Isn't a vector of vectors usually considered a matrix? So if you want to
vectorize a vector function you would normally rewrite it to operate on
matrices.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#.
On Thu, Aug 25, 2011 at 6:15 PM, markm0705 wrote:
> I've been building a ranked dot plot for several days now and am sorting the
> data using the reorder command. What I don't understand is how reorder
> works when mutiple varibles are plotted by grouping. In the example below
> I'm using re-ord
Hi All,
I have a data frame as follow:
user_id time age location gender
.
and I learn a logistic regression to learn the weights (glm with family= (link
= logit))), my response value is either zero or one. I would like to group the
users based on user_id and time and see the y values and
Perhaps it could help if you could assume a more flexible model for [T];
for instance, a PH model with a piecewise-constant baseline hazard,
which you can simulate as a Poisson variable.
Best,
Dimitris
On 8/26/2011 10:45 PM, Ravi Varadhan wrote:
Hi,
Here is an update. I implemented this bi
On Aug 26, 2011, at 4:52 PM, Newbie wrote:
Thank you for the quick response! I think you are on the right track
- but is
there any way of "calling" (is that the word for it) the function
price_call
in the mapply, so that this price_call function is changed to handle
vectors. I believe that
Dear All,
I have two hidden classes for my data set that I would like to learn them based
on the Mixture of Binomial regression model. I was wondering if there is any
package that I can use for that purpose. Has any one tried any package for the
Mixture models?
Thanks a lot,
Andra
__
Hi, please help me,
I want to have a functional Rcmdr but after install as indicated in:
http://socserv.mcmaster.ca/jfox/Misc/Rcmdr/installation-notes.html
obtain the following:
Loading Tcl/Tk interface ... done
Loading required package: car
Loading required package: MASS
Loading required package
I am familiar with pairwise t-tests, corrections for multiple testing, etc.
however I have a problem whose answer I have not found after extensive R-help
archive and Google searching.
What I have done in the past:
I have N items which are measured, exposed to a condition, and then measured
agai
Thank you David. I was indeed looking at rms, DAAG and caret.
I didn´t find a way to extract with predict the probabilities of the
crossalidated model in rms, so I will try with DAAG. But DAAG only gives one
set of crossvalidatet proabilities and I wanted the probabilities for all
crossvalidate
Thank you for the quick response! I think you are on the right track - but is
there any way of "calling" (is that the word for it) the function price_call
in the mapply, so that this price_call function is changed to handle
vectors. I believe that this should, in theory if it is correct, make the
r
hello all,
I am a beginner at R and not exactly a statistician either. I'm messing
around with a probit model on an enconomic time series. I can get the model
estimated but I have not been able to get it to give me predictions out of
sample data. I'm using the predict function but getting errors.
In my R learning I've come across a situation in which a piece of code that
works on the work space outside a function does not work inside the function.
WARNING THIS EMAIL CONTAINES THE CODE:#rm(list=ls()) THIS WILL CLEAR ALL
OBJECTS FROM YOUR WORKSPACE! When I use rm(list=ls()) and then l
Hi.
On Fri, Aug 26, 2011 at 8:58 AM, Ben qant wrote:
> If someone is able, can you tell me if there is a better way to do this?
> More specifically, do I have to rewrite all of the data members stuff and
> extend stuff of parent class in the child class?
No.
> See below. Thanks in
> advance!
>
Hi,
basehaz() in survival package is said to estimate the baseline cumulative
hazard from coxph(), but it actually calculate cumulative hazard by
-log(survfit(coxph.object)). But survfit() calculate survival based on MEAN
covariate value, not covariate value of 0. I thought baseline cumulativ
I thought I would post an example of using R.oo with inheritance and pass by
reference for future searches. With PerMore I am inheriting Person. Then I
am changing an object data memeber of an object of PerMore class using pass
by reference ('mimiced' by R.oo) in an object of AgeMultiplier. I chang
On Aug 26, 2011, at 4:58 PM, David Winsemius wrote:
On Aug 26, 2011, at 2:35 PM, Jim Silverton wrote:
H all,
I have a 6 x 3 matrix. The last column is simply the sum of of the
first two
rows.
x1 x2 x3
1 23
2 13
1 01
2 24
2 13
2 13
0
par(mfrow = c(2, 1))
plot(density(rnorm(100)^2))
plot(density(rnorm(100)^2))
?
