sir i could not find the plant breeding libraray in Rgui3.0.0
[[alternative HTML version deleted]]
__
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
Hello,
Which package is it?
Maybe you need to install it. And after installing it, you need to load
it in the R session. You do this with the following commands.
install.package(pkgname) # do this only once
library(pkgname) # do this every new R session
Hope this helps,
Rui Barradas
Em
Sorry,
It's the plural:
install.packages(pkgname)
Rui Barradas
Em 14-07-2013 11:20, Rui Barradas escreveu:
Hello,
Which package is it?
Maybe you need to install it. And after installing it, you need to load
it in the R session. You do this with the following commands.
install.package(pkgna
Haha, good point!
So, the correct code is:
any(names(formals(getOption("device"))) == "title") {
dev.new(title = "title")
}
Thanks for correcting my approach Rolf!
Best
Simon
On Jul 14, 2013, at 1:02 AM, Rolf Turner wrote:
>
>
> I think you ought to take a look at
>
>fortun
Hello everyone,
I have a dataset which includes the first three variables from the demo
data below (year, id and var). I need to create the new variable ans as
follows
If var=1, then for each year (where var=1), i need to create a new dummy
ans which takes the value of 1 for all corresponding id'
Dear William,
thanks a lot. I've found another nice alternative:
A <- matrix(t(expand.grid(c(1,2,3,4,5), 15, 16)), nrow = 3)
B <- combn(16, 3)
B.n <- B[, -which(duplicated(t(cbind(A, B - ncol(A)]
Best wishes,
Alrik
-Ursprüngliche Nachricht-
Von: arun [mailto:smartpink...@yahoo.com
Dear list,
I have a matrix M.1 (30x2) into which I would like to paste another matrix M.2
(10x2) three times. However, the columns get flipped in every odd-numbered
recycle run. How can I avoid this behavior?
M.1 <- matrix(numeric(30*2), ncol = 2)
M.2 <- t(combn(1:5, 2))
M.1[, 1:2] <- M.2
Many
Hi,
I am trying to use a power law y=bx^a as a nls model as below, however I
keep getting 'singular gradient' error. I have tried multiple different
starting values but always get an error.
> x2 <-
> c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,
> 31,32,33,3
Hello,
Your data seems to be of class 'matrix'. The following code needs it to
be a data.frame.
dat <- as.data.frame(your input matrix)
res <- do.call(rbind, lapply(split(dat, list(dat$id, dat$year)),
function(x){
x$ans <- if(any(x$var == 1)) 1 else 0
x}))
rownames(res) <- N
On 14/07/2013 14:30, JenPool wrote:
Hi,
I am trying to use a power law y=bx^a as a nls model as below, however I
keep getting 'singular gradient' error. I have tried multiple different
starting values but always get an error.
That is not the model you tried to fit. b*x*exp(a) is always
over-p
library(plyr)
M.1[,1:2]<-do.call(rbind,alply(replicate(3,M.2),3,function(x) x))
#or
M.1[,1:2]<-matrix(aperm(replicate(3,M.2),c(1,3,2)),ncol=2)
A.K.
- Original Message -
From: Thiem Alrik
To: "mailman, r-help"
Cc:
Sent: Sunday, July 14, 2013 9:48 AM
Subject: [R] Matrix column flip
Only vectors recycle. If you don't want the behaviour if the matrix to reflect
that of the underlying vector then don't use recycling. Instead use indexing.
M.1 <- M.2[rep(1:5,3),]
---
Jeff NewmillerTh
Hi,
You could try this: (if I understand it correctly)
dat1<- read.table(text="
year id var ans
2010 1 1 1
2010 2 0 0
2010 1 0 1
2010 1 0 1
2011 2 1 1
2011 2 0 1
2011 1 0 0
2011 1 0 0
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat1$newres<-with(dat1,ave(var,id,ye
Hi,
Excuse me for the indistinctness
Le 13/07/2013 17:18, arun a écrit :
> Hi,
> "when the value of Debut of lines i = value Fin of lines i-1"
> That part is not clear esp. when it is looked upon with the expected output
> (df2).
