Hi
>
> Re: [R] two question about plot
>
> The help for boxplot offers suggestions for both those things. You may
be
> particularly interested in:
>
>names: group labels which will be printed under each boxplot. Can
> be a character vector or an expression (see plotmath).
>
>
Andra Isan yahoo.com> writes:
>
> Hi All,
> In order to compare two different logistic regressions,
> I think I need to compare them based on their BIC
> values, but I am not sure if the smaller BIC would mean a better
> model or the reverse is true?
> Thanks a lot,Andra
Smaller (i.e. lowe
- This mail is in HTML. Some elements may be ommited in plain text. -
Meu nome é Frank P, eu trabalhei com o Banco SNS em Holland.I habe Você
contactado por conta de parente falecido com os nossos Banj e que foi deixado
sem pretensões.
Por favor, me envia um email com um número de telefone e ta
Dear R help group
I've been working on this plot for a while now and now getting around to the
minor adjusments. I would like to be able to put a border and background
fill around the legend in this plot.
I understand the legend 'bty' should do this have this capablity but not
sure how the syn
Friends
I am starting on a (section of the) project where I need to build a matrix
with on the order of 5 million rows and 200 columns
I am wondering if I can stay in R.
I need to do rollapply type operations on the columns, including some that
will be functions of (windows of) two columns.
I h
?try
Sent from my iPad
On Sep 1, 2011, at 22:46, Jie TANG wrote:
> hi ,when i read a lots of files
>
> for (i in 1:totnum)
> {
> cop_x_data<-read.table(flnm[i],skip=2)
> if(i==1) {cop_data=cop_x_data}
> else {cop_data=rbind(cop_data,cop_x_data)}
> }
>
> some of the files are missing . so this
On Sep 1, 2011, at 10:46 PM, Jie TANG wrote:
hi ,when i read a lots of files
for (i in 1:totnum)
{
cop_x_data<-read.table(flnm[i],skip=2)
if(i==1) {cop_data=cop_x_data}
else {cop_data=rbind(cop_data,cop_x_data)}
}
some of the files are missing . so this loop can not go on .I wonder
how can
hi ,when i read a lots of files
for (i in 1:totnum)
{
cop_x_data<-read.table(flnm[i],skip=2)
if(i==1) {cop_data=cop_x_data}
else {cop_data=rbind(cop_data,cop_x_data)}
}
some of the files are missing . so this loop can not go on .I wonder how can
I go on
the loop cycle while reading the files jus
Actually, it's very easy to integrate a function of two variables in a
single variable for a given value of the other variable.
Using your example:
MySum <- function(x,y) {
ans = x + y
return(ans)
}
Note a things about how I wrote this. One, I broke the function out and used
curly brac
Thanks for your reply Michael, it seems I have a lot of things to
learn yet but for sure, your response is being very helpful in this
proccess. I will try to explore every point you said:
A doubt I have is, if I define "func <- function(x,y) x + y" how can I
integrate it only in "x"? My solution f
Thanks David and Jorge for your comments!
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Hi,
I have created two objects: object c contains yearly "distances" between
cases and object g contains yearly interactions between cases. For each case
and every year I would like to calculate the following value:
Vit = sum(Dabt * Iait * Ibit)
Where Vit is the value of case i in year t, Dabt i
Anybody knows where I can find the MSVAR for ox package mentioned above? The
websited is not working anymore :(
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I'm not clear on what you would use that for, but you can use approx(original
prob from calibrate, calibrated prob from calibrate, xout=vector of original
predicted values)$y to get this.
Frank
yz wrote:
>
> Thanks Frank
>
> I got the predicted probability.
>
> But can I get the bootstrap corre
Hi All,
In order to compare two different logistic regressions, I think I need to
compare them based on their BIC values, but I am not sure if the smaller BIC
would mean a better model or the reverse is true?
Thanks a lot,Andra
[[alternative HTML version deleted]]
__
Hi,
Are you after this?
last_plot() + opts(aspect.ratio=1)
Also, see https://github.com/hadley/ggplot2/wiki/Themes for some
settings re: plot margins.
HTH,
baptiste
On 1 September 2011 05:18, Alaios wrote:
> Dear all,
> I am using ggplot with geom_tile to print as an image a matrix I have.
