Hi,
Try:
example <- read.table(text="ID Sex Location CL
1 F lake1 40
1 F lake1
1 F lake1 43
2 M lake1 30
3 M lake2 22
4 F lake2 25
4 F lake2 27",sep="",header=TRUE,stringsAsFactors=FALSE,fill=TRUE)
aggregate(CL~.,example,mean,na.rm=TRUE)
I'm pretty new to GLMMs and model averaging, but think I'm getting some
understanding of it all through lots of reading. However, I keep receiving
an error message when trying to average models that I don't understand and
can't find any resources about. I'm doing science education research trying
t
On 11/15/2013 12:53 PM, jpm miao wrote:
Hi,
I find a few Pearson correlation test functions like
fBasics::correlationTest or stats::cor.test
which give the p-value of the test result. Is there a function that
calculate the cutoff correlation value for a specific p-value , e.g., p =
0.0
Hi,
I find a few Pearson correlation test functions like
fBasics::correlationTest or stats::cor.test
which give the p-value of the test result. Is there a function that
calculate the cutoff correlation value for a specific p-value , e.g., p =
0.05?
I have a plot for the cross correlations
I'm using caret to assess classifier performance (and it's great!). However,
I've found that my results differ between R2.* and R3.* - reported accuracies
are reduced dramatically. I suspect that a code change to kernlab ksvm may be
responsible (see version 5.16-24 here:
http://cran.r-project.
Jean-Francois Chevalier bisnode.com> writes:
>
You have already given the answer yourself. You have binary variables x(j, i),
you need to set up the inequalities, and then apply one of the mixed-integer
linear programming solvers in R, for instance 'lpSolve', 'Rglpk', 'Rsymphony'.
Setting up th
Thanks Dave for your help on this. Waiting other suggestions from the list on
this.
Regards
Bander
> On 14 Nov 2013, at 09:29 pm, "David R Forrest" wrote:
>
> Hi Bander,
>
> I'm pushing this discussion back to the list, because I'm not sure of the
> shape/rate parameters for rpareto and rexp
During startup R gives a Warning message: "Setting LC_CTYPE=en_US gnumeric
& failed" What is the error and how can I fix it?
[[alternative HTML version deleted]]
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On 13-11-14 4:41 PM, Lopez, Dan wrote:
It's happening in the 64bit installation of R I have too (w/o Rstudio). This is
R version 2.15.1
That's kind of an old R version (from June 2012); there were two more
releases in the 2.15.x series. The current release is 3.0.2. Can you
upgrade?
Dunc
It's happening in the 64bit installation of R I have too (w/o Rstudio). This is
R version 2.15.1
Dan
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Thursday, November 14, 2013 1:38 PM
To: Lopez, Dan; R help (r-help@r-project.org)
Subject: Re: [R] Window
On 13-11-14 3:50 PM, Lopez, Dan wrote:> Windows 7, R 2.15.1 64bit,
RStudio 0.97.312, Rattle 2.6.26
>
> Hi,
> Please help.
>
> I removed rattle then reinstalled then loaded then attempted to open
rattle with rattle(). But each time I do that My R session in R studio
aborts and restarts.
>
> rem
Hi Bander,
I'm pushing this discussion back to the list, because I'm not sure of the
shape/rate parameters for rpareto and rexp and how they'd be applied across
this mix of typo'd papers.
# Reed Equation 6 http://www.math.uvic.ca/faculty/reed/dPlN.3.pdf exponentiated
per end of sec 3:
rdpln<-f
Windows 7, R 2.15.1 64bit, RStudio 0.97.312, Rattle 2.6.26
Hi,
Please help.
I removed rattle then reinstalled then loaded then attempted to open rattle
with rattle(). But each time I do that My R session in R studio aborts and
restarts.
remove.packages("rattle")
install.packages("rattle")
libr
Thanks, this was a useful pointer. Since the function I am trying to fit is
exponential, I decided to use nls. And I was able to reproduce exactly the
results and the plot in the URL I had posted. For future reference, here is the
R code I wrote:
require("gplots")
xx <- 1:10
yy <- c(1.56,1.20,
Please post the packages from which 'barabasi' and 'layout.fruch'
originate (not to mention whatever the plot() method is for whatever class
your 'g' is). Further, without seeing what your data look like we have no
way of knowing whether you've fed the appropriate elements of "L" to chull.
