On Jan 3, 2012, at 5:32 PM, Tal Galili wrote:
And I forgot to include the link to the image, here it is:
http://dl.dropbox.com/u/5371432/right-to-left-text%20example%202.png
I'm assuming this should be compared to the second of your three
examples.
The aleph ( ש ) is on the left end of b
It may be "mainstream" but it really isn't as easy to use or as universally
readable as plain text.
There are other forums to meet this "need", such as stackoverflow.com.
---
Jeff NewmillerThe ...
Hi
>
> hello sir,
> i m a student of bioinformatics,i hv a problem in R, i want
to
> do correlation analysis of microarray data,hw can i ignore the "NA"
value
> for the calculation or correlation.this is a very large data n many
empty
> fields they giving me NA values of correlatio
hello sir,
i m a student of bioinformatics,i hv a problem in R, i want to
do correlation analysis of microarray data,hw can i ignore the "NA" value
for the calculation or correlation.this is a very large data n many empty
fields they giving me NA values of correlation,wat i do??? pleas
Hi,
I am having trouble submitting a form using the RCurl package. My code runs
without errors but doesn't return the table I am after. I have a feeling it
is because there are 2 submit buttons on the page and I'm not hitting the
right one. I have tried everything I can think of to submit the form
Hi Aren,
Thanks for sending the data. I poked around a bit, and here was what
I came up with (a bit of a hack really, but perhaps acceptable
enough). lbl_formatter looks like another way to go.
require(ggplot2)
require(chron)
## hack to deal with non exported method
parse.format <- chron:::par
I imagine this is about as brief as one can be.
as.character(sapply(XXX, `[`, 1))
Michael
On Tue, Jan 3, 2012 at 9:44 PM, Dan Abner wrote:
> Hello everyone,
>
> Can someone please suggest the most succinct method for extracting the
> first value from each component of a list and saving the resu
Hello everyone,
Can someone please suggest the most succinct method for extracting the
first value from each component of a list and saving the resulting vector
as a character vector?
Thank you,
Dan
[[alternative HTML version deleted]]
__
R-h
Hello everyone,
I am also trying to do within subjects repeated measures anova followed by
the
test of sphericity (see sample dataset attached below) and I am facing some
problems.
I tried to assign the example to my research question, but could not really
solve it.
Here is the syntax:
> optio
Is it possible to allow HTML email on R-Help?
HTML email is mainstream and would really help with code markup and
embedded graphics.
Just a thought. It's frustrating dealing with the low usability of plain text.
Aren
__
R-help@r-project.org mailing li
Got it figured out. I found this post on the ggplot2 Google Group:
http://groups.google.com/group/ggplot2/browse_thread/thread/698e658b6dfec56c/5390824dab4a1cd7
It recommends you make this function:
lbl_formatter <- function(x) {
h <- floor(x/60)
m <- floor(x %% 60)
s <- round(60*(x %%
On Tue, Jan 3, 2012 at 4:03 PM, Carl Baribault wrote:
> Dear All,
>
> I've seen posts to the effect that..
> 1) choose.dir is only available for windows, and
> 2) tk_choose.dir would be the linux equivalent.
>
> I'm still having trouble with the subject package on linux/rhel6.
>
> I've specified -
Dear All,
I've seen posts to the effect that..
1) choose.dir is only available for windows, and
2) tk_choose.dir would be the linux equivalent.
I'm still having trouble with the subject package on linux/rhel6.
I've specified --with-tcltk during ./configure, and I still get the response...
> ins
> I want to make a histogram in R of the data in attached excel file called
> 'cbt'. However, I need the histogram to show a separation for Group 1 and
> Group 2, as in attached image.
>
> How do I do this in R? I know how to make a histogram for a single group,
> but how can I separate the 2 group
Hello Richard,
Thank you so much for getting back to me. In the ?glht example, the
confidence intervals are the same and the p-values are very similar.
I ran a 2-way ANOVA and compared the results for the glht code with
"Tukey" and TukeyHSD for "Treatment", which was a significant main
e
I was hoping for a solution without lattice.