Hello,
I am wanting to run a simulation study in R comparing several different
bandwidth selection methods for data simulated from several different
distribution types (normal, lognormal, bimodal, etc.) and wanted to know how
to calculate the mean integrated square errors for the optimal smoothin
The sys.call or match.call functions may be what you are looking for:
> tmpfun <- function(x,y,z,...) {
+ as.list( sys.call() )
+ }
>
> tmpfun( x=5, w=3, 1:10 )
[[1]]
tmpfun
$x
[1] 5
$w
[1] 3
[[4]]
1:10
> tmpfun2 <- function(x,y,z,...) {
+ as.list( match.call() )
+ }
> tmpfun2( x=5, w=3, 1:10
Dang Jim this looks to do the trick though I never heard of a data.table,
interesting, I will explore more. Thanks you very much.
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Thursday, September 01, 2011 11:20 AM
To: ROLL Josh F
Cc: r-help@r-project.org
Subj
On Sep 1, 2011, at 2:59 PM, Dan Abner wrote:
Hello everyone,
I have the following factor:
levels(pp_income)
[1] "" "1" "2" "3" "4" "5" "6" "7"
[9] "8" "9" "Renter"
I want to subset so that only values 1:9 are included. I have the
following:
Dropping all occurences of a factor does not drop that level. This actually
turns out to be much more useful than it first might appear, but if you
really need to get around it, it can be done.
Look at this toy example:
R> x = factor(c("A","B","C","A","B","C","C"))
R> x
[1] A B C A B C C
Levels:
Thanks, Don!
Appreciated your detailed explanation.
On Thu, Sep 1, 2011 at 11:28 AM, MacQueen, Don wrote:
> Even though it's not needed, here's a small followup.
>
> I usually use this
> split(x, paste(x$let,x$g))
>
> But since
>split(x, list(x$let,x$g))
> works, so does
> split(x, x
Hello everyone,
I have the following factor:
levels(pp_income)
[1] "" "1" "2" "3" "4" "5" "6" "7"
[9] "8" "9" "Renter"
I want to subset so that only values 1:9 are included. I have the following:
> income<-pp_income[pp_income %in% c(1:9)]
>
> leve
Even though it's not needed, here's a small followup.
I usually use this
split(x, paste(x$let,x$g))
But since
split(x, list(x$let,x$g))
works, so does
split(x, x[,c('let','g')])
> all.equal( split(x, x[,c('let','g')]) , split(x,list(x$let,x$g)))
[1] TRUE
As to which is the best, hard t
Is this close to what you are asking for:
> require(data.table)
> Dt.. <- data.table(Df..)
> R <- Dt..[,
+ list(
+ sum = sum(Volume)
+ , weight = sum(Volume * Mph) / sum(Volume)
+ )
+ , by = list(Min5Break, Day, Hour, Dir)
+ ]
> R
Min5Break Day Hour Dir
Thanks for the great helps from David, Jim and Liviu. It solved my problem.
Appreciated!
On Thu, Sep 1, 2011 at 11:01 AM, David Winsemius wrote:
>
> On Sep 1, 2011, at 1:53 PM, Changbin Du wrote:
>
> HI, Dear R community,
>>
>> I want to split a data frame by using two variables: let and g
>>
>
The help for boxplot offers suggestions for both those things. You may be
particularly interested in:
names: group labels which will be printed under each boxplot. Can
be a character vector or an expression (see plotmath).
add: logical, if true _add_ boxplot to current plot.
