http://r.789695.n4.nabble.com/file/n3312061/x_and_y_values.txt
x_and_y_values.txt
I have the absorbance values form HPLC Chromatogram. I need to find the
peaks in datapoints and area under each peak.
I am not sure how how to find the peaks, I tried couple of libraries and
peak function but the
Duncan,
I'm not sure how I missed your message. Sorry. What you describe is what I
do when (n-1)p is an integer so that R computes the sample quantile using a
single order statistic. My later posting has that exact binomial expression
in there as a special case. When (n-1)p is not an integer then
The data is in the attachment.
What I wanna get is:
, , Sex = Male
Eye
HairBrown Blue Hazel Green
Black32 1110 3
Brown53 502515
Red 10 10 7 7
Blond 3 30 5 8
, , Sex = Female
Eye
HairBrown Blue Hazel Green
Blac
On Fri, Feb 18, 2011 at 5:42 PM, Phil Spector wrote:
> Geoff -
> I think this will get you closer to a solution:
>
Yes, much closer, many thanks.
Geoff.
>
> newdf = reshape(df,varying=names(df)[-c(1,2)],direction='long',
>times=2000:2003,idvar=c('country','food'),v.names='X',
Geoff -
I think this will get you closer to a solution:
newdf = reshape(df,varying=names(df)[-c(1,2)],direction='long',
times=2000:2003,idvar=c('country','food'),v.names='X',
timevar='year')
xyplot(X~year|country*food,data=newdf)
Thanks again for your help. I was sniffing around this solution forever,
but that did it cleanly and without any problems.
On Thu, Feb 17, 2011 at 10:21 PM, wrote:
> The solution is probably to make the data-time columns POSIXct:
>
> > x <- read.table(textConnection("
> + ID event.date.ti
df=data.frame(country=c("A","A","A","B","B","B"),
food=rep(c("Apples","Pears","Bananas"),2),
X2000=c(4,5,6,7,6,8),
X2001=c(4,5,6,7,6,8),
X2002=c(4,5,6,7,6,8),
X2003=c(4,5,6,7,6,8));
I have data in the above form trying to get a plot of each fruit over time
year conditioned on c
The solution is probably to make the data-time columns POSIXct:
> x <- read.table(textConnection("
+ ID event.date.time
+ 1 '2009-07-23 00:20:00'
+ 2 '2009-08-18 16:25:00'
+ 3 '2009-08-13 08:30:00'
+ "), header = TRUE)
> y <- read.table(textConnection("
+ ID event.date.time
+ 4
I'm trying to rbind two data frames, both with the same columns names. One
of the columns is a variable with date-time and this variable is causing the
rbind to fail--giving the error
"Error in names(value[[jj]])[ri] <- nm : 'names' attribute [7568] must be
the same length as the vector [9]"
Is
Hi:
This is as far as I could get:
df <- read.table(textConnection("
Individual Parent1 Parent2 mark1 mark2
10 0 12 11
20 0 11 22
30 0 13 22
40 0 13 11
51 2
Dear Umesh,
I could not figure out exactly what your recoding scheme was, so I do
not have a specific solution for you. That said, the following
functions may help you get started.
?ifelse # vectorized and different from using if () statements
?if #
?Logic ## logical operators for your tests
##
Hi Dennis,
Thank you for your kind reply. Yes, essentially, we take integration
twice. However, I still have a few questions:
First, if we consider doing a double integration at the end, since the
first integration in within the indicator function,
it seems to be difficult.
Second, m1star is
Dear R users
The following question looks simple but I have spend alot of time to solve
it. I would highly appeciate your help.
I have following dataset from family dataset :
Here we have individuals and their two parents and their marker scores
(marker1, marker2,and so on). 0 means t
Mathijs,
To my eyes you seem to have repeated back what is already done.
More R and less English would help. In other words if it is not 2.5
you need, what is it? Please provide some input and state what the
output should be (and what you tried already).
