Hello,
For an experiment, I selected plots of land within a forest either with
honeysuckle or without honeysuckle. Thus, my main factor is fixed, with 2
levels: "honeysuckle present"(n=11) and "honeysuckle absent"(n=8).
Within each plot of land, I have a "trenched" subplot and an "untrenched"
sub
Hi,
I hope there is someone to help me with using 'meboot' package. I tried to
create an ensemble of 999 replicates of a vector. however, I am getting the
following error
> oil.ens <- meboot(x = panel, reps = jboot, colsubj = 1, coldata = 3)
Error in `[.data.frame`(x, order(x, na.last = na.last,
Good suggestion, and yes, the given examples work. Using a modification of
one of the supplied examples, where in the final line I only added the
parameter col=c(1,2,3):
data(dat.bcg)
dat <- escalc(measure="RR", ai=tpos, bi=tneg, ci=cpos, di=cneg,
data=dat.bcg, append=TRUE)
forest(dat$yi, dat$vi,
Dear All,
I made some headways on my ODE problem in R2OpenBUGS, but got stuck again.
Wonder if someone could help on the following:
I am running the following code:
library(R2OpenBUGS)
dosetotal <-c(500,250)
z <-c(4,2)
grid <-c(4,15)
n.grid <-2
tol <-0.001
dim <-2
T <-2
origin <-0
dosetime <
You really need to ask the developer for support. You haven't indicated where
you downloaded it or what instructions you followed, but I got a similar error
trying to install
http://pbil.univ-lyon1.fr/software/mareymap/MareyMap_1.0.zip, so it seems a
case of stale software. The DESCRIPTION fil
You have to understand the ``scope'' argument to add1. As Rui Barradas
points out, that argument expects a formula, but you pass it a data frame,
YET you only get two warnings. You should practice having that feel
strange to you, and having it make you investigate:
1- Why did you not have to giv
Hello,
I just downloaded the MareyMap: an R-based tool and followed the
instructions:
Option1: Click "Packages-Install package(s) from local zip files", it showed
the followed error messages:
> utils:::menuInstallLocal()
Error in read.dcf(file.path(pkgname, "DESCRIPTION"), c("Package", "Type")) :
On 01/05/2013 10:32 PM, mary wrote:
Thanks for the help:your reply has focused my problem
in particular I'm trying to do a converge process;
the robustm() function is:
function (x,z)
1# {eigen<-eigen(x)
2# d<-madmatrix(z)
##I have created this function to calculate "mad" of a whole matrix
3# e
Thanks all!
These 2 options worked great.
### OPTION 1
u<-with(occ.data,ftable(Rep~Año+Site+Especie+Pres))
u[1:5,]
library(reshape2)
flat.occ<-dcast(as.data.frame(u), Año + Site + Especie + Pres ~ Rep, fill =
0)
flat.occ[1:10,]
### OPTION 2
occ.data1<-occ.data[,-c(2:3)]
library(reshape2)
flat
Hi David,
Yes, it is - although the SO question was more directed at figuring out why
sapply seemed slower, the question to R-help is more nuanced in "is this
coercion really necessary for data.frames?", and I figured it might take
some more knowledge of R internals / the difference between lists
Given the data:
dat1<-read.table(header=TRUE,text="
ID V1 V2 V3 V4
16532
23221
36532
412 15 32
56832
63241
76533
812 15 31
96533
103275
116
On Jan 5, 2013, at 11:38 AM, Kevin Ushey wrote:
Hey guys,
I noticed something curious in the lapply call. I'll copy+paste the
function call here because it's short enough:
lapply <- function (X, FUN, ...)
{
FUN <- match.fun(FUN)
if (!is.vector(X) || is.object(X))
X <- as.list(X)
On Sat, Jan 5, 2013 at 7:38 PM, Kevin Ushey wrote:
> Hey guys,
>
> I noticed something curious in the lapply call. I'll copy+paste the
> function call here because it's short enough:
>
> lapply <- function (X, FUN, ...)