Sa
On Sep 1, 2011, at 1:53 PM, Changbin Du wrote:
HI, Dear R community,
I want to split a data frame by using two variables: let and g
x = data.frame(num =
c(10,11,12,43,23,14,52,52,12,23,21,23,32,31,24,45,56,56,76,45), let =
letters[1:5], g = 1:2)
x
num let g
1 10 a 1
2 11 b 2
3
On Thu, Sep 1, 2011 at 7:53 PM, Changbin Du wrote:
> HI, Dear R community,
>
> I want to split a data frame by using two variables: let and g
>
It's not clear what you want to do, but investigate the following:
> require(plyr)
Loading required package: plyr
> ddply(x, .(let, g), function(y) mean(
try this:
> split(x, list(x$let, x$g))
$a.1
num let g
1 10 a 1
11 21 a 1
$b.1
num let g
7 52 b 1
17 56 b 1
$c.1
num let g
3 12 c 1
13 32 c 1
$d.1
num let g
9 12 d 1
19 76 d 1
$e.1
num let g
5 23 e 1
15 24 e 1
On Thu, Sep 1, 2011 at 1:53 PM,
On Thu, Sep 1, 2011 at 5:48 PM, Jean V Adams wrote:
> Try this
>
> data <- eval(parse(text=paste(study, level, ".", population, sep="")))
>
..or this:
data <- get(paste(study, level, ".", population, sep=""))
Liviu
> Jean
>
> -
>
> dbateman wrote on 08/31/2011 17:44:44:
>
> I have several d
HI, Dear R community,
I want to split a data frame by using two variables: let and g
> x = data.frame(num =
c(10,11,12,43,23,14,52,52,12,23,21,23,32,31,24,45,56,56,76,45), let =
letters[1:5], g = 1:2)
> x
num let g
1 10 a 1
2 11 b 2
3 12 c 1
4 43 d 2
5 23 e 1
6 14 a 2
7
Leaving aside some other issues that this whole email chain has opened up,
I'd guess that your most immediate problem is that you are trying to
numerically integrate the PMF of a discrete distribution but you are
treating it as a continuous distribution. If you took the time to properly
debug (as
On Sep 1, 2011, at 10:59 AM, Li, Yan wrote:
Hi All,
Does anyone has experiences installing R in iSeries?
I was unable to find any postings to r-help that mentioned that
hardware line in particular. But Linux and (IBM's) AIX are capable of
supporting R and both those OSes run on iSeries b
I would suggest using a 'for' loop
rather than an apply function. The
advantage is that you will probably
understand the loop that you write,
and it will run in roughly the same
amount of time as a complicated call
to an apply function that you don't
understand.
On 01/09/2011 18:11, LCOG1 wrote:
On Thu, 2011-09-01 at 17:36 +0100, Prof Brian Ripley wrote:
> readBin is intended to read a few items at a time, not 10^9. You are
> probably getting 32-bit integer overflow inside your OS, since the
> number of bytes you are trying to read in one go exceeds 2GB.
>
> Don't do that: read say a m
1) how to modify the the tickment of x-axis or y-axis.
boxplot(data[,1:5])
the tickment in x-axis in V1 V2 V3 V4 V5 ,I want to be some name for
example
name<-c("1day","2day","3day","4day","5day")
2) how to overlap two plot into one figure?
plot(data[1:5])
boxplot(newdata[,1:5])
?
--
TAN
Hi guys,
I have a crap load of data to parse and have enjoyed creating a script that
takes this data and creates a number of useful graphics for our area. I am
unable to figure out one summary though and its all cause I dont fully
understand the apply family of functions. Consider the following:
You could also use match() directly instead
of going through factors. Any of the following
would map your inputs to small integers
> match(x, x)-1
[1] 0 1 0 3 0 5 0 7 8 9
> match(x, unique(x))-1
[1] 0 1 0 2 0 3 0 4 5 6
> match(x, sort(unique(x)))-1
[1] 3 4 3 6 3 2 3 0 5 1
Your num
On Thu, 1 Sep 2011, Prof Brian Ripley wrote:
readBin is intended to read a few items at a time, not 10^9. You are
probably getting 32-bit integer overflow inside your OS, since the number of
bytes you are trying to read in one go exceeds 2GB.
Don't do that: read say a million at time.
And B
So, please excuse me Michael, you are completely sure. I will try
describe I am trying to do, please let me know if I can provide more
info.
The idea is provide to "func" two probability density functions(PDFs)
and obtain another PDF that is a compound of them. In a final analysis
this characteriz
readBin is intended to read a few items at a time, not 10^9. You are
probably getting 32-bit integer overflow inside your OS, since the
number of bytes you are trying to read in one go exceeds 2GB.
Don't do that: read say a million at time.
And BTW, if these really are unsigned ints you will
This can be done using bwplot in lattice library. Also, it is better
to organize your data in 'long' format. Look at functions reshape or
melt in reshape library.