Matthew
--
View this message in contex
Well, if you really want to produce what SAS does:
ageval = read.table(textConnection('AgeCat FINWT
+ 1 98
+ 2 62
+ 1 75
+ 3 39
+ 4 28
+ 2 47
+ 2 66
+ 4 83
+ 1 19
+ 3 50
+ '),header=TRUE)
one = aggregate(ageval$FINWT,ageval['AgeCat'],sum)
two = prop.table(as.table(one$x)) * 100
three = cumsum(
On 2/17/2011 3:49 PM, William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Petr Savicky
Sent: Wednesday, February 16, 2011 10:46 PM
To: r-help@r-project.org
Subject: Re: [R] logsumexp function in R
On Wed, Feb 16,
I am new to R. I have a data set like this (given below is a fictional
dataset):
AgeCat FINWT
1 98
2 62
1 75
3 39
4 28
2 47
2 66
4 83
1 19
3 50
I need to calculate the weighted distribution of the variable AgeCat.
In SAS i can do:
proc freq data=ageval;
tables agecat;
weight finwt;
run;
Wha
According to [1] and [2], using RCurl to post a form with basic
authentication is done using the postForm method. I'm trying to post
generated interpolation data from R onto an HTTP form. The call I'm using is
page <- postForm('http://our.server.com/dbInt/new', opts =
curlOptions=(userpwd="test:tes
Yes, that is it. Thank you very much, Phil.
-Original Message-
From: Phil Spector [mailto:spec...@stat.berkeley.edu]
Sent: Thursday, February 17, 2011 3:36 PM
To: Jim Moon
Cc: r-help@r-project.org
Subject: Re: [R] sort by column and row names
Jim -
I believe
dat[order(rownames(d
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Petr Savicky
> Sent: Wednesday, February 16, 2011 10:46 PM
> To: r-help@r-project.org
> Subject: Re: [R] logsumexp function in R
>
> On Wed, Feb 16, 2011 at 04:14:38PM -0500, Bri
Jim -
I believe
dat[order(rownames(dat)),order(colnames(dat))]
is what you are looking for.
- Phil Spector
Statistical Computing Facility
Department of Statistics
Hello, All,
How can one sort on column and row names. For example:
How can this
X1 X3 X2
X1 1 0 0
X3 0 1 0
X2 0 0 1
become this?
X1 X2 X3
X1 1 0 0
X2 0 1 0
X3 0 0 1
Thank you for your time!
Jim
[[alternative H
Hi yf,
On Thu, Feb 17, 2011 at 2:35 PM, yf wrote:
>
> i am trying to convert sas file into csv. I used write.csv(a,
> file="cool.csv") but nothing come out. i don't know why. Thanks.
>
>
> library(Hmisc)
> a<- sasxport.get("C:\\Users\.")
Let's back up a little bit. What type of file are you
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of yf
> Sent: Thursday, February 17, 2011 2:35 PM
> To: r-help@r-project.org
> Subject: [R] convert the sas file into csv in R
>
>
> i am trying to convert sas file into csv. I use
i am trying to convert sas file into csv. I used write.csv(a,
file="cool.csv") but nothing come out. i don't know why. Thanks.
library(Hmisc)
a<- sasxport.get("C:\\Users\.")
write.csv(a, file="cool.csv")
--
View this message in context:
http://r.789695.n4.nabble.com/convert-the-sas-file-in
Hello,
I am working with a data frame containg character strings with many special
symbols from various European languages. When writing such character
strings to a file using the UTF-8 encoding, some of them are converted in a
strange way. See the following example, run in R 2.12.1 on Windows
Hi,
I am trying to do multinomial regression using the glmnet package, but the
following gives me an error (for no reason apparent to me):
library(glmnet)
cv.glmnet(x=matrix(c(1,2,3,4,5,6,1,2,3,4,5,6),
nrow=6),y=as.factor(c(1,2,1,2,3,3)),family='multinomial',alpha=0.5,
nfolds=2)
The error i get
Hi Hannah:
You have a few things going on, but the bottom line is that numer and denom
are both double integrals.
On Thu, Feb 17, 2011 at 1:06 PM, li li wrote:
> Hi all,
> I have some some problem with regard to finding the integral of a
> function containing an indicator function.
> please s
Works perfectly, thank you!