But it solves the problem, thanks Jean.
Cheers,
Marius
On 2012-01-03, at 19:06 , Jean V Adams wrote:
>
> Try the levelplot() function in package lattice.
>
> library(lattice)
>
> dat <- expand.grid(x=x, y=y)
> dat$z <- pmax(f(dat$x) + f(dat$y
Ah, yes I see. Thanks John and Michael.
On Tue, Jan 3, 2012 at 6:06 PM, R. Michael Weylandt
wrote:
> There's a coercion to character implicit (i.e., the number 999 gets
> converted to the string "999") and then comparison is done in lexical
> (dictionary) order in which digits are lower than char
On Jan 3, 2012, at 5:53 PM, Rich Shepard wrote:
On Tue, 3 Jan 2012, David Winsemius wrote:
Maybe we need to backtrack a bit.
Yes. I've been trying to do this but still have too little
experience with
R to be successful on my own.
You originally were complaining about an error that sai
There's a coercion to character implicit (i.e., the number 999 gets
converted to the string "999") and then comparison is done in lexical
(dictionary) order in which digits are lower than characters.
You'll also note you get apparently strange behavior like "34" < "9"
if you don't think about thin
Dear Ista,
This is a consequence of coercion of the numbers to character:
> c("Z", "a", 999, Inf)
[1] "Z" "a" "999" "Inf"
> sort(c("Z", "a", 999, Inf))
[1] "999" "a" "Inf" "Z"
I hope this helps,
John
On Tue, 3 Jan 2012 17:56:29 -0500
Ista Zahn wrote:
> Hi all,
>
> I just discovered
Hi all,
I just discovered that R considers characters to be really big:
> "a" > 999
[1] TRUE
> "a" > 9e307
[1] TRUE
> "a" > 9e308
[1] FALSE
and that some characters are literally infinitely big:
> "Z" >= Inf
[1] TRUE
although not all:
> "a" > Inf
[1] FALSE
This came as a surprise to me (alt
On Tue, 3 Jan 2012, David Winsemius wrote:
Maybe we need to backtrack a bit.
Yes. I've been trying to do this but still have too little experience with
R to be successful on my own.
You originally were complaining about an error that said you had
duplicated index entries as you attempted t
FYI, if you're looking for the technical term for this type of text it's bidi:
http://en.wikipedia.org/wiki/Bi-directional_text
Hadley
On Tue, Jan 3, 2012 at 4:32 PM, Tal Galili wrote:
> And I forgot to include the link to the image, here it is:
>
> http://dl.dropbox.com/u/5371432/right-to-left-
And I forgot to include the link to the image, here it is:
http://dl.dropbox.com/u/5371432/right-to-left-text%20example%202.png
On Wed, Jan 4, 2012 at 12:30 AM, Tal Galili wrote:
> Thanks to an e-mail from David, I realized that non-Hebrew speakers will
> not be able to know how a proper r
Thanks to an e-mail from David, I realized that non-Hebrew speakers will
not be able to know how a proper right-to-left output should look like
(sorry for not thinking about it myself...)
Here is some example code of how the output should look like vs how it
currently looks.
This is shown only for
try this:
> library(sqldf)
> table1 <- read.csv(text = "POSTAL | VALUE
+ 1000|49
+ 1010|100
+ 1020|50", sep="|")
> table2 <- read.csv(text = "INSEE | POSTAL
+ A|1000
+ B|1000
+ C|1010
+ D|1020", sep="|")
> table3 <- sqldf("
+ select table2.INSEE
+ , 1.0 * table1.VALUE / counts.nPos
I'm not sure I know yet what I'm looking for. :-)
I have a list of all traffic tickets written in Dallas, TX over a few
years. One of the first things I am doing is reviewing the data to
make sure the quality is good; I'm trying to see whether it just looks
right.