Weidong Gu
On Thu, Sep 1, 2011 at 12:10 PM, Jie TANG wrote:
> hi
>
> i have a dataframe with the name "obsdata"
> V1 V2 V3 V4
try this:
> x <- read.table('clipboard', header = TRUE)
> x
V1 V2 V3 V4 V5V6V7V8V9 V10 V11 V12 V13
1 1001 3 24 12 24.7 44.4 70.1 49.3 33.7 3.0 6.8 2.7 NA
2 1001 3 25 0 70.1 49.3 33.7 138.2 152.5 NA 4.2 6.9 17.5
3 1001 3 25 12 33.7 187.7 286.5 386.7N
On Sep 1, 2011, at 10:54 AM, Thomas Chesney wrote:
I have a text file full of numbers (it's a edgelist for a graph) and
I would like to recode the numbers as they are way too big to work
with. So for instance the following:
6765290986671000198767829
676529098667100867672856227
6765
hi
i have a dataframe with the name "obsdata"
V1 V2 V3 V4V5V6V7V8 V9 V10 V11 V12 V13
11001 3 24 12 24.7 44.4 70.1 49.3 33.7 3.0 6.8 2.7NA
21001 3 25 0 70.1 49.3 33.7 138.2 152.5NA 4.2 6.9 17.5
31001 3 25 12 33.7 187.7
On 01/09/2011 11:29 AM, aelmore wrote:
Hi,
For the first time, I am trying to call/run an R script from another
program, passing parameters and results back and forth. I have been
learning a little programming as I have progressed through the various
projects I have been working on, but I haven
I have a text file full of numbers (it's a edgelist for a graph) and I would
like to recode the numbers as they are way too big to work with. So for
instance the following:
6765290986671000198767829
676529098667100867672856227
67652909866791098726278
67652909866798928373
10928373
Kindly provide a reproducible example.
2011/9/1 Jim Maas :
> Intermittently I'm getting this error from the geweke.diag function of the
> coda package. Would anyone be kind enough to enlighten me as to the
> possible source of such an error, or how to debug/locate it?
>
> Error in { : task 22 fai
Hi Jim,
yes - it definitely is 64 bit. I call it with r64 and
.Platform$r_arch
[1] "x86_64"
It is on a apple snow leopard (10.6.8) with 16 GB of Ram - not windows
Florian
On 1 Sep 2011, at 16:22, jim holtman wrote:
> Are you running a 64-bit version of R? It sounds like your operating
> sy
it's me, Florian (just subscribed to R list, but my messages are still held for
approval )
fail means: It executes without an error, allocates the specified length (n=)
as zeroes, but does not actually read any data in.
florian
On 1 Sep 2011, at 16:30, Duncan Murdoch wrote:
On 01/09/2011 11:
Hi,
For the first time, I am trying to call/run an R script from another
program, passing parameters and results back and forth. I have been
learning a little programming as I have progressed through the various
projects I have been working on, but I haven't had any formal training and
am now stu
Can anyone suggest a package or code for modeling a hysteresis process in R?
I'm currently modeling a certain dataset with a GAM using mgcv, something
like
gam(y~ s(x, by=z) + z, family = Gamma(link=log),data=data)
and getting fits with about 9 estimated degrees of freedom in the smooth for
eac
Try this
data <- eval(parse(text=paste(study, level, ".", population, sep="")))
Jean
-
dbateman wrote on 08/31/2011 17:44:44:
I have several datasets that come from different studies (fv02 and fv03),
they represent different levels (patients and lesions), and they have
different patient
On 01/09/2011 11:13 AM, Benton, Paul wrote:
Posting for a friend
What does "fails" mean, i.e. what is the error message? (You might want
to get Florian online here.)
Duncan Murdoch
Begin forwarded message:
From: "Geier,
Florian"mailto:florian.geie...@imperial.ac.uk>>
Subject: Fwd: read
nice. thank you all.
R is a great tool and R users are also so kind .
2011/9/1 David Winsemius
>
> On Sep 1, 2011, at 3:45 AM, Jie TANG wrote:
>
> thank you. it works.
>> but further question is that if we can let the " flnm" to be a
>> 2-dimension
>> matrix [3,5]?
>> since
>> mtdno<-paste("da
On 01/09/2011 11:14 AM, Jannis wrote:
Dear list,
I am wondering whether there is an (easy) way to access all arguments and their
values passed to a function inside this function and (for example) store them
in a list object?
I could imagine using ls() inside this function and then looping th
Perfect! Thank you!
S
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Wednesday, August 31, 2011 4:44 PM
To: Shane Phillips
Cc: r-help@r-project.org
Subject: Re: [R] Converting anova/ancova summary to data frame
On 31.08.2011 22:33, Shane Phillips
Are you running a 64-bit version of R? It sounds like your operating
system is not giving you enough memory. It looks like this is not
under Windows in a native mode.