On Thu, Feb 17, 2011 at 4:11 PM, Erik Iverson wrote:
> Gene,
>
> ?rapply is a recursive version of ?lapply, and should work.
>
> rapply(masterlist, function(x) x*NA, how = "replace")
>
>
> --Erik
>
> Gene Leynes wrote:
>
>> Hello all,
>>
>> Maybe I'm being thick, but
I don't think tikz() can capture the plots in rgl. To export rgl
graphics, you need to use rgl.snapshot() or rgl.postscript().
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Wed, Feb 16, 201
My main question is: can the bootstrap procedure sometimes be less
precise than the non-bootstrap procedure when carrying out latent class
analysis?
I am asking this question because I carried out some analyses and when I
*did not* use the bootstrap procedure I found adequate fit for one of
Gene,
?rapply is a recursive version of ?lapply, and should work.
rapply(masterlist, function(x) x*NA, how = "replace")
--Erik
Gene Leynes wrote:
Hello all,
Maybe I'm being thick, but I was trying to figure out a simple way to create
a list with the same dimension of another list, but popul
Hi Henrique,
Thanks for the reply!
I posted the question in a hurry yesterday. After a bit research today, I
found
that back in 2009, Andrew actually asked the same question and there was a good
discussion about it:
http://r.789695.n4.nabble.com/how-do-you-know-which-functions-are-being-debu
Dear All,
I am running Debian testing on my system for the amd64 architecture,
When trying to install the RGtk package I get this error
> install.packages('RGtk2')
Installing package(s) into ‘/usr/local/lib/R/site-library’
(as ‘lib’ is unspecified)
trying URL
'http://rm.mirror.garr.it/mirrors/C
Hello all,
Maybe I'm being thick, but I was trying to figure out a simple way to create
a list with the same dimension of another list, but populated with NA
values.
masterlist = list(
aa=list(
a=matrix(rnorm(100),10,10),
b=matrix(rnorm(130)
Aldi,
Thanks for the reference to the R News article. That seems like a lifetime
ago... :-)
As an exercise in learning to use R's base graphics and some of the concepts of
manipulating and annotating base R plots, it is a decent intro.
That being said, unless pressured (eg. weapon pointing at
Hi Julie,
Here is a little example. First I make a histogram (it looks odd
because there isn't much data), then add the x axis in manually
specifying at what points to place tick marks.
## histogram suppressing x axis
hist(c(0, 200), xaxt = "n")
## add axis with custom ticks
axis(1, at = seq(0,
Hi:
A couple more approaches to consider:
# Utility function to extract two rows from a data frame
# Meant to be applied to each data subset
sampler <- function(d) if(nrow(d) > 2) d[sample(1:nrow(d), 2, replace =
FALSE), ] else d
library(plyr)
> ddply(x, 'id', sampler)
id v1 V2
1 1 2 13
2 1
I haven't been able to get any code to make more tick marks than the default by
50.
Thanks! Code:
hist(OMY$FL, main = "2010 Oncorhynchus mykiss Fork Length Frequencies at Buck
Creek Reach 1", include.lowest = TRUE, col = "blue", border = "white", breaks =
140,xlab = "Fork Length in mm", ylab
Sorry, I have read other posts on tick marks but nothing is working.
Here is my R code. I was able to turn off the x axis at one point (I
removed that code), but never got the tick marks on by 20s. It always goes
to the default of 50, 100, 150, 200.
Thanks.
R code: hist(OMY$FL, main = "2010 O
Dear all,
I am using the BAS package to make predictions using the predict.bma
function. Does anyone know how to calculate credible intervals for
predictions from objects of class bma? It seems like all of the
required information is available from the bma class objects and/or
output from the pr
On Thu, Feb 17, 2011 at 3:39 PM, Stephen Sefick wrote:
> Janet:
>
> The zoo package and aggregate.zoo should do the trick. I have done this
> many times with these tools.
Here is some code. Suppose precip is the data. We convert it to a zoo
object with date/times assuming the first one is at th
As the FAQ
(http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-put-error-bars-or-confidence-bands-on-my-plot_003)
says,
there are about 5 different packages that will draw error bars.