> {
> FUN <- match.fun(FUN)
> if (!is.vector(X) || is.object(X))
>
HI,
Sorry, there was a mistake, which I noticed after seeing David's post.
dat1<-read.table(text="
ID V1 V2 V3 V4
1 6 5 3 2
2 3 2 2 1
3 6 5 3 2
4 12 15 3 2
5 6 8 3 2
6 3 2 4 1
7 6 5 3 3
8 12 15 3 1
9 6
Hey guys,
I noticed something curious in the lapply call. I'll copy+paste the
function call here because it's short enough:
lapply <- function (X, FUN, ...)
{
FUN <- match.fun(FUN)
if (!is.vector(X) || is.object(X))
X <- as.list(X)
.Internal(lapply(X, FUN))
}
Notice that lapp
Hi,
Just to add:
As you have more columns of data, it would be better to do in this way:
Time1<-as.POSIXct("2013-01-01 00:00:00")
tseq<-seq(Time1,length.out=50,by="secs")
dat2<-data.frame(TIME=tseq,matrix(NA,nrow=50,ncol=3)) #change nrow and ncol
names(dat2)[-1]<-paste("Data",1:3,sep="")
dat
HI Fares,
date1<-
seq.Date(as.Date("1jan2003",format="%d%b%Y"),as.Date("1jan2013",format="%d%b%Y"),by="day")
length(date1)
#[1] 3654F
set.seed(51)
donation<-sample(1000:300,3654,replace=FALSE)
dat1<-data.frame(date1,donation)
library(chron)
nrow(dat1[is.weekend(dat1[,1]),])
#[1] 1044
HI,
You could try this:
Time1<-as.POSIXct("2013-01-01 00:00:00")
tseq<-seq(Time1,length.out=50,by="secs")
dat2<-data.frame(TIME=tseq,Data1=NA,Data2=NA,Data3=NA)
dat3<-read.table(text="
ID Date TIME Data1 Data2 Data3
1 2013-01-01 00:00:00 34
HI,
May be this helps:
dat1<-read.table(text="
ID V1 V2 V3 V4
1 6 5 3 2
2 3 2 2 1
3 6 5 3 2
4 12 15 3 2
5 6 8 3 2
6 3 2 4 1
7 6 5 3 3
8 12 15 3 1
9 6 5 3 3
10 3 2 7 5
Hi,
One more way:
dat1<-read.table(text="
ID V1 V2 V3 V4
1 6 5 3 2
2 3 2 2 1
3 6 5 3 2
4 12 15 3 2
5 6 8 3 2
6 3 2 4 1
7 6 5 3 3
8 12 15 3 1
9 6 5 3 3
10 3 2 7 5
11 6 5
On Jan 5, 2013, at 20:30 , Rolf Turner wrote:
> On 01/06/2013 07:42 AM, David Arnold wrote:
>> Hi,
>>
>> Can someone explain this:
>>
>>> options(digits=20)
>>> 1/3
>> [1] 0.1483
>>
>> Why the 1483 at the end?
>
> There are 10 sorts of people; those who understand binary
> ari
On 01/06/2013 07:42 AM, David Arnold wrote:
> Hi,
>
> Can someone explain this:
>
>> options(digits=20)
>> 1/3
> [1] 0.1483
>
> Why the 1483 at the end?
There are 10 sorts of people; those who understand binary
arithmetic and those who don't. See also FAQ 7.31.
cheers,
Rolf
This is a slight modification of John's approach using 6 variables and 28
observations:
set.seed(42)
xx <- data.frame(aa = 1:28, matrix(sample(1:6, 6*28,
replace = TRUE), nrow= 28))
dd <- ((1:nrow(xx)-1) %/% 7) +1
result <- aggregate(xx[,-1], by=list(dd), FUN=mean)[dd,-1]
result <- data.f
On Sat, Jan 5, 2013 at 6:48 PM, R. Michael Weylandt
wrote:
> To be curt, "it's complicated," but it basically comes down to the
> fact that 1/3 is not expressible by a finite sequence of powers of 2
> so it can't be perfectly represented in binary.
