
dat - as.matrix(cbind(before,after))
rownames(dat) - rep(ids,3)
position - c(rep(10,3),rep(13,3),rep(19,3))
mdat - cbind(melt(dat),position)
colnames(mdat) -
* round(a data frame with numeric and factor columns)
rounds the numeric columns and leaves the factor columns unchanged, rather
than failing.
Tim Hesterberg
NEW! Mathematical Statistics with Resampling and R, Chihara Hesterberg
http://www.amazon.com/Mathematical-Statistics-Resampling-Laura
to, at least for specific situations,
produce results with artifacts? It is very likely that the R-implementation, as
opposed to the alternative algorithm described above and in the attachment, has
a very good statistical explanation, but one that unfortunately is not dawning
on me.
Sincerely,
Tim
5
What is the suggested method to do this for 1 million rows, with 12 variables
and ~ 4,000 unique users?
Thank you,
Tim Stutt
t...@ischool.berkeley.edu
[[alternative HTML version deleted]]
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https
Unfortunately, this won't help for expression arrays. Last time I checked,
those IDATs appeared to be encrypted.
crlmm, methylumi, and minfi can all read IDAT files... *if* they are
version 3 or later (e.g. genotyping, methylation, etc).
Otherwise you are probably stuck with GenomeStudio if it
a reasonable level of
homogeneity. Then from there, you can do a basic inference test for group means
to detect whether there are significant differences detected between groups.
Cheers,
Tim
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
is the most efficient way to resample, with replacement, and generate
means for each grouping variable?
Thanks in advance,
Tim
sessionInfo()
R version 2.12.2 (2011-02-25)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252
Answering my own question.
?sample (!)
y - by(x, x$TrSeasYr, function(x) mean(sample(x[,1], size=999, replace =
TRUE)))
Tim Howard 10/13/2011 9:42 AM
All -
I have an uneven set of replicates and would like to sample from this set X
number of times to generate a mean for each grouping
,
The permutation test answers the question - given that there is exactly
1 outlier in my combined data, what is the probability that random chance
would give a difference as large as I observed. The bootstrap would
answer some other question.
Tim Hesterberg
NEW! Mathematical Statistics with Resampling and R
We are contemplating the installation of R on a Windows server with Hyper-V but
we can't find any information on whether it can be done.
This transmission is intended solely for the person or organization to whom it
is addressed and it may contain
Hi,
I was trying to overlay/combine two freqpoly plots. The sample code below
illustrates the problem. Essentially, I want to do is:
1. Have the same colour for all the lines in 'Plot 1' (and 'Plot 2').
Currently, all the lines in Plot 1 have different colours and all the
lines in Plot 2 have
53 11 7.7
What I want to have is this:
Nr FraSand FraSilt FraClay pH
(kg.kg-1) (kg.kg-1) (kg.kg-1) (-)
1 19 60 21 7.1
2 35 53 11 7.7
Thank you.
TIM
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R-help@r-project.org
' is not an exported object from 'namespace:boot').
Tim Hesterberg
Do
names(bootObj)
to find out what the components are, and use $ or [[ to extract
components.
Do
help(boot)
for a description of components of the object (look in the Value section).
That is general advice in R, applying to all kinds
(), return lists with
a class added, and you can operate on the object as a list using
names(), $, etc.
Tim Hesterberg
Dear R user,
I used the following to do a bootstrap.
bootObj-boot(data=DAT, statistic=Lp.est,
R=1000,x0=3)
I have the following output from the above bootstrap. How
can I extract
Agree - I started with RMySQL but eventually changed over to RODBC when I
realised some of the additional functionality it contains and the fact that I
needed to access different types of RDBMSs (e.g. acquire data from a Microsoft
access database file).
Cheers,
Tim
-Original Message
William,
I think to convert to numeric, you might need to do something like:
as.numeric(as.character()) ## and not just as.numeric()
As it stands, it would appear that it is still being read as a character string.
From: William Armstrong
for the bootstrap biases, see e.g.
Hesterberg, Tim C. (2004), Unbiasing the Bootstrap-Bootknife Sampling
vs. Smoothing, Proceedings of the Section on Statistics and the
Environment, American Statistical Association, 2924-2930.
http://home.comcast.net/~timhesterberg/articles/JSM04-bootknife.pdf
And other
estimated from the original data.
And, you can compute the model matrix once and resample rows of that
along with y, rather than computing a model matrix from scratch each time.
Tim Hesterberg
The only reason the boot package will take more memory for 2000
replications than 10 is that it needs
.
Can anyone see a more efficient way to get the same results? Or is there
existing function which does this?
Thanks for your help
Tim
Function:
miss - function (data)
{
miss.list - list(NA)
for (i in 1:length(data)) {
miss.list[[i]] - table(is.na(data[i
Dear Petr
Thanks so much. That is a LOT more efficient.
Tim
-Original Message-
From: Petr PIKAL [mailto:petr.pi...@precheza.cz]
Sent: Tuesday, April 19, 2011 3:37 PM
To: tesutton
Cc: r-help@r-project.org
Subject: Odp: [R] Simple Missing cases Function
Hi
Hi
try
colSums(is.na(data.m
Thanks Tyler
This function has some useful features
Tim
From: Tyler Rinker [mailto:tyler_rin...@hotmail.com]
Sent: Tuesday, April 19, 2011 3:52 PM
To: tesutton; r-help@r-project.org
Subject: RE: [R] Simple Missing cases Function
I use the following code/function which gives me some
is the range of the middle 95% of
the recorded differences.
Tim Hesterberg
P.S. I think you're mixing up the response and explanatory variables.
I'd think of eating hot dogs as the cause (explanatory variable),
and waistline as the effect (response, or outcome).
P.P.S. I don't like the terms independent
]] - factor(y[[x]] , levels=
my.factor.defs[[x]])})
str(y)
'data.frame': 3 obs. of 3 variables:
$ colOne : Factor w/ 6 levels 1,2,3,4,..: 1 2 3
$ colTwo : Factor w/ 5 levels apple,pear,..: 1 2 3
$ colThree: num 4 5 6
Thank you again!
Best,
Tim Howard
William Dunlap wdun...@tibco.com 2
= factor.defs[2][[1]]$lvl)))
##Error in function (x = character(), levels, labels = levels, exclude = NA, :
## object 'y$colTwo' not found
Any help or perspective (or better way from the beginning!) would be greatly
appreciated.
Thanks in advance!
Tim
[[alternative HTML version
Hi,
I had a function that looked like:
diff - lm(x ~ y + z)
How can I pass the argument to the 'lm' function on the fly? E.g., if I pass it
in as a string (e.g. x ~ y + z), then the lm function treats it as a string
and not a proper argument.
many thanks
[[alternative HTML
files.
Thanks,
Tim
library(plotrix)
Satelite.Palette -
colorRampPalette(c(blue3,cyan,aquamarine,yellow,orange,red))
mycol-Satelite.Palette(ceiling(5000+1))#Max relative angle multiplied by
100 to give larger range. Max is 3.1415, rounded up to 3.15 plus one.
col.labels
Thanks, I will take it up with MS. I just downloaded their latest converter
and that hasn't fixed the issue. Hopefully they will have additional advice.
Aloha,
Tim
Tim Clark
Marine Ecologist
National Park of American Samoa
Pago Pago, AS 96799
From
6 10
3 12 7 11
#the result should be
a b c
1 2.5 3.5 6
2 5 4.5 7
3 7.5 5.5 8
Can anyone direct me down the right path?
Thanks in advance
Tim Howard
[[alternative HTML version deleted]]
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