At 2:38 PM -0600 2/17/11, Aldi Kraja wrote:
Hi Toby and Maria,
I did a check on Toby's suggestion and i
Hi all,
I have some some problem with regard to finding the integral of a
function containing an indicator function.
please see the code below:
func1 <- function(x, mu){
(mu^2)*dnorm(x, mean = mu, sd = 1)*dgamma(x, shape=2)}
m1star <- function(x){
integrate(func1, lower = 0, upper = Inf,x)
Hi all,
I have some some problem with regard to finding the integral of a
function containing an indicator function.
please see the code below:
func1 <- function(x, mu){
(mu^2)*dnorm(x, mean = mu, sd = 1)*dgamma(x, shape=2)}
m1star <- function(x){
integrate(func1, lower = 0, upper = Inf,x
R users,
I have been trying to write a program in R that will extract rows from a data
frame and combine the rows
into a new smaller data frame while meeting several criteria. I would greatly
appreciate any advice
that could help me get started down the right path. What I want to do is to
ext
A new version of rms is now available on CRAN for Linux/UNIX. I expect
Mac and Windows versions to be available in a day or so. This version
works with and requires the newest version of Therneau's survival package.
More information is at http://biostat.mc.vanderbilt.edu/Rrms
Changes in ve
The reason that it's easy to produce dotcharts or barplots in R with
data like yours, and difficult to produce a line plot, is because
dotcharts and barplots are appropriate for your data, whereas a line
plot is not. Since the values on the x-axis represent names,
and those names are not measur
Hi Toby and Maria,
I did a check on Toby's suggestion and it is not there:
R version 2.12.1 (2010-12-16)
> ??barploterrbar
No help files found with alias or concept or title matching
'barploterrbar' using fuzzy matching.
Also I went to the following location which does not exist.
http://www.dkf
dat<-c(150,152,155,134,141)
plot(dat,type="o",ylim=c(100,160),xlab="Names",ylab="Height")
On Thu, Feb 17, 2011 at 7:45 PM, world peace wrote:
> Hi All
>
> I have data like this
>
> tom randy mike dan doug
> height 150 152 155 134 141
>
> I am trying to create a line pl
On Feb 17, 2011, at 1:33 PM, andrija djurovic wrote:
This is, maybe, not the best solution but I hope it will help you:
x<-data.frame(id=c(1,1,1,2,2,2,2,3,3,3,4,4), v1=c(1:12), V2=c(12:23))
do.call("rbind",by(x,x$id,function(x) x[c(sample(nrow(x),2)),]))
Andrija
Another way (and note that
Janet:
The zoo package and aggregate.zoo should do the trick. I have done this
many times with these tools.
HTH,
Stephen
On Thu, 2011-02-17 at 11:56 -0800, Janet Choate wrote:
> Hi all,
> i'm sure there is an easy way to do this, but i'm stumped, so any help would
> be appreciated.
>
> i have
muhammad mohsin yahoo.com> writes:
> Hope you will be fine. I am student of Ph.D and doing '
> some work on distribution.
> I developed a new distribution and having some problems in estimating their
> parameters by MLE. I used R-program and used "maxLik" function (maxLik: A
> Package for Max
After ctree builds a tree, how would I determine the direction missing values
follow by examining the BinaryTree-class object? For instance in the example
below Bare.nuclei has 16 missing values and is used for the first split, but
the missing values are not listed in either set of factors. (
Hi Janet,
One relatively simple way would be to transofrm the data into a 96 x
Ndays matrix and use colSums(). Of course, lets say on one day, the
measurement tool had technical difficulties and missed two
observations, then you only have 94 observations for that day, you
will need a fancier solu
Hi all,
i'm sure there is an easy way to do this, but i'm stumped, so any help would
be appreciated.
i have a single column of data for precipitation every 15 minutes over a
year. i want to sum the precip to daily data.
so the first 96 records = the first day, the second 96 records = the second
d
Hi All
I have data like this
tom randy mike dan doug
height 150 152 155 134 141
I am trying to create a line plot, with names on X-axis and height measure
on Y. how can i get it through R.
I could get several versions which are close (dotchart, bargraph), but not
qui
Dear R users,
I would like to know how to indentify the polygons that are located on the
border of a map (i.e.shapefile).