But of course, that in turn leads to gems like t
On Sat, 5 Jan 2013, David Arnold wrote:
Hi,
Can someone explain this:
options(digits=20)
1/3
[1] 0.1483
Why the 1483 at the end?
Due to floating-point arithmetic, see FAQ 7.31.
Thanks,
David.
david-arnolds-macbook-pro-2:~ darnold$ R --version
R version 2.15.2 (2012-10-
On Sat, Jan 5, 2013 at 6:42 PM, David Arnold wrote:
> Hi,
>
> Can someone explain this:
>
>> options(digits=20)
>> 1/3
> [1] 0.1483
>
> Why the 1483 at the end?
To be curt, "it's complicated," but it basically comes down to the
fact that 1/3 is not expressible by a finite sequence
Hi,
Can someone explain this:
> options(digits=20)
> 1/3
[1] 0.1483
Why the 1483 at the end?
Thanks,
David.
david-arnolds-macbook-pro-2:~ darnold$ R --version
R version 2.15.2 (2012-10-26) -- "Trick or Treat"
Copyright (C) 2012 The R Foundation for Statistical Computing
ISBN 3
On Jan 4, 2013, at 9:13 PM, Brian Z Ring wrote:
I am using metafor to create forest plots. This code gives me the
expected
plot (setting x axis tick marks):
forest(forest$OR, ci.lb=forest$Low, ci.ub=forest$High, at=log(c(.
05, .25,
1, 10)), slab=forest$SNP, atransf=exp)
As does this (set
Well, a rather simple-minded, brute force approach would be to add a factor
variable to the data frame and use aggregate on it.
I am sure there are better ways but this will work.
EXAMPLE
###
xx <- data.frame(aa =1:24,
b = matrix(sample(c(1,2,3,4,5,6), 72, replace = TRUE
On Sat, 5 Jan 2013, Simonas Kecorius wrote:
Dear R users,
Could you share your knowledge on following problem:
Suppose we have dataframe:
ID TIME Data1 Data2Data3
... Data700
1.2013-01-01 00:00:00 3453 6
Dear R users, I came up to a problem by taking means (or other summary
statistics) of a big dataframe.
Suppose we do have a dataframe:
ID V1 V2 V3 V4 V71
16 53 2 3
23 22 1 1
3
Dear R users,
Could you share your knowledge on following problem:
Suppose we have dataframe:
ID TIME Data1 Data2Data3
... Data700
1.2013-01-01 00:00:00 3453 66
55
2.2013-01-01 00:00:01
Homework? There is a no homework policy on this list.
-- Bert
On Sat, Jan 5, 2013 at 3:06 AM, Katarzyna Nurzynska
wrote:
> Hi
>
> I am trying to build a regression model. My data looks like this:
>
> A B C DE
> 1 1 12.575
> 2 0 0 1.64
Jan Hornych gmail.com> writes:
>
> Hi,
>
> I would like to do coloring of map regions based on the region values
> "weight". The approach I am taking is first to break regions into
> equal intervals,
>
> classIntervals(spdf$weight,4)$brks #4 intervals in this case
>
> and coloring all regions
Hello,
The "error" is just a warning, these are not the same thing. And the
warning says that you're giving a wrong scope argument, one that
includes the response. See ?add1. Argument 'scope' is "a formula giving
the terms to be considered for adding or dropping." And you're passing
the entir
Hi
I am trying to build a regression model. My data looks like this:
A B C D E
1 1 1 2.57 5
2 0 0 1.64 3
0 5 1 4.8 1
1 3 0 3.56 168
1 1 1 2.13 1
0 3 1 5
Thanks for the help:your reply has focused my problem
in particular I'm trying to do a converge process;
the robustm() function is:
function (x,z)
1# {eigen<-eigen(x)
2# d<-madmatrix(z)
##I have created this function
Thanks for the help,
For Mario: you're absolutely right in fact the process never stopped!!!
in particular I'm trying to do a converge process;
the robustm() function is:
function (x,z)
1# {eigen<-eigen(x)
2# d<-madmatrix(z)
I'm sorry
I read it again the answers and I realized that the condition
"if(abs(b2[i,i]-b[i,i])<=0.001)" must be tested for the entire matrix and
not only for the diagonal, I misspelled.
thanks
--
View this message in context:
http://r.789695.n4.nabble.com/Iterative-loop-using-repeat-tp46546
Exactly Jim!