I want to group the lines which have the same caracteristics (Matric
On Jul 14, 2013, at 3:42 AM, Simon Zehnder wrote:
> Haha, good point!
>
> So, the correct code is:
>
> any(names(formals(getOption("device"))) == "title") {
> dev.new(title = "title")
> }
>
This might be easier to read:
"title" %in% names(formals(getOption("device")) )
--
David.
> T
Hi,
I have several really big data files in csv format like this: the first line is
the header, the second to fourth lines have info about the file and are the
lines I need to skip (data in 2-4th lines are not correspoding to variable
names in the hearder), from the fifth line, real data begins
On 08.07.2013 11:42, sridhar srinivasan wrote:
Dear R Developers,
I have two doubts related to R
1. i try to install R package 3.0 in my linux system ./configure.
it gives Error as
configure: error: --with-readline=yes (default) and headers/libs are not
available
Read the R Installation man
hi Arun,
Thanks for this. This solution works great.
Knid Regards
Anup
On Sun, Jul 14, 2013 at 8:07 PM, arun wrote:
> Hi,
> You could try this: (if I understand it correctly)
> dat1<- read.table(text="
> yearid var ans
> 2010 1 1 1
> 2010 2 0 0
> 2010 1 0 1
> 2010 1 0 1
>
On Jul 14, 2013, at 9:48 AM, Houhou Li wrote:
> Hi,
>
> I have several really big data files in csv format like this: the first line
> is the header, the second to fourth lines have info about the file and are
> the lines I need to skip (data in 2-4th lines are not correspoding to
> variable
An other remark :
If I calculate
df1$D<- as.numeric(as.Date(paste0(substr(df1$Debut,7,10),"-",
substr(df1$Debut,4,5),"-",substr(df1$Debut,1,2
and
df1$F <- as.numeric(as.Date(paste0(substr(df1$Fin,7,10),"-",
substr(df1$Fin,4,5),"-",substr(df1$Fin,1,2
and if there is no interruption of time f
It looks like match() (and relatives like %in% and is.element) act a bit
unpredictably
on lists when the list elements are vectors of numbers of different types. If
you match
integers to integers or doubles to doubles it works as expected, but when the
types
don't match the results vary. I wou
On Jul 14, 2013, at 10:57 AM, David Winsemius wrote:
>
> On Jul 14, 2013, at 9:48 AM, Houhou Li wrote:
>
>> Hi,
>>
>> I have several really big data files in csv format like this: the first line
>> is the header, the second to fourth lines have info about the file and are
>> the lines I need
Hi,
May be this helps you.
df1$contrat[grep("^CDD",df1$contrat)]<- "CDD détaché ext. Cirad"
df1[48,8]
[1] "31/12/4712" #strange value
df1[48,8]<- "31/12/2013" #changed
indx<-as.numeric(interaction(df1[,1:6],drop=TRUE))
res<-do.call(rbind,lapply(split(df1,indx),function(x) {x1<-
as.Date(x$Debut
Dear,
I need a book about repeated measurements analisys with R.
In Amazon, I found this one:
Models for Repeated Measurements (Oxford Statistical Science Series)
J. K. Lindsey 1999 2ed.
I would like a book with examples, data and R code. I work with trees
(forest breeding).
Could you recomend
Look before you post -- specifically at the "Books" subpage of CRAN's
R homepage:
http://www.r-project.org/
-- Bert
On Sun, Jul 14, 2013 at 12:56 PM, Marcelo Laia wrote:
> Dear,
>
> I need a book about repeated measurements analisys with R.
>
> In Amazon, I found this one:
>
> Models for Repeat
Yes!
I have a look there before I post!
Thank you very much!
2013/7/14 Bert Gunter :
> Look before you post -- specifically at the "Books" subpage of CRAN's
> R homepage:
>
> http://www.r-project.org/
>
> -- Bert
>
> On Sun, Jul 14, 2013 at 12:56 PM, Marcelo Laia wrote:
>> Dear,
>>
>> I need a
On 7/14/2013 1:20 PM, Marcelo Laia wrote:
Yes!
I have a look there before I post!
You may know that "mixed effects" is another term for "repeated
measurements".
Spencer Graves
Thank you very much!