On Fri, Aug 26, 2011 at 10:20 AM, Benjamin Polidore wrote:
> I have two distributions. Both are left skewed. Is there a good
> statistical approach to determining if the skew of distribution 1 is
> statistically sim
On Aug 26, 2011, at 2:35 PM, Jim Silverton wrote:
H all,
I have a 6 x 3 matrix. The last column is simply the sum of of the
first two
rows.
x1 x2 x3
1 23
2 13
1 01
2 24
2 13
2 13
0 00
I want to create a column of probabilities p,
Hi,
Here is an update. I implemented this bivariate lognormal approach. It works
well in simulations when I generated the marginal [T] from a lognormal
distribution and independently censored it. It, however, does not do well when
I generate from a marginal [T] that is Weibull or a distribut
On Aug 26, 2011, at 2:43 PM, Newbie wrote:
Dear R-users
I am trying to "vectorize" a function so that it can handle two
vectors of
inputs. I want the function to use phi (a function), k[1] and t[1]
for the
first price, and so on for the second, third and fourth price.
? mapply
I trie
Jay wrote:
While I'm very pleased with the results I get with rpart and
rpart.plot, I would like to change the scientific notation of the
dependent variable in the plots into integers. Right now all my 5 or
more digit numbers are displayed using scientific notation.
One way of getting rpart.pl
Try this:
file.info("your_file")$mtime
On Fri, Aug 26, 2011 at 4:48 PM, alan wu wrote:
> Hi All,
>
> I want to add the last modified time of the dataset I used in the
> analysis. Any idea about how to get it from R in windows XP system?
>
> Thanks
>
>
> Alan
>
> _
Hi All,
I want to add the last modified time of the dataset I used in the
analysis. Any idea about how to get it from R in windows XP system?
Thanks
Alan
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Hello all,
I'm trying to run a gird parameter search for a svm.
Therefore I'M using the ksvm function from the kernlab package.
svp <- ksvm(Ktrain,ytrain,type="nu-svc",nu=C)
The problem is that the optimization algorithm does not return
for certain parameters.
I tried to use setTimeL
No, it's not much faster. I'd say it's faster about 10-15% in my case.
I dont want neither plyr or data.table package because our software on the
server does not support R version over 2.10 and both of them have
dependency for R >= 2.12. Also I do not want to use old archives because I
did not ha
Thanks
I substituted
points(inactT, inactR, pch="|")
with
rug(inactT, side=1, ticksize = 0.03)
and now it is perfect
Thanks again.
Claudio
2011/8/26 David Winsemius
>
> On Aug 26, 2011, at 11:31 AM, heverkuhn wrote:
>
>
>> No attachment. No code.
>>
>> I apology for that. I provided below t
Hi Ravi,
are you saying to apply the idea of a trapezoidal method "again", because I
was expecting the integrate() already do it for me. Is not the case?
I am also dealing with troubles related to integrate. Just to say, I tried
to figure out the problem in Mathematica and it gives me the
expecte
I have two distributions. Both are left skewed. Is there a good
statistical approach to determining if the skew of distribution 1 is
statistically similar to the skew of distribution 2?
Thanks,
bp
[[alternative HTML version deleted]]
__
R-hel
Dear R-users
I am trying to "vectorize" a function so that it can handle two vectors of
inputs. I want the function to use phi (a function), k[1] and t[1] for the
first price, and so on for the second, third and fourth price. I tried to do
the mapply, but I dont know how to specify to R what input
On Thu, Aug 25, 2011 at 4:45 AM, JPF wrote:
> This is for coxph:
>
> The cluster term is used to compute a robust variance for the model. The
> term + cluster(id) where each value of id is unique is equivalent to
> specifying the robust=T argument, and produces an approximate jackknife
> estimate
Hi,
I have a right-censored (positive) random variable (e.g. failure times subject
to right censoring) that is observed for N subjects: Y_i, I = 1, 2, ..., N.
Note that Y_i = min(T_i, C_i), where T_i is the true failure time and C_i is
the censored time. Let us assume that C_i is independen
On Aug 26, 2011, at 2:09 PM, Jon Toledo wrote:
Dear experts,
I am looking for a package that does logistic regression with
corssvalidation and gives me the probabilites of all the
corssvalidations so that I can plot them in ROCR.
Would also like to know if the corssvalidate model would giv
H all,
I have a 6 x 3 matrix. The last column is simply the sum of of the first two
rows.
x1 x2 x3
1 23
2 13
1 01
2 24
2 13
2 13
0 00
I want to create a column of probabilities p, such that for each row, I want
to find the probability of
On Thu, Aug 25, 2011 at 08:25:02AM -0500, Giovanni Petris wrote:
> Hello!
