On 25.01.2012 22:20, Tal Galili wrote:
Hello dear list,
I just noticed that:
Package ‘doSMP’ was removed from the CRAN repository.
http://cran.r-project.org/web/packages/doSMP/index.html
Does any one know the reason for this?
Is this a technical or a legal (e.g: license) issue?
If legal
On Thu, Jan 26, 2012 at 8:02 AM, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
On 25.01.2012 22:20, Tal Galili wrote:
Does any one know the reason for this?
Is this a technical or a legal (e.g: license) issue?
If legal issues were the reason, you had not found it in the archives
On 26.01.2012 09:39, Barry Rowlingson wrote:
On Thu, Jan 26, 2012 at 8:02 AM, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
On 25.01.2012 22:20, Tal Galili wrote:
Does any one know the reason for this?
Is this a technical or a legal (e.g: license) issue?
If legal issues were the
I have a dataset like this (dput for this below) which represents user
computer sessions:
username machine start end
1 user1 D5599.domain.com 2011-01-03 09:44:18 2011-01-03 09:47:27
2 user1 D5599.domain.com 2011-01-03 09:46:29 2011-01-03 10:09:16
Workshop on Bayesian methods and WinBUGS
***
A two-day workshop on Bayesian methods is being held on Friday 3 - Saturday 4
February 2012 at the University of Sydney.
This course is suitable for graduate students, academics, researchers and
professionals who
Thank you everyone for your dedication to improving 'R' - its function to
statistical analysis and comments.
I have now 48 models (unique combinations of 1 to 6 variables) and have put
them into a list and gained the results for all models. Below is a sample of
my script results:
m$model48 -
I am new to R, and I am trying to cut a continuous variable BMI into
different categories and can't figure out how to use it. I would like to cut
it into four groups: 20, 20-25, 25-30 and = 30. I am having difficulty
figuring the code for 20 and =30? Please help. Thank you.
--
View this message
On 01/26/2012 07:23 PM, citadel wrote:
I am new to R, and I am trying to cut a continuous variable BMI into
different categories and can't figure out how to use it. I would like to cut
it into four groups:20, 20-25, 25-30 and= 30. I am having difficulty
figuring the code for20 and=30? Please
Hi,
I have a data frame:
class(B27.vec)
[1] data.frame
head(B27.vec)
AGE Gend B27 AgeOn DD uveitis psoriasis IBD CD UC InI BASDAI BASFI Smok UV
1 571 119 38 2 1 1 1 1 1 5.40 8.08 NA 1
2 351 133 2 2 1 1 1 1 1 1.69 2.28
Hi, Philip,
counter-questions:
1. Which/where is the grouping variable for the test of differences in
survival?
2. Assume the grouping variable is Gend in B27.vec. Then, why aren't you
using
survdiff( Surv( AgeOn, UV) ~ Gend, rho = 0, data = B27.vec)
?
Hth -- Gerrit
On Thu, 26 Jan
Dear list,
I would like to find data points that at least should be checked one more
time before I process them further.
I've had a look at the outliers package for this, and the outliers function
in that package, but it appears to only return one value.
An example:
That's not how to do it. Please follow the link in the footer and the correct
way to subscribe should become clear.
-pd
On Jan 25, 2012, at 21:49 , Paul Johnston wrote:
subscribe
__
R-help@r-project.org mailing list
According to the help file for 'outlier' , (quoting)
x a data sample, vector in most cases. If argument is a dataframe, then
outlier is
calculated for each column by sapply. The same behavior is applied by apply
when the matrix is given. (endquote)
Looks like you could create a matrix that
It is not valid to categorize BMI. This will result in major loss of
information and residual confounding. Plus there is huge heterogeneity in
the BMI = 30 group. Details are at
http://biostat.mc.vanderbilt.edu/CatContinuous and see these articles:
@Article{fil07cat,
author =
tsippel wrote on 01/25/2012 01:23:05 PM:
Simple question (I hope?). How does one extract the data from a trellis
object? I need to get the data from a call to histogram().
A simple example below...
dat-data.frame(year=as.factor(round(runif(500, 1990, 2010))),
x=rnorm(500,
75, 35),
Are there other packages besides CRF available on R, for Conditional
Random Fields ?
Thanks in advance.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Dear All,
I would like to ask for help on how to read different files automatically and
do analysis using scripts.