e
layout.fruchterman.reingold(g) returns a random result, so you
want to call it once and use the one return value. Also, I think
you need to avoid the rescaling that plot.igraph does. It looks
like you need to explicitly specify xlim and ylim if you do that,
but I may not have looked long enough a
On Thu, Nov 14, 2013 at 12:06 PM, Pooya Lalehzari
wrote:
> Hello,
> I noticed an unexpected behavior when using the xts object and I was
> wondering if anyone knows why that happens. I have a code to create a new
> column and copy one of the columns to the new column (please see below):
>
> libr
It would be more clear if you tell, what you want to do instead of what you do
not want to do.
If you start with a usual cost matrix (whatever cost function you have) and you
have to assign N to N this reduces to the well-known Munkre’s algorithm (see
for example: http://gallery.rcpp.org/artic
Hi:
I want to create a polygon encircling a graph. For this i use convex
hull to get the coordinate points for polygon.
g <- barabasi.game(10)
L<-layout.fruchterman.reingold(g)
temp1 <- chull(L)
temp1 <- c(temp1, temp1[1])
plot(g, layout=layout.fruchterman.reingold)
But when i plot the polygon
Hello,
I'm trying to solve a multiple assignment problem.
I found a package Adagio and its function mknapsack which
maximize vstar = p(1)*(x(1,1) + ... + x(m,1)) + ... ... + p(n)*(x(1,n) + ... +
x(m,n))
subject to w(1)*x(i,1) + ... + w(n)*x(i,n) <= k(i) for i=1,...,m
x(1,j) + ... + x(m,j) <= 1
Hello,
I noticed an unexpected behavior when using the xts object and I was wondering
if anyone knows why that happens. I have a code to create a new column and copy
one of the columns to the new column (please see below):
library(xts)
df = data.frame(stringsAsFactors=FALSE)
df[1:3,"date"] = c("
HI,
A more general form would be:
D[rep(1:nrow(D),D[,3]),-3]
A.K.
On Thursday, November 14, 2013 12:27 PM, arun wrote:
Hi,
Try:
D1 <- D[rep(row.names(D),D[,3]),-3] ##assuming rownames(D) are from 1:nrow(D)
row.names(D1) <- 1:nrow(D1)
A.K.
On Thursday, November 14, 2013 5:32 AM, peron
do.call (rbind, testlist) and Reduce (rbind, testlist) are perfect.
Thanks
hermann
2013/11/14 arun
>
>
> Hi Hermann,
>
> You may try:
> do.call(rbind,testlist)
> #or
> Reduce(rbind,testlist)
> #or
> library(plyr)
> ldply(testlist)
> A.K.
>
>
>
>
> On Thursday, November 14, 2013 8:27 AM, Herman
Hi,
I am trying to estimate a 2sls using panel data (random effect model). I
tried the same estimation in STATA using the ivtreg2 command. However STATA
and R are giving me two different results. I figure there is something with
my R code:
iv=plm(formula=wecon~fdistockgdp +trade + polrightsrevers
Also,
library(stringr)
str_sub(s2,-5,-1)
str_extract(s1,"[[:alpha:]]+")
A.K.
On Thursday, November 14, 2013 10:50 AM, arun wrote:
Try:
s1 <- "hello.world"
s2 <- c("GLEm0045", "GLEn0042", "GLEz0048")
substr(s1,1,5)
#or
gsub("\\..*","",s1)
n <- 5
substr(s2,nchar(s2)-n +1,nchar(s2))
A.K.
Try:
s1 <- "hello.world"
s2 <- c("GLEm0045", "GLEn0042", "GLEz0048")
substr(s1,1,5)
#or
gsub("\\..*","",s1)
n <- 5
substr(s2,nchar(s2)-n +1,nchar(s2))
A.K.
If I have a column that has "hello.world" how could I extract hello?
In addition in a column how can I extract the last 5 characte
Hi,
I think you used a column that doesn't exist in the dataset.
Targetstation <- read.table(text="V1 V2 V3 V4 V5 V6 V7
0 0 0 1.2 0 0 0.259
0 0 12.8 0 23.7 0 8.495
6 0 81.7 0.2 0 20 19.937
0 1.5 60.9 0 0 15.5 13.900
1 13 56.8 17.5 32.8 6.4 27.654
4 3 66.4 2 0.3 NA 17.145",sep="",header=T
Hi,
Try:
D1 <- D[rep(row.names(D),D[,3]),-3] ##assuming rownames(D) are from 1:nrow(D)
row.names(D1) <- 1:nrow(D1)
A.K.