One measure is whether the volum
right. replace dbetas with pbetas.
albyn
Quoting Duncan Murdoch :
On 03/01/2012 1:33 PM, Albyn Jones wrote:
What do quantiles mean here? If you have a mixture density, say
myf<- function(x,p0) p0*dbeta(x,2,6) + (1-p0)*dbeta(x,6,2)
then I know what quantiles mean. To find the Pth quan
The latest AmStatNews caught my eye with a list of student internships,
including 6 positions with Google
requiring 'high proficiency with R' among other things.
www.google.com/intl/en/jobs/students/tech/internships/uscanada
--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psy
Hi,
I have following 2 tables:
Table 1:
POSTAL | VALUE
1000|49
1010|100
1020|50
Table 2:
INSEE | POSTAL
A|1000
B|1000
C|1010
D|1020
I would like to convert this to the following:
INSEE | VALUE_SPREAD
A|24.5
B|24.5
C|100
D|50
I can achieve this with a nested SQL query (through counting the
num
Hello,
Thank you very much for your prompt replies!
@Jean V Adams: yes, that looks a lot like the graph that I want! I'm
wondering though if there isn't an easier way (i.e. less code) to do this?
Like, for instance, using the stackpoly function (but modified, so that a
vertically stacked plot can
Thanks, Rolf, Justin and Uwe!
Actually I wanted to run .R file as a script, just like what people do
for bash scripts or python scripts. It seems to me that adding
"#!/usr/bin/R -f" at the first line is what I need. Is this true?
Li Sun
2012/1/2 Rolf Turner :
> On 03/01/12 17:02, Li SUN wrote:
Hi,
I am curious if you know about any book that is dealing with the web
analytics / customer analytics subject and is referencing R as the main
statistical tool. I am particularly interested into using R in the real
production environment and not only as the analytical tool.
Thank you
Jan
If one attempts to install RODBC (via install.packages('RODBC'))
without having an ODBC driver installed, this error message results:
checking sqlext.h presence... no
checking for sqlext.h... no
configure: error: "ODBC headers sql.h and sqlext.h not found"
ERROR: configuration failed for package '
Dear Spencer, Dear Ted,
thanks for your quick and detailled answers.
I should have indicated that I use the Ryacas package - for example:
library(Ryacas)
x <- Sym("x")
Simplify(deriv(x^3),x,2))
Just now I have realized that the deriv function is translated
to the Deriv function in yacas, which
Am Dienstag, 3. Januar 2012, 08:50:44 schrieb VictorDelgado:
> VictorDelgado wrote
>
> > quantile(x)
>
> Correcting to
>
> quantile(q)
>
> -
Dear Victor,
thank you for your answer.
Best,
Gerhard
> Victor Delgado
> cedeplar.ufmg.br P.H.D. student
> www.fjp.mg.gov.br reseacher
> --
> Vie
Maybe we need to backtrack a bit. You originally were complaining
about an error that said you had duplicated index entries as you
attempted to make a zoo object. I assumed, incorrectly it now
appears, that you understood that an index in a zoo object was a
vector. You now seem to be admit
Dear Gerhard,
you could also use package "distr"; e.g.
library(distr)
## use generating function "AbscontDistribution"
D <- AbscontDistribution(d = function(x) dbeta(x, 2, 6) + dbeta(x,6,2),
low = 0, up = 1, withStand = TRUE)
## quantiles
q(D)(seq(0,1,0.1))
Best
Matthias
On 03.01.2012 19:3
On Tue, 3 Jan 2012, David Winsemius wrote:
That's rather unconvincing. What does this show:
burns.tds[ !duplicated(burns.tds) ]
I saw that in the help page but assumed it was the opposite of duplicated;
apparently not.
burns.tds[ !duplicated(burns.tds) ]
Error in .data.frame(burns.tds, !du
On 03/01/2012 1:33 PM, Albyn Jones wrote:
What do quantiles mean here? If you have a mixture density, say
myf<- function(x,p0) p0*dbeta(x,2,6) + (1-p0)*dbeta(x,6,2)
then I know what quantiles mean. To find the Pth quantile use uniroot
to solve for the x such that myf(x,p0) - P =0.