1. Description of the data
1.1. there are 5 text files, each of which contains cleaned data for the same
100 SNPs. Observations (e.g., position on gnome, alelle type, ...) for
On 12-01-25 3:46 PM, Grant Anderson wrote:
Whenever I open Microsoft Word (2003), I get this warning two times---and thenÂ
must manually 'OK' it twice before I can proceed in Word (a pain to do):
Â
The drive or network connection that the shortcut 'R.lnk' refers to is unavailable.
Make
Michael Wither wrote on 01/26/2012 12:08:19 AM:
Hi, I have a question about running multiple in regressions in R and
then
storing the coefficients. I have a large dataset with several
variables,
one of which is a state variable, coded 1-50 for each state. I'd like to
run a regression of 28
Hello everyone,
I have a character vector of 24 hour time values in the format hm
without the delimiting :. How can I insert the : immediately to
the left of the second digit from the right?
mytimes-scan(what=)
1457
1457
1310
1158
137
1855
Thanks!
Dan
Dear R-help,
I must be missing something very obvious, but I am confused as to why
the null distribution for p values generated by binom.test() appears to
be non-uniform. The histogram generated below has a trend towards
values closer to 1 than 0. I expected it to be flat.
When I try to create a Rasch simulation of data using the sim.rasch
function, I get more items than I intend
#My code
library(eRm)
#Number of items
k - 20
#Number of participants
n - 100
#Create Rasch Data
#sim.rasch(persons, items, seed = NULL, cutpoint = randomized)
r.simulation -
I'm trying to kick off some of the rust and learn some of the R packages
for Rasch modeling. When I tried using the eRm package, I get item
difficulty estimates for all my items accept the first (in terms of order)
item.
#Begin code
library(eRm)
r.simulation - sim.rasch(20,100)
r.data -
Hi R- listeners,
I should add that I would like also to compare my field data to an index
model. The index was created by using the following script:
devel.index - function(values, weights=c(1, 2, 3, 4, 5, 6)) {
foo - values*weights
return(apply(foo, 1, sum) / apply(values, 1, sum))
}
Hi Rui,
Thanks for your reply to my post. My code still has various shortcomings but at
least now it is fully functional.
It may be that, as I transition to using R, I'll have to live with some less
than ideal code, at least at the outset. I'll just have to write and re-write
my code as I
Hello, Dan,
you could probably use a combination of nchar(), substr() (or substring())
and paste(). Take a look at their online help pages.
Hth -- Gerrit
Hello everyone,
I have a character vector of 24 hour time values in the format hm
without the delimiting :. How can I insert the :
sub(([[:digit:]]{2,2})$, :\\1, mytimes)
[1] 14:57 14:57 13:10 11:58 1:37 18:55
That will convert 05 to :05 and will do nothing
to 5. Pad with 0's before calling sub if that is
required.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From:
Sorry, sent this earlier but forgot to add an informative subject line. Am
resending, in the hopes of getting further replies. My apologies. Hope this is
OK.
Paul
Hi Rui,
Thanks for your reply to my post. My code still has various shortcomings but at
least now it is fully functional.
It
Barry, thanks a lot!
I was able to read in Candian data set from gadm:
library(raster)
# Finding ISO3 code for Canada
getData('ISO3') # Canada's code is CAN
# Reading in data at different levels
can0-getData('GADM', country=CAN, level=0)
can1-getData('GADM', country=CAN, level=1)
I believe that what you are seeing is due to the discrete nature of the
binomial test. When I run your code below I see the bar between 0.9 and 1.0 is
about twice as tall as the bar between 0.0 and 0.1, but the bar between 0.8 and
0.9 is not there (height 0), if you average the top 2 bars
Hello
I will appreciate your help with the following.
Running a script in R 2.14.1 under windows vista I get the following error
message:
Error in ifelse(append, a, w) :
(list) object cannot be coerced to type 'logical'
However, the very same script runs perfectly well
Hi,
I am a beginner with R, and I think the answer to my question will
seem obvious, but after searching and trying without success I've
decided to post to the list.
I am working with data loaded from a csv filewith these fields:
order_id, item_value
As an order can have multiple items, an
what if I don't need to store the combination results, I just want to get the
combination result for a given row.
e.g. for the 5 elements divided into 3 groups , and if I give another
parameter which is the row number of the results, in petr's example, say if
I set row number to 1, then I get
On Jan 26, 2012, at 10:17 AM, Juan Antonio Balbuena wrote:
Hello
I will appreciate your help with the following.