2013/7/14 Bert Gunter :
Look before you post -- specifically at the "Books"
Hi,
Have any of you ever encountered a situation where R stops processing an
instruction but does not give a "not responding" message?
The reason I ask is I am working in RStudio (Mac OS/X 10.7, 1.8 Ghz i7, 4
GB DDR3) and the instruction I entered in the command line pane is still
being processed
On 15/07/13 08:57, Spencer Graves wrote:
You may know that "mixed effects" is another term for "repeated
measurements".
I must of course preface this comment with an "I am no expert"
disclaimer, but I do not
believe that this assertion is correct. It would be more correct, I
beli
HI,
Try this:
Geo<- read.table(text="
long lat.comp confianza
9.31 -42.72 3
11.66 -40.63 9
10.88 -38.60 9
10.72 -37.86 9
13.06 -39.04 9
16.02 -38.51 6
",sep="",header=TRUE)
col1<- as.numeric(factor(Geo$confianza))
wi
On Jul 14, 2013, at 2:40 PM, L S wrote:
> Hi,
>
> Have any of you ever encountered a situation where R stops processing an
> instruction but does not give a "not responding" message?
>
> The reason I ask is I am working in RStudio (Mac OS/X 10.7, 1.8 Ghz i7, 4
> GB DDR3) and the instruction I e
If you are writing a script that you know will take a long time to process,
"pepper" it with "progress" reports so you know what part of the script it
is in and when it is going around loop. On some of my long scripts, I will
print out a message every n'th time through the loop so that I know if i
On 7/14/2013 3:05 PM, Rolf Turner wrote:
On 15/07/13 08:57, Spencer Graves wrote:
You may know that "mixed effects" is another term for "repeated
measurements".
I must of course preface this comment with an "I am no expert"
disclaimer, but I do not
believe that this assertion is c
HI Michel,
This gives the same order as that of df2.
df1$contrat[grep("^CDD",df1$contrat)]<- "CDD détaché ext. Cirad"
df1[48,8]<- "31/12/2013"
indx<-as.numeric(interaction(df1[,1:6],drop=TRUE))
lst1<-split(df1,indx)
lst2<-lst1[match(unique(indx),names(lst1))]
res<-do.call(rbind,lapply(lst2,functio
Thanks Jim and David for your helpful feedback. I still have not
terminated RStudio (and it still has not gone to completion). A few
observations I forgot to mention is that the red "stop" icon is showing in
RStudio so I am unable to enter any new commands. Also there is a blinking
cursor under
Super !!!
Thank you very much Arun
Michel
Le 15/07/2013 03:47, arun a écrit :
HI Michel,
This gives the same order as that of df2.
df1$contrat[grep("^CDD",df1$contrat)]<- "CDD détaché ext. Cirad"
df1[48,8]<- "31/12/2013"
indx<-as.numeric(interaction(df1[,1:6],drop=TRUE))
lst1<-split(df1,indx)
l
Hello all,
I have been trying to figure out how to replace non-zero values of a matrix
with non-zero values from corresponding rows from another matrix. More
specifically, let say we have the following original matrix:
original <-
matrix(c(0,0,4,0,0,1,2,2,12,1,0,2,0,5,0,0,10,1,3,1,0,5,0,4),byrow=
Hi,
I am a researcher in chemistry. I have to do pls-da for some of my samples.
The number of variables for my sample is 4000 . The categorical variable
has two classes. Some of the values are missing (cant find out).
I s there any package for R (or any illustration that would be helpful) to
perfor
On Jul 14, 2013, at 8:24 PM, L S wrote:
> Thanks Jim and David for your helpful feedback. I still have not terminated
> RStudio (and it still has not gone to completion). A few observations I
> forgot to mention is that the red "stop" icon is showing in RStudio so I am
> unable to enter any
For small matrix you could use a for-loop.
for (i in 1:nrow(randomized)){
randomized[i,randomized[i,]!=0] <- sample(original[i,original[i,]!=0])
}
randomized
If you have a larger matrix sapply is probably faster
randomized <- t(sapply(1:nrow(randomized), function(i) {
randomized[i,randomized[
40 matches
Mail list logo