You
--- begin included messge ---
Assume that we collect below data : -
subjects = 20 males + 20 females, every single individual is
independence,
and difference
events = 1, 2, 3... n
covariates = 4 blood types A, B, AB, O
http://r.789695.n4.nabble.com/file/n4245397/CodeCogsEqn.jpeg
?m = hazards ra
What do quantiles mean here? If you have a mixture density, say
myf <- function(x,p0) p0*dbeta(x,2,6) + (1-p0)*dbeta(x,6,2)
then I know what quantiles mean. To find the Pth quantile use uniroot
to solve for the x such that myf(x,p0) - P =0.
albyn
Quoting VictorDelgado :
Gerhard wrot
Lixia:
This is not fundamentally an R issue. You seem to have some confusion
about what IC50's and ED50's are in the statistical modeling context.
I suggest you consult one of the fine statisticians at MD Anderson to
help you resolve your confusion.
Cheers,
Bert
On Tue, Jan 3, 2012 at 10:11 AM,
Hi,
I am trying to use drc package to calculate IC50 value. The ED50 calculated in
some models (LL4 for example) as a response half-way between the upper and
lower limit, which is the definition of the relative IC50 value. Does that mean
the ED50 in drc package is IC50? How the ED function in d
Brett Magill sbcglobal.net> writes:
>
> Hello all,
>
> To anyone who is interested, I'm trying to learn a bit more about
> developing applications in R with user interfaces. I've been playing
> around with gWidgets to develop a model building interface.
>
> I'd appreciate any comments, sugge
Try the levelplot() function in package lattice.
library(lattice)
dat <- expand.grid(x=x, y=y)
dat$z <- pmax(f(dat$x) + f(dat$y) - 10, 0)
levelplot(z ~ x * y, dat,
at=c(-1, 0.02, 1, 5, 10, 20, 50, 500, 900),
labels=TRUE, contour=TRUE, colorkey=FALSE,
col.regions=gr
On Jan 3, 2012, at 12:48 PM, Rich Shepard wrote:
On Tue, 3 Jan 2012, Rich Shepard wrote:
I _think_ the problem comes from a duplicated factor column in the
data
frame. Now I need to figure out how subset() generated that
additional
column.
Nope. That's not it.
Running 'duplicated(bur
On Jan 3, 2012, at 12:26 PM, Rich Shepard wrote:
On Tue, 3 Jan 2012, David Winsemius wrote:
How can I identify the non-unique index entries within R?
?duplicated
Thank you, David.
I _think_ the problem comes from a duplated factor column in the data
frame. Now I need to figure out how
On Tue, 3 Jan 2012, Rich Shepard wrote:
I _think_ the problem comes from a duplicated factor column in the data
frame. Now I need to figure out how subset() generated that additional
column.
Nope. That's not it.
Running 'duplicated(burns.tds, incomparables = FALSE)' produces a listing
of
I have had clients who also wanted to make little changes to the graphs (mostly
changing colors or line widths). Most after doing this a couple of times have
been happy to give be better descriptions of what they want so I can just do it
correctly the first time.
I mostly give them the graph
Hello Majid,
When you say the text renders correctly on one but not the other, you mean
the fonts, or the directionality?
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew)
VictorDelgado wrote
>
>
> quantile(x)
>
>
Correcting to
quantile(q)
-
Victor Delgado
cedeplar.ufmg.br P.H.D. student
www.fjp.mg.gov.br reseacher
--
View this message in context:
http://r.789695.n4.nabble.com/calculate-quantiles-of-a-custom-function-tp4256887p4257575.html
Sent from the R
Gerhard wrote
>
>
> Suppose I create a custom function, consisting of two beta-distributions:
>
> myfunction <- function(x) {
> dbeta(x,2,6) + dbeta(x,6,2)
> }
>
> How can I calculate the quantiles of myfunction?
>
> Thank you in advance,
>
> Gerhard
>
>
Gehard, if do you want to know
Hello,
thanks for the quick answers! Some answers were more useful to me than
others but I really appreciate everybodys effort, reading and
answering my question.