On Thu, Sep 1, 2011 at 11:13 AM, Benton, Paul
wrote:
> Posting for a friend
>
> Begin forwarded message:
>
> From: "Geier, Flori
Intermittently I'm getting this error from the geweke.diag function of
the coda package. Would anyone be kind enough to enlighten me as to the
possible source of such an error, or how to debug/locate it?
Error in { : task 22 failed - "inner loop 1; cannot correct step size"
Thanks a bunch.
J
Posting for a friend
Begin forwarded message:
From: "Geier, Florian"
mailto:florian.geie...@imperial.ac.uk>>
Subject: Fwd: readBin fails to read large files
Date: September 1, 2011 4:10:53 PM GMT+01:00
To:
Begin forwarded message:
Date: 1 September 2011 16:01:45 GMT+01:00
Subject: readBin fa
Dear list,
I am wondering whether there is an (easy) way to access all arguments and their
values passed to a function inside this function and (for example) store them
in a list object?
I could imagine using ls() inside this function and then looping through all
names and assigning list entr
Hi
I do not know sp package therefore I can not help you. Anyway without
knowing "what is wrong" and without some reproducible example you hardly
get any reasonable answer.
It seems to me that your d list especially x and y is not what you expect
to be.
see
str(d)
430071.4887:460006.8067
Hi All,
Does anyone has experiences installing R in iSeries? Does R supports iSeries?
Any documentation on this topic? Thank you very much!
Regards,
Yan
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
https://stat.et
I'm going to try to put this nicely:
What you provided is not a problem with integrate. Instead, you provided a
rather unintelligible and badly-written piece of code that (miraculously)
seems to work, though it's not well documented so I have no idea if 1.3e-21
is what you want to get.
Let's try
Dear list,
I just tried to do the same thing, and did not find anything on a
weighted qqplot. My weights are actually counts (positive integers).
Here is a modification of qqplot, following Duncan Murdoch's
suggestion. Any feedback would be welcome!
Thanks,
Jean-Christophe
weighted.qqplot <- func
Thank you
I want to make a map from my spatial data that I can show it in GIS.
I used the package sp and
d<- list(x=430071.4887:460006.8067, y =3390040.0591:3420006.2701, z =
matrix(mat, 970, 960))
gt<- GridTopology(cellcentre.offset = c(d$x[1], d$y[1]),
cellsize=c(diff(d$x[1:2]), diff(d$y[1
Hi Michael,
This is the problem:
func <- Vectorize(function(x, a, sad, samp="pois", trunc=0, ...) {
result <- function(x) {
f1 <- function(n) {
f <- function() {
dcom <- paste("d", sad, sep="")
dots <- c(as.name("n"), list(...))
do.call(dcom,
Hi
> Thank you Petr
>
> It work.
>
> Now I have a matrix 970*960. If I want to convert to spatial grid (each
> pixel has x and y coordinate).
> How can I do?
I do not understand. What do you want to do with your data? Maybe you
could consult spatial package or CRAN Task views. One option c
Bioconductor has the pcaMethods package, with multiple options for PCA,
including a robust L1-norm SVD.
Kevin Wright
On Thu, Sep 1, 2011 at 5:16 AM, Igor Carron wrote:
> Hi,
>
> I am not sure if this should go to r-help or r-dev list. I have looked
> at some archives of R libraries but cannot
try this:
> myDate <- as.POSIXct(day, format = "%y%m%d")
> myDate
[1] "2011-08-09 EDT"
> myDate <- myDate + 0:3600
> str(myDate)
POSIXct[1:3601], format: "2011-08-09 00:00:00" "2011-08-09 00:00:01"
"2011-08-09 00:00:02" ...
>
On Thu, Sep 1, 2011 at 7:18 AM, J. Augusiak wrote:
> Dear list,
>
>
>
If you format it a little differently, it is easier to read:
bob<-read.csv('shi.csv', header=T)
newmean<-matrix(0, test, dim(bob)[2]-6)
a<-0
for (i in c(4,8:(dim(bob)[2]))){
a<-a+1
newmean[,a]<-tapply(bob[,i], bob$Exam, mean)
}
colnames(newmean)<-colnames(bob)[c(4,8:(dim(bob)[2]))]
It lo
see ?mean
Then avoid other peoples code.
cenae27 wrote:
>
> bob<-read.csv('shi.csv', header=T)
>
> newmean<-matrix(0, test, dim(bob)[2]-6);a<-0; for (i in
> c(4,8:(dim(bob)[2])))
> {a<-a+1;newmean[,a]<-tapply(bob[,i], bob$Exam, mean)}
> colnames(newmean)<-colnames(bob)[c(4,8:(dim(bob)[2]))]
>
Mr ". .",
MASS::area comes to mind but it may be more helpful if you could say what
you are looking for / why integrate is not appropriate it is for whatever
you are doing.