On Thursday, November 14, 2013 5:32 AM, peron wrote:
Hello
I have data in following form : 100 ind describe by two variables x and y.
D<-data .frame( x=rnorm(3), y=r
Hi Hermann,
You may try:
do.call(rbind,testlist)
#or
Reduce(rbind,testlist)
#or
library(plyr)
ldply(testlist)
A.K.
On Thursday, November 14, 2013 8:27 AM, Hermann Norpois
wrote:
Hello,
having a list like testlist I would like to transform it in dataframe. How
does it work?
Thanks
Herman
Hi,
the output of lapply() is a list; see ?lapply and ?sapply.
# if you know the length of your list in advance,
# this definition is better:
uu <- vector("list", 2)
# list elements
uu[[1]] <- c(1,2,3)
uu[[2]] <- c(3,4,5)
# some options to achieve what you want:
matrix(unlist(uu), 2, 3, T)
do
Carl Witthoft wrote
> Well, given that an upvoted question on math.stackexchange got no answers,
> I'd say you're asking a very difficult question. Perhaps this paper
> http://www.geometrictools.com/Documentation/IntersectionOfEllipsoids.pdf
>
> will be of some help. It's possible that you c
Hi, Carl Witthoft
yes, it looks like a mathematical question. I will try based on your
suggestion to calculate the volume of the intersection. But I still want to
know whether there are some functions in R which can calculate the volume of
an ellipsoid(area for p=2, hypervolume for p>3) containing
Hi,
Thanks for your response.
On Thu, Nov 14, 2013 at 4:44 PM, Terry Therneau wrote:
> I think that your data is censored, not truncated.
> For a fault introduced 1/2005 and erased 2/2006, duration = 13 months
> For a fault introduced 4/2010 and still in existence at the last
> observation 1
Hi Nina,
I think the following code does what you want (thanks to Jim Lemon for showing
me what you wanted in terms of x-axis tick labels)
However I think that barcharts are generally evil so I changed your geom_bar
to geom_point. Feel free to change it back if your discipline requires it bu
Thanks Martin, for making this clear to me, I thought of Pearson’s
Method-of-Moments.
On 14 Nov 2013, at 16:08, Martin Maechler wrote:
>> "SZ" == Simon Zehnder
>>on Thu, 14 Nov 2013 08:52:16 +0100 writes:
>
>SZ> Check the gmm package with a weighting matrix equal to
>SZ> th
Thanks all! So many ways
On Thu, Nov 14, 2013 at 10:35 AM, Rui Barradas wrote:
> Hello,
>
> You are applying rbind to each element of the list, not rbinding it with
> the others. Try instead
>
> do.call(rbind, uu)
>
> Hope this helps,
>
> Rui Barradas
>
> Em 14-11-2013 15:20, Brian Smith es
With a sample data frame:
d = data.frame(id = rep(c(1,2,3,4),5), x = round(rexp(20), digits=2), y =
round(runif(20), digits=2))
I would like to generate 100 bootstrap data with replacement and save each
bootstrap data as b1, b2, ..., b100. I attempted the sample function but
could not make it work
I think that your data is censored, not truncated.
For a fault introduced 1/2005 and erased 2/2006, duration = 13 months
For a fault introduced 4/2010 and still in existence at the last observation 12/2010,
duration> 8 months.
For a fault introduced before 2004, erased 3/2005, in a machine
Hello,
You are applying rbind to each element of the list, not rbinding it with
the others. Try instead
do.call(rbind, uu)
Hope this helps,
Rui Barradas
Em 14-11-2013 15:20, Brian Smith escreveu:
Hi,
I was trying to use lapply to create a matrix from a list:
uu <- list()
uu[[1]] <- c(1,2
On 14-11-2013, at 16:20, Brian Smith wrote:
> Hi,
>
> I was trying to use lapply to create a matrix from a list:
>
> uu <- list()
> uu[[1]] <- c(1,2,3)
> uu[[2]] <- c(3,4,5)
>
> The output I desire is a matrix with 2 rows and 3 columns, so I try:
>
> xx <- lapply(uu,rbind)
>
> Obviously, I'
Thanks for the reproducable example. I can confirm that it fails on my machine using
survival 2-37.5, the next soon-to-be-released version,
The issue is with NextMethod, and my assumption that the called routine inherited
everything from the parent, including the environment chain. A simple t
Hi,
I was trying to use lapply to create a matrix from a list:
uu <- list()
uu[[1]] <- c(1,2,3)
uu[[2]] <- c(3,4,5)
The output I desire is a matrix with 2 rows and 3 columns, so I try:
xx <- lapply(uu,rbind)
Obviously, I'm not doing something right, but what!?