Do you have any suggestion on how to do it?
Thank you very much,
Leo Monasterio.
[[alternative HTML version deleted]]
__
*Dear All,*
*I'm trying to do Multivariate BLUP in R. I'd highly appreciate if someone
can share R code and data for Multivariate BLUP. Thanks*
*
*
*Regards!
*
--
*
Muhammad Yaseen
*
[[alternative HTML version deleted]]
__
R-help@r-project.org
This is a presenation from the R Users Group (Bay Area)
Covers building predictive analytic models in R, exporting in PMML and
using ADAPA for model deployment and execution. Introduction to the
Predictive Model Markup Language (PMML) standard and how it helps to
overcome memory and speed limitat
On 17/02/2011 11:02 AM, Alex F. Bokov wrote:
Motivation: during each iteration, my code needs to collect tabular data (and
use it only during that iteration), but the rows of data may vary. I thought I
would speed it up by preinitializing the matrix that collects the data with
zeros to what I
This is, maybe, not the best solution but I hope it will help you:
x<-data.frame(id=c(1,1,1,2,2,2,2,3,3,3,4,4), v1=c(1:12), V2=c(12:23))
do.call("rbind",by(x,x$id,function(x) x[c(sample(nrow(x),2)),]))
Andrija
On Thu, Feb 17, 2011 at 6:39 PM, yf wrote:
>
> But i need for each id have two data
On Thu, Feb 17, 2011 at 10:02 AM, Alex F. Bokov
wrote:
> Motivation: during each iteration, my code needs to collect tabular data (and
> use it only during that iteration), but the rows of data may vary. I thought
> I would speed it up by preinitializing the matrix that collects the data with
>
Hi, when ploting Kaplan-Meier estimate curves as below, the censoring symbols
(crosses) to not change thickness along the lines
plot(survfit(surv ~ I(x>=cut.off) ),lty=c(1,2), lwd=2)
is there any strightforward way to make it happen? thanks
robert
--
View this message in context:
http://r.789
Dear Colleagues,
Hope you will be fine. I am student of Ph.D and doing some work on
distribution.
I developed a new distribution and having some problems in estimating their
parameters by MLE. I used R-program and used "maxLik" function (maxLik: A
Package for Maximum Likelihood Estimation in R
On Thu, Feb 17, 2011 at 12:35 AM, mipplor wrote:
>
> thx for your suggestions , i have made it csv file,and it looks:
>
>
>> data<‐read.table('E:/my documents/r/1.csv',header=TRUE)
It looks like you could benefit from (re)reading my previous email.
You either need to specify the sep = argument in
I believe that this code will work (...for very small) samples, but let some
correct me if there is something wrong.
require(logistf);require(combinat)
permY<-permn(data$y)
ntimes<-length(permY)
results<-matrix(nrows=ntimes,ncols=number_of_coefficients)
for(i in 1:ntimes){
results[i,]<-logistf(unl
But i need for each id have two data.
Like...
> x
id v1 V2
1 1 1 12
2 1 2 13
4 2 4 15
5 2 5 16
8 3 8 19
9 3 9 20
11 4 11 22
12 4 12 23
So should write sample( if sample id >2 ,2). I don't know how to write (if
sample id >2). Thanks.
--
View this message in context:
Simona -
I don't think preallocating your random variables would make
the code run any faster.
A very simple change that would speed things up a little would
be to replace
simPD.vec[i]=length(R.vec[R.vec
Dear all,
Does anyone have any idea on how to speed up the for() loop below.
Current
You may write as this:
for (i in 1:nrow(x)){
for (j in 1:ncol(x)){
if (!is.na(x[i, j])) {
if(x[i, j] == 'A') {x2[i, j] <- 'A/A'} else{
if(x[i, j] == 'T') {x2[i, j] <- 'T/T'} else{
if(x[i, j] == 'G') {x2[i, j] <- 'G/G'} else{
In addition to the other answers that you received you can also do:
library(TeachingDemos)
i[ quantile(i,.25) %<% i %<% quantile(i,.75) ]
This may or may not be more readable than the others. Also note that
precomputing both quantiles in one step may be faster than calling quantile
twice.