Strictly speaking, I suppose there are all sorts of "alternatives" to
integrate() if you are willing to be really creative a
package "caTools"
see ?trapz
. wrote:
>
> Hi all,
>
> is there any alternative to the function integrate?
>
> Any comments are welcome.
>
> Thanks in advance.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
Hi all,
is there any alternative to the function integrate?
Any comments are welcome.
Thanks in advance.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/postin
Good Morning,
I'm trying to install the rJava package on a local (work) machine and having
some trouble. The following occurred in an RGui session.
> sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=E
This should be due to the fact that "110809 60 0" is not interpreted as a
valid time value.
You probably want to have
time.t <- 0:3599
rather than time.t <- 1:3600 if you want one value for each second in the
hour, starting from 00:00:00 and running to 00:59:59. 00:60:00 is not a
proper time val
Thanks Frank
I got the predicted probability.
But can I get the bootstrap corrected probability for individual subject.
for instance, I can get predicted probability from predict(fit,
type="fitted"). Is there similar one to retrieve the bootstrap corrected
probability for individual subject.
TH
Hi,
I have a dataset with proportions that vary around a fixed mean, is it
possible to use betareg to look at variance in the dispersion parameter
while keeping the mean fixed?
I am very new to R but have tried the following:
svec<-c(qlogis(mean(data1$scaled)),0,0,0)
f<-betareg(scaled~-1 | expt
The survreg function uses case weights. That is, if a subject is given
a weight of 2, the result is the same as if there were a second
observation (exactly the same).
Early in my career data sets that contained only categorical variables
were often collapsed in just this way, in order to save on
Hi:
On Wed, Aug 31, 2011 at 11:58 PM, Alaios wrote:
> Dear Dennis,
> I would like to thank you for your reply.
> I also checked the web sites that you gave me, it is hard to find everything
> about ggplot2 at one place with concrete examples that help you understand
> directly what you are plotti
Thank you Petr
It work.
Now I have a matrix 970*960. If I want to convert to spatial grid (each pixel
has x and y coordinate).
How can I do?
Thanks
From: Petr PIKAL
To: azam jaafari
Cc: R-help
Sent: Thursday, September 1, 2011 8:59 AM
Subject: Re: [R] convert to grid file
Hi
>
>
Hi all,
I am using the negative binomila glm in MASS. This is my data alled data1:
Reps Treats counts
HSM11 0 21
HSM22 0 34
HSM33 0 27
PTM11 1 32
PTM22 1 20
PTM33 1 23
I goal is to do a GLM to extract the p-value
bob<-read.csv('shi.csv', header=T)
newmean<-matrix(0, test, dim(bob)[2]-6);a<-0; for (i in
c(4,8:(dim(bob)[2])))
{a<-a+1;newmean[,a]<-tapply(bob[,i], bob$Exam, mean)}
colnames(newmean)<-colnames(bob)[c(4,8:(dim(bob)[2]))]
Could anyone please help me what does the above code does ... I want to fin
On Sep 1, 2011, at 3:45 AM, Jie TANG wrote:
thank you. it works.
but further question is that if we can let the " flnm" to be a 2-
dimension
matrix [3,5]?
since
mtdno<-paste("data",1:3,sep="")
tyno<-paste("obs",1:5,sep="")
flnm<-paste(mtdno,tyno,"_err.dat",sep="")
flnm would be a 1-dimension
Hello,
I have a dataset with proportions that vary around a fixed mean, is it
possible to use betareg to look at variance in the dispersion parameter
while keeping the mean fixed?
I am very new to R but have tried the following:
svec<-c(qlogis(mean(data1$scaled)),0,0,0)
f<-betareg(scaled~-1 | exp
Hi,
I am using the survival package to perform a Cox's regression analysis on
multiple events of myocardial infarctions. I have been using the Andersen
and Gill model:
coxph(Surv(time1,time2,status)~factor(treatment)+age+sex+cluster(id).