Running a script in R 2.14.1 under windows vista I get the
following error
message:
Error in ifelse(append, a, w) :
(list) object cannot be coerced to type
I have a problem with including extra data in a lattice graphic. I am
trying to put some calculated values in an extra column of a boxplot. A
minimal example should show what I am trying to do:
foo - data.frame(
Treatment=rnorm(1:12,2),
Variant=c(A,A,B,C,D,C,B,D,D,B,B,A),
Please include the context of the discussion in your responses. See
inline below.
On 1/24/2012 11:33 PM, Ajay Askoolum wrote:
Thanks you,
I can get the length of aa with length(unlist(aa)). If aa has 4
dimensions, I imagine I'd need to do
max(sapply(aa,sapply,sapply,length)
Correct.
How
On Thu, 26 Jan 2012, James Holland wrote:
When I try to create a Rasch simulation of data using the sim.rasch
function, I get more items than I intend
#My code
library(eRm)
#Number of items
k - 20
#Number of participants
n - 100
#Create Rasch Data
#sim.rasch(persons, items, seed = NULL,
On Thu, 26 Jan 2012, James Holland wrote:
I'm trying to kick off some of the rust and learn some of the R packages
for Rasch modeling. When I tried using the eRm package, I get item
difficulty estimates for all my items accept the first (in terms of order)
item.
#Begin code
library(eRm)
Hi Michael,
I am using the same version contributed by Martin Backer. I have been in touch
with him via email as well. He said yesterday that he can help me with it but
is busy for some time.
This distribution function does not behave properly as can be seen in the
attached spreadsheet. Check
Greg, thanks for the reply.
Unfortunately, I remain unconvinced!
I ran a longer simulation, 100,000 reps. The size of the test is
consistently too small (see below) and the histogram shows increasing
bars even within the parts of the histogram with even bar spacing. See
Dear Rhelp,
When I try to plot a barplot, I get the following message:
Error in plot.new() : figure margins too large
What does this mean and how can I fix it?
I appreciate your help,
Horatio Gates
[[alternative HTML version deleted]]
Yes that is due to the discreteness of the distribution, consider the following:
binom.test(39,100,.5)
Exact binomial test
data: 39 and 100
number of successes = 39, number of trials = 100, p-value = 0.0352
alternative hypothesis: true probability of success is not equal to 0.5
95
Hi Bill,
Thanks very much for your response.
Can you suggest an approach for the pre-padding? Here is a more
respresentative sample of the values:
mytimes-scan(what=)
1334
2310
39
2300
1556
3
404
37
1320
4
211
2320
Thanks!
Dan
On Thu, Jan 26, 2012 at 10:41 AM, William Dunlap
On Thu, Jan 26, 2012 at 4:03 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Barry, thanks a lot!
I was able to read in Candian data set from gadm:
library(raster)
# Finding ISO3 code for Canada
getData('ISO3') # Canada's code is CAN
# Reading in data at different levels
I usually get that error when I'm replotting on a window/device that's
already really small:
e.g., on my Mac
plot(1:5)
## Make window really small
plot(1:6) # Throws error
Michael
On Thu, Jan 26, 2012 at 11:32 AM, Harish Eswaran hra...@att.net wrote:
Dear Rhelp,
When I
On 1/25/2012 10:09 AM, patzoul wrote:
I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?
A combination of Reduce and Map will work. Map to stitch together the a
and b vectors, Reduce
One way to pad with initial zeros is to convert your
strings to integers and format the integers:
sprintf(%04d, as.integer(mytimes))
[1] 1334 2310 0039 2300 1556 0003 0404
[8] 0037 1320 0004 0211 2320
It has the added benefit that the call to as.integer
will warn you if any supposed
This might get more traction on the R-SIG-Ecology lists.
And best of luck to you; I quite like turtles.
Michael
On Thu, Jan 26, 2012 at 4:37 AM, Jhope jeanwaij...@gmail.com wrote:
Hi R- listeners,
I should add that I would like also to compare my field data to an index
model. The index was
Thank you very much, Barry!
Dimitri
On Thu, Jan 26, 2012 at 12:00 PM, Barry Rowlingson
b.rowling...@lancaster.ac.uk wrote:
On Thu, Jan 26, 2012 at 4:03 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Barry, thanks a lot!
I was able to read in Candian data set from gadm:
To pretend that AIC solves this problem is to ignore that AIC is just a
restatement of P-values.
Frank
Rubén Roa wrote
I think we have gone through this before.
First, the destruction of all aspects of statistical inference is not at
stake, Frank Harrell's post notwithstanding.