>
> I am using R on two different machines (under Ubuntu and OS X, but this
> is probably irrelevant) and I would like to keep the two installations
> 'synchronized', in particular in terms of installed packages. For
> examp
Dear experts,
I am looking for a package that does logistic regression with corssvalidation
and gives me the probabilites of all the corssvalidations so that I can plot
them in ROCR.
Would also like to know if the corssvalidate model would give me a summary
coefficient for the intercept and my
To save anyone else 30 seconds, here's how google translates the below:
Ladies and Gentlemen,
I would like to inform me whether there is the possibility of security
certificates for R packages are.
Sincerely,
Christopher W. Weinberger
2011/8/26 Christoph W. Weinberger
> Sehr geehrte Damen un
This was an example using package KFAS, which needs to be loaded before
running the code ('kf' is the function that computes Kalman filter in
package KFAS)
Giovanni
On Fri, 2011-08-26 at 07:11 -0700, quantguy wrote:
> Thanks! However, I added the line of code and receive an error during the
> opt
Ramnath:
With my apologies if I'm wrong, it does not look like you have made
much of an effort to learn R's basics, e.g. by working thru the
"Introduction to R" tutorial distributed with R. If that is the case,
why do you expect us to help?
-- Bert
On Fri, Aug 26, 2011 at 8:52 AM, Ramnath wrote
On Aug 26, 2011, at 11:35 AM, chirleu wrote:
Hi. I'm trying to add some vertical lines to an xyplot in which x
axis is a
temporal variable (class=dates) and y axis a factor variable.
This is my code:
xyplot(factor(Abacus$Emisor)~Abacus$Dia,xlab="Date",
ylab="Fish",pch=124,
scales=list(x=
Hi All,
Has anyone coded up the OptiGrid clustering algorithm for high dimensional
space?
If so is anyone willing to share?
Many thanks in Advance
Mike
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.eth
On Aug 26, 2011, at 11:31 AM, heverkuhn wrote:
No attachment. No code.
I apology for that. I provided below the code and the vectors,
Whatever that mean. Perhaps it means "below"?
Yes, below, I mean out of the axis...
And I think, that as you suggested, the xpd is what I was looking for.
Dear Sarah and Rolf,
Thanks for your suggestions, yes I was looking for something that'd execute
strings eval(parse) was a solution. It's extremely helpful when you want to do
something on variables generated with loops.
Juta
I lied, that was not my last question: how can I add two arrows at the
bottom with the words "in favor of A / B"? This is not specified in the pdf
and with "text" I have the impression that I can't add text below the
x-axis.
2011/8/26 Paola Tellaroli
> Dear Prof. Viechtbauer,
> thank you so muc
Hi All,
I am a beginner in programming in r and please do forgive me if my question
seems to be silly and sometimes not understandable.
1. we have a list of elements in a list say:
ls<-list("N","E","E","N","P","E","M","Q","E","M")
2. We have an another list of tables in a list say:
n <- list(
Thank you for your answers. The problem persists always (without ".pdf" or
using "print" or "plot"). The code runs, but the probem is to save the
figure with pdf format in order to load it directly in word.
Howerver, I will read the link indicated.
--
View this message in context:
http://r.789
Thanks! However, I added the line of code and receive an error during the
optim() procedure:
rror in fn(par, ...) : could not find function "kf"
The update code is here:
tmp <- ts(read.table("http://shazam.econ.ubc.ca/intro/P.txt";,
header=T), start=c(1978,1), frequency=12) * 100
y <- tmp[,1:4
I am working with data from the USGS with data every 30 minutes from
4/27/2011 to 8/25/2011.
I am having trouble with setting the frequency.
My R script is below:
> shavers=read.csv("shavers.csv")
> names(shavers)
[1] "agency_cd""site_no" "datetime" "tz_cd""Temp"
i can't comment on gtools but to read in a data file (eg csv) these commands
will help you and are provided in R without a need for extra packages.
read.table
read.csv
to get help on these use this
?read.table
if you type installed.packages() do gtools/gdata come up?
please describe what steps
I'm not familiar with eRm but subscript out of bounds often occurs when the
data isn't the structure you think it is. Perhaps your input is not what is
expected - might not be as long/wide as required. Depending on how allb is
stored, you might be flipping rows/columns around? try checking the
dime
Dear Forum,
I haven't programmed in a while, and I'm new to R, so please keep your
answers simple.