I will try Dons approach because as he said, it looks quite readable.
I'll need to take a closer look on the nested ifelse statements
Dear community,
I'm trying to model growth with this function: Yi = A* exp(-k*(1/ti^m)) ; A
asymptote, k rate of decrease of the relative growth rate, m shape
parameter.
I don't have variable time so, finally, following some papers, I try to fit
Yi+a = A*exp(-k* (1/(-k/(log(Yi/A)))^(1/m)+a)^
On Tue, 3 Jan 2012, David Winsemius wrote:
How can I identify the non-unique index entries within R?
?duplicated
Thank you, David.
I _think_ the problem comes from a duplated factor column in the data
frame. Now I need to figure out how subset() generated that additional
column.
Regard
On Jan 3, 2012, at 11:45 AM, Rich Shepard wrote:
I have a situation that I cannot resolve by myself. When I try to
create a
zoo object (with read.zoo() ) I get this error:
Error in merge.zoo(`BC-0.5 = c(" 0.000", " 0.010", " 0.010", "
0.060", :
series cannot be merged with non-uniqu
I have a situation that I cannot resolve by myself. When I try to create a
zoo object (with read.zoo() ) I get this error:
Error in merge.zoo(`BC-0.5 = c(" 0.000", " 0.010", " 0.010", "
0.060", :
series cannot be merged with non-unique index entries in a series
This suggests that th
Ok, so just for anyone's interest, I managed to create the calibration plot for
the glmnet object using the val.prob() function from the rms package.
Now, my question moves slightly, how can I superimpose calibration curves from
two models, so that they can be graphically compared?
This is wha
1. Are you organizing your project into an R package? If no, I
suggest you do so, because doing so typically improves your software
development productivity by providing a structured way to organize your
documentation and testing -- which also means you tend to get more
trustworthy sof
I suggest you look at the help pages for "expression" and "evel", then
consider the following example in addition to the function "DD" in the
"examples" for "deriv":
Dxy <- deriv( ~x*y*z, c('x', 'y'))
x <- 2; y <- 3; z <- 4
eval(Dxy)
Dxy.fn <- deriv(expression(x*y*z), c('x', 'y'), TRUE)
Dxy.fn
Am Dienstag, 3. Januar 2012, 11:05:11 schrieben Sie:
>
> The quick way is to look at the structure with 'str':
>
> str(integrate(myfunction,0,.9))
> List of 5
> $ value : num 1.85
> $ abs.error : num 2.05e-14
> $ subdivisions: int 1
> $ message : chr "OK"
> $ call: l
Sorry, that should've been sum(diag(D)) or max(eigen(D)$values) in stead of
max(diag(D)).
Tsjerk
On Jan 3, 2012 4:52 PM, "Tsjerk Wassenaar" wrote:
Hi Riccardo,
Would it be possible to use max(diag(D))*diag(ncol(D)) - D ? That also
reverses the order of eigenvalues/-vectors.
Cheers,
Tsjerk
>
Hi Riccardo,
Would it be possible to use max(diag(D))*diag(ncol(D)) - D ? That also
reverses the order of eigenvalues/-vectors.
Cheers,
Tsjerk
On Jan 2, 2012 4:35 PM, "riccardo24" wrote:
Hi, I need to maximize a quadratic function under constraints in R.
For minimization I used solve.QP but f
>...(1000,0,1), I presume.
Yes thank you.
...z <- nls(y ~ rbinom(1000,1,a+b*x),data=d,start= list(a =0.1,b=0.2),trace=T);
>This makes no sense. Random numbers in a model specification
And maybe that model spec is nonsense, although it seems to me that it
correctly incorporates the id
On Jan 3, 2012, at 7:24 AM, Gerhard wrote:
Hi,
I guess that my problem has an obvious answer, but I have not been
able to
find it.
Suppose I create a custom function, consisting of two beta-
distributions:
myfunction <- function(x) {
dbeta(x,2,6) + dbeta(x,6,2)
}
Given the symmetry
Gerhard:
Strictly speaking, it's quantiles of a custom "distribution", not function.