Second,
On Thu, Jan 26, 2012 at 08:29:22AM -0800, yan wrote:
what if I don't need to store the combination results, I just want to get the
combination result for a given row.
e.g. for the 5 elements divided into 3 groups , and if I give another
parameter which is the row number of the results, in
Cheers Bill!!
On Thu, Jan 26, 2012 at 12:03 PM, William Dunlap wdun...@tibco.com wrote:
One way to pad with initial zeros is to convert your
strings to integers and format the integers:
sprintf(%04d, as.integer(mytimes))
[1] 1334 2310 0039 2300 1556 0003 0404
[8] 0037 1320 0004 0211
Paul,
I have a partial solution for you. It is partial in that I have not quite
figured out the correct incantation to convert a 5 digit year (eg. 11/23/21931)
properly using the R date functions. According to various sources (eg. man
strptime and man strftime) as well as the R help for both
On Tue, Jan 24, 2012 at 11:54 AM, Paul Miller pjmiller...@yahoo.com wrote:
Hello Everyone,
Still new to R. Wrote some code that finds and prints invalid dates (see
below). This code works but I suspect it's not very good. If someone could
show me a better way, I'd greatly appreciate it.
On 26-01-2012, at 17:58, Brian Diggs wrote:
On 1/25/2012 10:09 AM, patzoul wrote:
I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?
A combination of Reduce and Map will work.
I know that [] is used for indexing.
I know that [[]] is used for reference to a property of a COM object.
I cannot find any explanation of what [[1]] does or, more pertinently, where it
should be used.
Thank you.
[[alternative HTML version deleted]]
On 26-01-2012, at 19:10, Berend Hasselman wrote:
On 26-01-2012, at 17:58, Brian Diggs wrote:
On 1/25/2012 10:09 AM, patzoul wrote:
I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a
Have you read ?[[ ?
The short answer is that you can use both [] and [[]] on lists, the []
construct will return a subset of the list (which will be a list) while [[]]
will return a single element of the list (which could be a list or a vector or
whatever that element may be): compare:
tmp
This is my reproducible example (three data frames: a, b, c)
a-structure(list(date = structure(1:6, .Label = c(2012-01-03,
2012-01-04, 2012-01-05, 2012-01-06, 2012-01-07, 2012-01-08,
2012-01-09, 2012-01-10, 2012-01-11, 2012-01-12, 2012-01-13,
2012-01-14, 2012-01-15, 2012-01-16, 2012-01-17,
Hello,
I'm quite new to R and want to make a Weibull-regression with the survival
package. I know how to build my Surv-object and how to make a
standard-weibull regression with survreg.
However, I want to fit a translated or 3-parametric weibull dist to account for
a failure-free time.
I think
I might do something like this:
mergeAll - function(..., by = date, all = TRUE) {
dotArgs - list(...)
Reduce(function(x, y)
merge(x, y, by = by, all = all, suffixes=paste(., names(dotArgs),
sep = )),
dotArgs)}
mergeAll(a = a, b = b, c = c)
str(.Last.value)
You also might be able to set
Here is a page that should help:
http://www.burns-stat.com/pages/Tutor/more_R_subscript.html
Also Circle 8.1.54 of 'The R Inferno'
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf
On 26/01/2012 18:27, Ajay Askoolum wrote:
I know that [] is used for indexing.
I know that [[]] is used for
On Fri, Jan 27, 2012 at 5:36 AM, Chris Wallace
chris.wall...@cimr.cam.ac.uk wrote:
Greg, thanks for the reply.
Unfortunately, I remain unconvinced!
I ran a longer simulation, 100,000 reps. The size of the test is
consistently too small (see below) and the histogram shows increasing bars
On 1/26/2012 10:33 AM, Berend Hasselman wrote:
On 26-01-2012, at 19:10, Berend Hasselman wrote:
On 26-01-2012, at 17:58, Brian Diggs wrote:
On 1/25/2012 10:09 AM, patzoul wrote:
I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] ,
Dear all,
I am trying to analyze some non-linear data to which I have fit a curve of
the following form:
dum - nls(y~(A + (B*x)/(C+x)), start = list(A=370,B=100,C=23000))
I am wondering if there is any way to determine meaningful quality of fit
statistics from the nls function?
A summary yields
Hi,
I want to apply a function (in my case SDF; package “sapa”) repeatedly over
discrete sections of a daily time series object by sliding a time window of
constant length (e.g. 10 consecutive years or 1825 days) over the entire ts at
increments of 1 time unit (e.g. 1 year or 365 days). So
Thank you for the explanation Uwe.