I'm trying to analyse data in an xml file. I successfully loaded the data
into lists using:
doc <- xmlTreeParse("PSparsed.xml", getDTD = F)
I was going to write a script to traverse the tree and flat
Hi. I'm trying to add some vertical lines to an xyplot in which x axis is a
temporal variable (class=dates) and y axis a factor variable.
This is my code:
xyplot(factor(Abacus$Emisor)~Abacus$Dia,xlab="Date", ylab="Fish",pch=124,
scales=list(x=list(format="%b %Y",tick.number=30))
, panel=function(
Sehr geehrte Damen und Herren,
ich würde mich gerne informieren, ob es die Möglichkeit gibt
Sicherheitszertifikate für R-Pakete gibt.
Mit freundlichen Grüßen
Christoph W. Weinberger
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/
No attachment. No code.
I apology for that. I provided below the code and the vectors,
Whatever that mean. Perhaps it means "below"?
Yes, below, I mean out of the axis...
And I think, that as you suggested, the xpd is what I was looking for.
I need something like this
http://www.springerima
If someone is able, can you tell me if there is a better way to do this?
More specifically, do I have to rewrite all of the data members stuff and
extend stuff of parent class in the child class? See below. Thanks in
advance!
Example 1:
setConstructorS3("ClassA", function(A,x) {
if(missing(A))
I just saw this old post, but it seems that nobody replied, so let me try.
If you can assume that also U[t] evelves as a random walk, I would build a
DLM by taking the state vector to be
x[t] = (U[t], UN[t], pi[t])'
By plugging in the equation for pi[t] the random walk expressions for U[t]
and
Slight change; forgot that 'all.equal' only accepts a single value:
(and it is a little shorter
f.x <- function(x) any(all.equal(x, 56.36) || all.equal(59.91))
mask <- with(vaslinks4, (yearsep < 1988) &
(sapply(proc1, f.x)
| sapply(proc2, f.x)
On Aug 26, 2011, at 10:16 AM, Joanne Demmler wrote:
Dear all,
I'm trying to rerun some data linkage exercises in R (they are
designed to be done in SPSS, SAS or STATA)
The exercise in question is to relabel the column "treat" to "1", if
"yearsep" is smaller than 1988 and columns "proc1"-"p
You solution is not bad since it tries to make use of vectorized
operations, but there are some problems in it.
1) you should be using "&" instead of "&&" and "|" instead of "||"
(look at the help page to understand the difference.
2) FAQ 7.31: You are checking against floating point values (...
On Aug 26, 2011, at 10:06 AM, Rainer M Krug wrote:
On Fri, Aug 26, 2011 at 3:37 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
I don't have the FactoMineR library so I can't replicate, but it
seems that
your problem is that you never plot anything.
Effectively, your code runs
Dear all,
I'm trying to rerun some data linkage exercises in R (they are designed
to be done in SPSS, SAS or STATA)
The exercise in question is to relabel the column "treat" to "1", if
"yearsep" is smaller than 1988 and columns "proc1"-"proc3" contain the
values 56.36 or 59.81.
My pathetic s
On Fri, Aug 26, 2011 at 3:37 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> I don't have the FactoMineR library so I can't replicate, but it seems that
> your problem is that you never plot anything.
>
> Effectively, your code runs
>
> windows()
> x = 1:5
> x
>
> and then wondering
On Fri, Aug 26, 2011 at 3:45 PM, Iain Gallagher <
iaingallag...@btopenworld.com> wrote:
> Hi Giovanni
>
> Using Ubuntu and MacOSX may not be irrelevant. I use Ubuntu and if I carry
> out a fresh install (e.g. after a new release - although I've stuck with
> 10.04 so far) then I always have to mess
That's not the way to use logistic regression.
See
http://stats.stackexchange.com/questions/14803/graphically-presenting-model-fits-of-logistic-regression
for some other ideas.
Frank
Weidong Gu-2 wrote:
>
> Andra,
>
> You can transfer the probabilities into class membership by setting up
> a cu
Hi Giovanni
Using Ubuntu and MacOSX may not be irrelevant. I use Ubuntu and if I carry out
a fresh install (e.g. after a new release - although I've stuck with 10.04 so
far) then I always have to mess around, check the web etc to install external
packages that R libraries I want to use rely on.