You
Dear all
when I call color.scale like this:
require('plotrix')
colcolor<-color.scale(c(range_sr,sr),extremes=c("red","blue"))
Error in if (min(reds) < 0 || max(reds) > 1) reds <- rescale(reds, c(0, :
missing value where TRUE/FALSE needed
range_sr
[1] -10.00 44.02977813958366
Hi,
You could use car::recode to change the levels of the factors,
library(car)
transform(x, locus1 = recode(locus1, "'A' = 'A/A' ; else = 'T/T'"),
locus2 = recode(locus2, "'T'='T/T' ; 'C' = 'C/C'"),
locus3 = recode(locus3, "'C'='C/C' ; 'G' = 'G/G'"))
HTH
Try this:
xNew <- as.data.frame(mapply(paste, x, x, sep = "/"))
xNew[is.na(x)] <- NA
xNew
On Thu, Feb 17, 2011 at 2:54 PM, Josh B wrote:
> Hi all,
>
> I'm having a problem once again, trying to do something very simple.
> Consider
> the following data frame:
>
> x <- read.table(textConnection("
Josh, you've made it far too complicated. Here's one simpler way (note
that I changed your read.table statement to make the values NOT factors,
since I wouldn't think you want that).
> x <- read.table(textConnection("locus1 locus2 locus3
+ A T C
+ A T NA
+ T C C
+ A T G"), header = TRUE, as.is=TRU
Hi all,
I'm having a problem once again, trying to do something very simple. Consider
the following data frame:
x <- read.table(textConnection("locus1 locus2 locus3
A T C
A T NA
T C C
A T G"), header = TRUE)
closeAllConnections()
I am trying to make a new data frame, replacing "A" with "A/A", "
I hate to sound like David "Have You Read The Posting Guide?" Winsemius, but
there's no way for anyone to know what you are trying to accomplish here
without a lot more information. You don't show us the output you expect and
the output you got. I would expected "relatedness" to be on a scale
Thanks for helping me out so generously. After reading the vignettes and the
other info I still have a question (sorry I am a R novice):
I am not so much trying to construct time series (although it comes very
close). Rather for each pair (Bi,Bj) in project (An) I am trying to sum up
the values o
If you google "barplot with error bars" you immediately find
http://svitsrv25.epfl.ch/R-doc/library/prada/html/barploterrbar.html .
Toby.
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Lathouri, Maria [m.lathour...@imperial
On Thu, 17 Feb 2011, Uwe Ligges wrote:
On 17.02.2011 15:29, Damian Betebenner wrote:
In building a package, is it possible to make the package depend upon
another package that is only available on R-Forge (not CRAN). For example,
by doing something in the DESCRIPTION file
I'd like to add a d
Thank you for your hint!
I see that there was a thread discussing implementation of what I
wanted to do (Newey-West standard errors in a VAR context), but that
there is a conflict due to how the type = "const" is defined in the VAR
command: https://stat.ethz.ch/pipermail/r-sig-finance/2009q2/00427
Dear MCMCglmm users,
I am currently struggling with the specification of a proper prior and model
formula for a multi-response MCMCglmm with two of the three response variables
being Gaussian and the third being za-poisson. The model includes several fixed
effects and three nested random effect
Dear all,
Does anyone have any idea on how to speed up the for() loop below.
Currently it takes approximately 2 minutes and 30 seconds.
Because of the size of Nsim and N, simulating a multivariate normal
(instead of simulating Nsim times a vector of N normal distributions)
would require too much
I am using VGAM package to model a multinormial outcome with spline
function. But from the following result, I only can get p-value for variable
with spline function, such as s(As). However, there is no p-value available
for variable with linear function, such as SEX, Parity. Does anybody know
why?
Dear R-list
I'm working with with geom_errorbar; specifically I'm trying to
reproduce the example Hadley Wickham have on
http://had.co.nz/ggplot2/geom_errorbar.html (all in the button of the
page) where he makes an nice plot with errorbars and then draw lines
between the points.