I was just wondering if this model should satisfy proport
Dear list,
I want to create a POSIX time vector as follows:
day<- as.character("110809")
time.t <- 1:3600
t.min <- time.t %/% 60
t.sec <- time.t-t.min*60
DATE <- as.POSIXct(strptime(paste(day,t.min,t.sec),"%y%m%d %M%S"))
Tail(DATE)
The problem is that the last element (3
Hi
>
> Hi
>
> I computed probability in each cell.
> I have:
>
> [99883,] -0.0062412957690
> [99884,] -0.0062412957690
> [99885,] -0.0062412957690
> [99886,] -0.0062412957690
> [99887,] -0.0062412957690
> [99888,] -0.0062412957690
> [99889,] 0.9909126638948
> [99890,] 0.990912663894
Hi,
I am not sure if this should go to r-help or r-dev list. I have looked
at some archives of R libraries but cannot seem to see a project that
focuses on the new matrix factorization techniques that are showing up
in the literature. I have made a list of them:
https://sites.google.com/site/igor
cal <- calibrate(fit, ...); note that cal is a matrix. colnames(cal) will
tell you what to pick, in this case cal[,'calibrated.corrected'].
Be sure to follow the posting guide.
Frank
yz wrote:
>
> Dear R users:
>
> In Prof. Harrell's library rms, calibrate.rms plot the Bias-corrected
> Probab
Hi
I computed probability in each cell.
I have:
[99883,] -0.0062412957690
[99884,] -0.0062412957690
[99885,] -0.0062412957690
[99886,] -0.0062412957690
[99887,] -0.0062412957690
[99888,] -0.0062412957690
[99889,] 0.9909126638948
[99890,] 0.9909126638948
[99891,] 0.9909126638948
[99
Do a little debugging on your code (put print(f1)) and you will see
that you keep adding to the length of the expression to be evaluated
and the results you see are correct. Learn how to debug your
functions.
On Thu, Sep 1, 2011 at 8:10 AM, . . wrote:
> Why time is increasing for the same operat
On Thu, Sep 1, 2011 at 8:10 AM, . . wrote:
> Why time is increasing for the same operation?
>
> I was expecting +/- the same time for each n.
>
> Thanks in advance.
>
> bench <- function(f1, n, ...) {
> t <- 0
> for(i in 1:n) {
> func <- function(x) x^2
> expr <- list(...)[1]
> f1 <- c(
Hi:
Here's one approach:
X1 <- sample(1:4, 10, replace = TRUE, prob = c(0.4, 0.2, 0.2, 0.2))
foo <- function(x) {
m <- matrix(NA, nrow = length(x), ncol = length(x))
m[, 1] <- x
idx <- seq_len(length(x))
for(j in idx[-1]) {
k <- sample(idx, 2)
x <- replace(x, k, 5)
On 9/1/2011 1:04 PM, Eran Eidinger wrote:
>>
>>> Hello,
>>>
>>> I wonder how I might create a package that only reveals some of the
>>> function
>>> in the package to the user.
>>>
>>> I've tried creating an R package using the following:
>>> f<- function(x,y) x+y
>>> g<- function(x,y) x-y
>>> h<
Why time is increasing for the same operation?
I was expecting +/- the same time for each n.
Thanks in advance.
bench <- function(f1, n, ...) {
t <- 0
for(i in 1:n) {
func <- function(x) x^2
expr <- list(...)[1]
f1 <- c("system.time(y <- ", gsub("XXX",expr,f1),")[3]")
t1 <- e
On 11-09-01 7:04 AM, Eran Eidinger wrote:
Yes, the package works fine without the NAMESPACE file, and all 3 functions
are visible.
When you include the NAMESPACE file, what does R CMD check tell you?
Duncan Murdoch
On Thu, Sep 1, 2011 at 2:02 PM, Duncan Murdochwrote:
On 11-09-01 2:27 AM,
Yes, the package works fine without the NAMESPACE file, and all 3 functions
are visible.
On Thu, Sep 1, 2011 at 2:02 PM, Duncan Murdoch wrote:
> On 11-09-01 2:27 AM, Eran Eidinger wrote:
>
>> Hello,
>>
>> I wonder how I might create a package that only reveals some of the
>> function
>> in the p
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