On Aug 26, 2011, at 9:12 AM, David Winsemius wrote:
This arrived without a refers to in its headers and was a follow up
to:
On Aug 25, 2011, at 9:33 PM, Jim Silverton wrote:
Hi all,
I have some data. I want to fit a smooth cdf to the data. Then I
want to
find both the value of x and the
Oops..
You need to add the following line, right after the "m <- NCOL(y)"
statement:
k <- m * (m+1) / 2
'k' is the number of independent parameters in an m-by-m covariance
matrix, and two such matrices are estimated (Ht and Qt, both time
invariant). Hence the unknown parameter theta has length
I don't have the FactoMineR library so I can't replicate, but it seems that
your problem is that you never plot anything.
Effectively, your code runs
windows()
x = 1:5
x
and then wondering where your graph is.
Add plot(res.pca) and i think your problem will be solved.
Michael Weylandt
On Fri,
Hi,
I created a figure with R and I want to save it in .pdf. I used this code:
> pdf("res.pca.pdf",width=10,height=8)
> library(FactoMineR)
> res.pca<-PCA(acp)
> res.pca
> dev.off()
When I go in my folder, I find an empty file ( 0 Ko).
Do you know where is the problem.
Thank you in advance
On Fri, Aug 26, 2011 at 3:05 PM, Giovanni Petris wrote:
> Hi Rainer,
>
> This certainly helps, but it still requires to do some work by hand. I
> was hoping for something more automatic - but so far nobody has
> suggested a better approach.
>
Well - you could define two scripts, one which is exe
This arrived without a refers to in its headers and was a follow up to:
On Aug 25, 2011, at 9:33 PM, Jim Silverton wrote:
Hi all,
I have some data. I want to fit a smooth cdf to the data. Then I
want to
find both the value of x and the % on the y axis for which the the
slope is
1 ( or the p
Dear Prof. Viechtbauer,
thank you so much for your help and kindness.
Clearly graphs are the minor problem in our work, and the parameters and
options that can vary in R are so many that it is obvious that you can
not expect
to change everything you want!
Your suggestions are very helpuf, but I h
This seems to be the easiest way to handle the problem:
> a = xts(coredata(a), time(a))
> b = xts(coredata(b), time(b))
> merge(a,b)
ZWD.UGX SCHB.Close
2010-03-31 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03 7.5458 NA
2010-04-04 7
Hi Rainer,
This certainly helps, but it still requires to do some work by hand. I
was hoping for something more automatic - but so far nobody has
suggested a better approach.
Thank you,
Giovanni
On Thu, 2011-08-25 at 15:43 +0200, Rainer M Krug wrote:
>
>
> On Thu, Aug 25, 2011 at 3:25 PM, Gio
On Fri, Aug 26, 2011 at 7:27 AM, Jeff Newmiller
wrote:
> ".*" is greedy... might want regex "number[^0-9]*([0-9] {4})" to avoid
> getting 1999 from "I want the number 2000, not the number 1999."
If such inputs are possible we could also do this where we have added
a ? after the * to make the repe
Andra,
You can transfer the probabilities into class membership by setting up
a cut-off value, usually 0.5.
set.seed(3)
x<-rnorm(100)
y<-rbinom(100,prob=1/(1+exp(x)),size=1)
model<-glm(y~x,family=binomial(link=logit))
pred<-ifelse(predict(model,type='response')>0.5,1,0)
table(pred,y)
Weidong Gu
Diaz-Escamilla, Rafael E saclink.csus.edu> writes:
>
This question is really best for the r-sig-mixed-mod...@r-project.org
mailing list: please direct any follow-ups there.
> Data Collected My data consist of three levels: level 1 is four
> setting for each student (setting nested within s
I was rather too quick
It has probably something to do with versions of zoo and xts
after updating to zoo 1.7.4 and xts 0.8.2 I got with your examples
> merge(a,b)
ZWD.UGX SCHB.Close
2010-04-01 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03
Hi:
Try this:
require('xts')
merge.zoo(zoo(a), zoo(b), all = c(TRUE, TRUE))
ZWD.UGX SCHB.Close
2010-03-31 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03 7.5458 NA
2010-04-04 7.4532 28.30
2010-04-05 7.4040 28.38
2010-04-06
Hi
>
> On 26 August 2011 03:37, R. Michael Weylandt
wrote:
> > If you could, dput() them so we can see everything about them. You
also
> > might see if merge() gives you more expected behavior
>
> Ok...