I am getting some inspiration from papers like these.
http://www.cs.dal.ca/sites/default/files/technical_reports/CS-2011-04.pdf
In short, I am trying to do a correlation across logs, rather than just
visualizing them.
On Sat, Jan 5, 2013 at 4:55 PM, Jim Holtman wrote:
> I have do
I have done a lot of processing of performance logs from UNIX/ Linux systems
using vmstat,iostat, process accounting, etc. R is very useful in this area.
Is this what you are trying to process?
Sent from my iPad
On Jan 5, 2013, at 1:43, Ramprakash Ramamoorthy
wrote:
> I thought mailing lis
On 5 January 2013 10:04, peter dalgaard wrote:
>> So it looks like some assembly is required, but puzzle pieces are available.
Concerning MT940 spec, I don't know who is regulating this. There is
no RFC or similar.
The closest thing I found was from DB:
https://deutschebank.nl/nl/docs/MT94042_EN.
Dear Matteo,
as I stated in my previous email, if the random effects variance is set to zero
then they are identically zero (as mean=0, sd=0) hence you just need to use
'plm(<...> , model="pooling")'.
Best,
Giovanni
-Messaggio originale-
Da: matteo ognibene [mailto:ognibenemat...@hotmail
On Jan 5, 2013, at 03:23 , Ben Bolker wrote:
> Suzen, Mehmet gmail.com> writes:
>
>>
>> Hello List,
>>
>> Are there any R package that can process MT940/942?
>>
>> Thanks
>>
>> mem
>
> I find it hard to say because I have no idea what those formats are
> (although googling suggests they'r
HI,
May be this helps:
occ.data1<-occ.data[,-c(2:3)]
res<-reshape(occ.data1,direction="wide",idvar=c("specie","Site","Pres","year"),timevar="Rep",v.names="Rep")
res<-res[,c(1:4,7:8,5:6,9)]
res[grep("Rep",names(res))]<-apply(res[grep("Rep",names(res))],2, function(x)
ifelse(is.na(x),0,1))
nam
Hi,
You could also use ?dcast()
occ.data1<-occ.data[,-c(2:3)]
library(reshape2)
res1<-dcast(occ.data1,year+Site+specie+Pres~Rep,value.var="Rep")
names(res1)[grep("[0-9]",names(res1))]<-paste("Rep",1:5,sep="")
res1[,-c(1:4)]<-sapply(res1[,-c(1:4)],function(x) as.integer(is.na(x)))
res1
# year Sit
Hi,
For the second part of your question, use ?merge() or ?join() from library(plyr)
dat1<-read.table(text="
2011-02-01 15.29130
2011-02-08 17.60278
2011-02-15 17.99737
2011-02-22 25.43690
",sep="",header=FALSE,stringsAsFactors=FALSE)
dat2<-read.table(text="
2011-02-01 342.34
2011-02-02 68.45
20
I am using metafor to create forest plots. This code gives me the expected
plot (setting x axis tick marks):
forest(forest$OR, ci.lb=forest$Low, ci.ub=forest$High, at=log(c(.05, .25,
1, 10)), slab=forest$SNP, atransf=exp)
As does this (setting colors):
forest(forest$OR, ci.lb=forest$Low, ci.ub=
Hi
I am trying to build a regression model. My data looks like this:
A B C D E
1 1 1 2.575
2 0 0 1.643
0 5 1 4.8 1
1 3 0 3.56168
1 1 1 2.131
0
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