There may be some way to handle your example easily, but, in general,
you would need to solve the resulting integral equation. This is hard
-- closed form solutions rarely exist; good approximations require
work.
It's below. And I meant offense_time, not offense_hour.
> dput(dallas[1:40, "offense_time", drop = FALSE])
structure(list(offense_time = structure(c(0.3895833,
0.336, 0.4270833, 0.4097222, 0.4375,
0.6486111, 0.4715278, 0.561, 0.01
Hi,
i tried to find the answer but didn't so my apologies if the question is
obvious !
I'm trying to parallelize the following R code :
pk2test =
c(1:16,(12*16+1):(12*16+16),(16*16+1):(16*16+16),(20*16+1):(20*16+16))
score.mat = matrix(nc=16*4,nr=16*4)
for(i in 1:(16*4)) {
Hi,
I guess that my problem has an obvious answer, but I have not been able to
find it.
Suppose I create a custom function, consisting of two beta-distributions:
myfunction <- function(x) {
dbeta(x,2,6) + dbeta(x,6,2)
}
How can I calculate the quantiles of myfunction?
I have not seen any c
Dear Jorge,
I appreciate your help.
I didn't forget this arguments, I just didn't know it ;)
Now I adjust my code and it works well.
Thanks a lot!
If I can have one more question:
g <- rep(c(1, 2), each = 1000)
what if the number of values is not equal?
Przemek
--
View this message in conte
hello barry,
thank you very much for your help!
we managed to do what you said and it basicaly worked - there only is a
problem with our file, but we will create the file again and then it should
work completely. i will keep you posted how it turned out. thanks again!
marion
2012/1/2 Barry Ro
See in-line below.
On 03-Jan-2012 alexander16 wrote:
> Dear everyone,
> the following is obviously used to compute the nth derivative,
> which seems to work
> (deriv(sqrt(1 - x^2),x,n))
Well, it doesn't seem to work for me! In fact:
n <- 2
(deriv(sqrt(1 - x^2),x,n))
# Error in deriv(sqrt(1
On 02.01.2012 23:11, Adedoyin-Olowe Mariam wrote:
Hello,
Can you please, as a matter of urgency, tell me which R version support
tm.plugin.sentiment
This question should go to the maintainer, since this is not a CRAN nor
a BioC package, as far as I can see ...
and how I can install the
Dear everyone,
the following is obviously used to compute the nth derivative, which seems
to work
(deriv(sqrt(1 - x^2),x,n))
However, before using this, I wanted to make sure it does what I think it
does
but can't figure it out when reading the ?deriv info or any other
documentation on deriv for
On 03.01.2012 07:50, Rolf Turner wrote:
On 03/01/12 17:02, Li SUN wrote:
Hello,
I am a beginner to the R language and find it fantastic and
well-designed, quite different from other programming languages.
What a refreshingly sensible attitude!!! :-)
This is the first time I post on the r-he
Dear all,
seasons greetings,
I had made a list of elements that I want to have
# list generator
Agent<-list(id=NA, position=list(x=NA,y=NA))
# Make list of Agents
AgentList<-rep(list(CRAgent),5)
# Initialization. Is it shorter to do that with a lapply statement?
for (i in c(1:length(CRAgentList
Nameless,
http://had.co.nz/ggplot2/geom_histogram.html has a few examples
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
thierry.
I want to make a histogram in R of the data in attached excel file called
'cbt'. However, I need the histogram to show a separation for Group 1 and
Group 2, as in attached image.
How do I do this in R? I know how to make a histogram for a single group,
but how can I separate the 2 groups?
http:/
On Jan 3, 2012, at 05:25 , G Vishwanath wrote:
> I am trying to learn nls using a simple simulation. I assumed that the
> binomial prob varies linearly as 0.2 + 0.3*x in x {0,1},
> and the objective is to recover the known parameters a=0.2, b=0.3
>
> ..data frame d has 1000 rows...
>
> d$x<-
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