> > dput(a)
> structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class =
c("
Hello,
I want to create 2 variables with normal distribution that the correlation
between them is equal to 0.8 and also each one has a serial correlation equal
to 0.6.
To generate variables with correlation between them I use:
t1<-10
N<-45
sigma2<-matrix(c(1,0.8*sqrt(1),0.8*sqrt(1),1),
".*" is greedy... might want regex "number[^0-9]*([0-9] {4})" to avoid getting
1999 from "I want the number 2000, not the number 1999."
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO
On 08/26/2011 09:14 AM, AdamMarczak wrote:
> Thank you all for suggestions, they were great and informative.
> I will surely use data.tables in future when our server will be upgraded for
> now this is solution that I used. This solution performs exactly same task
> and produces exact same result
Hello all,
I am currently writing a thesis in Value at Risk prediction and have to
model a data set of four stocks. I want to simulate 1 times from the
used D-vine and predict 300 new datapoints, but am not quiet sure how to do
this correctly. The produced matrix should have dimension N x numb
Thank you all for suggestions, they were great and informative.
I will surely use data.tables in future when our server will be upgraded for
now this is solution that I used. This solution performs exactly same task
and produces exact same results at ddply.
s <- split(past, paste(past$"CNTRY_N
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Coghlan, Avril
> Sent: 25 August 2011 16:02
> To: r-help@r-project.org
> Cc: Coghlan, Avril
> Subject: [R] within-groups variance and between-groups variance
>
> Hello,
>
> I ha
On 26 August 2011 03:37, R. Michael Weylandt wrote:
> If you could, dput() them so we can see everything about them. You also
> might see if merge() gives you more expected behavior
Ok...
> dput(a)
structure(c(7.6343, 7.6343, 7.5458, 7.4532, 7.404, 7.3317), class = c("xts",
"zoo"), .indexCLAS
Hmm, that's quite puzzling. I don't know but I'd willing to guess the
time/date stamps on a,b are more different than the output is leading us to
believe. My experience is that there's always more to a time/date object to
trick us up than one would expect.
If you could, dput() them so we can see e
I have two xts objects, call them "a" and "b", and am trying to merge them...
> class(a)
[1] "xts" "zoo"
> class(b)
[1] "xts" "zoo"
> head(a)
2010-04-01 7.6343
2010-04-02 7.6343
2010-04-03 7.5458
2010-04-04 7.4532
2010-04-05 7.4040
2010-04-06 7.3317
> head(b)
2010-04-01 568.80
2010-04-05
Thanks Jean.
This is really helpful; it does the job (almost) at least for the initial
analyses. However, I think I will need to do more research on the choropleth
after my initial analyses.
Best, Reza
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R-h
ISOweek is a small package which contains functions to substitute the %V and
%u formats which are not implemented on Windows.
In addition, the package offers functions to convert from standard calender
format -mm-dd to ISO 8601 week format -Www-d and vice versa.
Uwe Block
[[altern
Hi, I am running the non-parametric rasch model tests using eRm. I have a
reasonably large dataset for this type of exercise (110 items, 248 persons).
I run:
> allb2=as.matrix(allb)
> rsample <- rsampler(allb2, ctr)
> t102<-NPtest(rsample, method="T10") #global test, subgroup inv
and receive erro
Hi
I use this kind of output to excel if the table is not too big
write.table(tab, "clipboard", sep = "\t", row.names = F)
In excel sheet I press ctrl - V to copy from clipboard
Regards
Petr
> Using write.table i would like to save data as an excel file to a
folder. I
> am not too sure how
Hi
>
> Thanks, Jim. quote='' works. And then I found a single quote in each of
> these lines:
> 3262
> 10403
> 17544
> 24685
> 31826
> 38967
>
> None of them near the position the table got truncated. Why is it?
>
> And read.table is a great function. Is it possible for it to give a
warning
>
On Thu, Aug 25, 2011 at 9:51 PM, Lorenzo Cattarino
wrote:
> Apologies for confusion. What I meant was the following:
>
> mytext <- "I want the number 2000, not the number two thousand"
>
> and the problem is to select "2000" as the first four digits after the word
> "number". The position of 2000
You might want to take a look at 'regexpr' and/or 'gregexpr':
mytext <- "I want the number 2000, not the number two thousand"
idx <- regexpr("\\d{4}", mytext)
idx <- c(idx, (idx + attributes(idx)$match.length)-1)
substr(start=idx[1], stop=idx[2], mytext)
HTH,
Janko
On 26.08.2011 03:51, Lorenzo
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