What confuses me
Try putting d,e,f in a list:
Xxx = list(d,e,f)
For (I in 1:length(xxx))
For (j in 1:length(xxx))
If (i!=j) bigfunction(xxx[[i]],xxx[[j]])
(bad indentation, caps thanks to outlook)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of s
Hi everybody,
I tried to analyze a custom Affymetrix 3'-biased Array. So I wanted to make
a cdf package. (My CDF file size is 1.12Go).
I tried several methods but the same error occured
Method 1
> #Set the working directory
> setwd("D:/Analyse R/Cel files")
> #library to create cdf env
>
Im am curerntly trying to use R software to extract uncompressed waveform
values with I have acquired from IDLreadGLAS tool.
I am very lost with this & know I have to assess the GLA01 product data,
however, any advice on the matter atall would be brilliant, particularly on
define thresholds and co
Try the function ctree() in the package party or earth() in earth. You can
use factor variable as is, or you can transform the factor to binary variables
(i.e., is_P is 0 or 1, is_D is 0 or 1). In the second case, you can use any
algorithm, and earth() automatically transforms factors to bina
Motivation: during each iteration, my code needs to collect tabular data (and
use it only during that iteration), but the rows of data may vary. I thought I
would speed it up by preinitializing the matrix that collects the data with
zeros to what I know to be the maximum number of rows. I was su
Hello Marta,
arrg, sorry, I have not carefully enough read your message. Well, if you follow
the cited thread further down, you will find:
(and hereby directly quoting from:
https://stat.ethz.ch/pipermail/r-sig-finance/2009q2/004274.html)
<>
On Tue, 9 Jun 2009, Matthieu Stigler wrote:
> Hi
Hi Maria,
Look at barplot function
? barplot for help
Regards
M
Le 17/02/11 17:00, Lathouri, Maria a écrit :
> Dear all
>
> I have six variables of the average metal concentrations
>
> Var1 4.77
> Var2 23.5
> Var3 5.2
> Var4 12.3
> Var5 42.1
> Var6 121.2
>
> I want to plot them as a barplot wi
Dear all
I have six variables of the average metal concentrations
Var1 4.77
Var2 23.5
Var3 5.2
Var4 12.3
Var5 42.1
Var6 121.2
I want to plot them as a barplot with error bars. Could you help me?
Cheers
Maria
[[alternative HTML version deleted]]
On Feb 17, 2011, at 10:36 AM, Maas James Dr (MED) wrote:
I'm using the range command to get the minimum and maximum values of
an array as in
x <- range(array_y)
which gives me two values such as
[1] -2 9
I need to be able to test if this range of values includes a third
value. For ex
On 2011-02-17 07:36, Maas James Dr (MED) wrote:
I'm using the range command to get the minimum and maximum values of an array
as in
x<- range(array_y)
which gives me two values such as
[1] -2 9
I need to be able to test if this range of values includes a third value. For
example I'd like
I'm using the range command to get the minimum and maximum values of an array
as in
x <- range(array_y)
which gives me two values such as
[1] -2 9
I need to be able to test if this range of values includes a third value. For
example I'd like to query
1) does the range of -2 to 9 include 3,
On 17.02.2011 15:29, Damian Betebenner wrote:
In building a package, is it possible to make the package depend upon another
package that is only available on R-Forge (not CRAN). For example, by doing
something in the DESCRIPTION file
I'd like to add a dependency to my package such that when t
Dear Bill,
Thank you very much for your careful discussion of the issue.
It is not surprising that the deviance is the same whether you fit the
model using a factor response with weights or individual 0/1 responses.
I think this happens because the fitted probabilities in the saturated
models are
I'm using the range command to get the minimum and maximum values of an array
as in
x <- range(array_y)
which gives me two values such as
[1] -2 9
I need to be able to test if this range of values includes a third value. For
example I'd like to query
1) does the range of -2 to 9 include 3
On 16.02.2011 22:38, poisontonic wrote:
Uwe Ligges-3 wrote:
If the available space got too fragmented, there is not single 3.8 block
of memory available any more
Is there anything I can do to prevent this?
If you did it after a fresh reboot: I don't see a way to prevent it.
Neverth
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