With regards,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
I'm not sure if it's easily doable with a ts class, but the rollapply
function in the zoo package will do this easily. (Also, I find zoo to
be a much more natural time-series workflow than ts so it might make
the rest of your life easier as well)
Michael
On Thu, Jan 26, 2012 at 2:24 PM, Jorge
On Thu, Jan 26, 2012 at 4:00 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
I'm not sure if it's easily doable with a ts class, but the rollapply
function in the zoo package will do this easily. (Also, I find zoo to
be a much more natural time-series workflow than ts so it might make
thank you for your reply
I'll study and test your code (which is a bit obscure to me up to now);
by the way do you think that merge_all is a wrong way to hit?
thanks again
m
--
View this message in context:
http://r.789695.n4.nabble.com/merge-multiple-data-frames-tp4331089p4331830.html
Sent
I ask the question about when to stop adding another variable even though it
lowers the AIC because each time I add a variable the AIC is lower. How do I
know when the model is a good fit? When to stop adding variables, keeping
the model simple?
Thanks, J
--
View this message in context:
johannes rara wrote on 01/26/2012 02:46:57 AM:
I have a dataset like this (dput for this below) which represents user
computer sessions:
username machine start end
1 user1 D5599.domain.com 2011-01-03 09:44:18 2011-01-03 09:47:27
2 user1
Inline below.
-- Bert
On Thu, Jan 26, 2012 at 12:16 PM, Max Brondfield
max.brondfi...@gmail.com wrote:
Dear all,
I am trying to analyze some non-linear data to which I have fit a curve of
the following form:
dum - nls(y~(A + (B*x)/(C+x)), start = list(A=370,B=100,C=23000))
I am wondering
Simple question. 8 million pages in the statistical literature of
answers. What, indeed, is the secret to life?
Post on a statistical help list (e.g. stats.stackexchange.com). This
has almost nothing to do with R. Be prepared for an onslaught of often
conflicting wisdom.
-- Bert
On Thu, Jan 26,
Nevil Amos wrote:
I am getting the above warning following loading of Geneland 3.1.5 on
unix , while a simple plot sends output to the pdf file ( see attached
code) no output results from Geneland functions, resulting in empty pdf
files
That message is saying it can't find an X11
On Thu, Jan 26, 2012 at 08:29:22AM -0800, yan wrote:
what if I don't need to store the combination results, I just want to get the
combination result for a given row.
e.g. for the 5 elements divided into 3 groups , and if I give another
parameter which is the row number of the results, in
Dear List,
I'll appreciate your help on this. I'm trying to create a variable as in
cumsum_y.cond1 below, which should compute the cumulative sum of y
conditional on the value of cond==1.
set.seed(1)
d - data.frame(y= sample(c(0,1), 10, replace= T),
cond= sample(c(0,1), 10,
Axel Urbiz wrote
Dear List,
I'll appreciate your help on this. I'm trying to create a variable as in
cumsum_y.cond1 below, which should compute the cumulative sum of y
conditional on the value of cond==1.
set.seed(1)
d - data.frame(y= sample(c(0,1), 10, replace= T),
Hello, again.
I now have a more complete answer to your points.
1. It's too long. My understanding is that skilled programmers can usually
or often complete tasks like this in a few lines.
It's not very shorter but it's more readable. (The programmer is always
suspect)
2. It's not
Hi al,
I am looking for a R command to test the difference of two linear
regressoon betas.
Lets say I have data x1, x2...x(nï¼1).
beta1 is obtained from regressing x1 to xn onto 1 to n.
beta2 is obtained from regressing x2 to x(nï¼1) onto 1 to n.
Is there a way in R to test whether beta1 and
I don't know what is available in R but a relevant paper that provides a
test is by
Hotelling, H ( September, 1940 )
The Selection of Variates For Use in Prediction With Some Comments On The
General Problem of Nuisance Parameters.
Annals of Mathematical Statistics, 11, 271-283.
On Thu, Jan
Hello,
I have a very large content analysis project, which I've just begun to
collect training data on. I have three coders, who are entering data on up
to 95 measurements. Traditionally, I've used Excel to check coder agreement
(e.g., percentage agreement), by lining up each coder's measurements
http://r.789695.n4.nabble.com/file/n4332721/Рисунок1.jpg
Ultimately, I want, when I will set the limits of height's variation from
*min* to *max*, get into the file a set of clusters (*A, B, C, ... ..*) in
the form of:
l1; 32, 33, 34, 30, 31, 55, 60, 58, 59, 57, 61
l2; 50, 51, 52, 54, 47, 53,
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