[[alternative HTML versi
> "SZ" == Simon Zehnder
> on Thu, 14 Nov 2013 08:52:16 +0100 writes:
SZ> Check the gmm package with a weighting matrix equal to
SZ> the identity. Best
SZ> Simon
No."MM" is ambigous: The gmm package is about
"Generalized Method of Moments"
but Izhak is reall
This is not an R question per se. It really seems like an RStudio question.
They have their own help forum and it is probably best to ask there.
My first thought was that you had just closed the output window in RStudio but
in my version, 0.97.449 under Ubuntu 13.10 the window automatically
On Thu, Nov 14, 2013 at 1:13 AM, Frans Marcelissen
wrote:
> R purists forbid the use of the for loop,
That is utter nonsense. Please do not make such statements when you
have no idea what you're talking about. It just promulgates confusion.
-- Bert
> simple solution:
>
> for (i in 1:(length(V6)
On Thu, Nov 14, 2013 at 2:26 AM, Tstudent wrote:
> I have the following two xts object:
>
> https://dl.dropboxusercontent.com/u/102669/series1.rdata
>
> https://dl.dropboxusercontent.com/u/102669/series2.rdata
>
> With str i see that they are both xts objects
>
> I can't understand why it's imposs
You should cc r-help on all correspondence, so others can follow the thread.
Here is a very simple example of a home-made smooth function. Perhaps you
can modify it to suit your needs.
Jean
# simple smooth function, using a weighted mean
smoothfunction <- function(allxs, allys, centerindex, ha
Hi. Here is one way:
l <- list(structure(list(BP_A = c(27001689L, 27001689L, 27001689L,
27001689L, 27001689L, 27001689L), SNP_A = c("rs4822747", "rs4822747",
"rs4822747", "rs4822747", "rs4822747", "rs4822747"), BP_B = c(27002392L,
27004298L, 27004902L, 27004964L, 27005122L, 27005158L), SNP_B =
c("r
Dear All,
recent desktop versions of Adobe Acrobat and Adobe Reader have
built-in support for 3D models in PDF.
You can take a look at http://www.pdf3d.com/gallery.php for a gallery
of professional results achieved with a commercial tool.
For an overview of what 3D PDF can be used for I (natural
Hello,
having a list like testlist I would like to transform it in dataframe. How
does it work?
Thanks
Hermann
> testlist
[[1]]
BP_A SNP_A BP_B SNP_B R2
2 27001689 rs4822747 27002392 rs4820690 0.695642
3 27001689 rs4822747 27004298 rs5761627 0.695642
4 27001689 rs4822747
I have the following two xts object:
https://dl.dropboxusercontent.com/u/102669/series1.rdata
https://dl.dropboxusercontent.com/u/102669/series2.rdata
With str i see that they are both xts objects
I can't understand why it's impossible to compare each element.
For example: series1 > series2
Wh
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
R purists forbid the use of the for loop, but I am afraid this is the most
simple solution:
for (i in 1:(length(V6)-1)) if(is.na(V6[i])) V6[i]<-V6[i+1]
2013/11/14 Jim Lemon
> On 11/14/2013 04:02 PM, dila radi wrote:
>
>> Hi all,
>>
>> I have a data set which treat missing value as NA and now
Well, given that an upvoted question on math.stackexchange got no answers,
I'd say you're asking a very difficult question. Perhaps this paper
http://www.geometrictools.com/Documentation/IntersectionOfEllipsoids.pdf
will be of some help. It's possible that you could do:
1) find the ellipse o
On 13.11.2013 22:45, Elisa Frutos Bernal wrote:
Hi!
I need to print a graph that I have in a window. Previously I used:
try(win.print(), silent = TRUE)
if (geterrmessage() != "Error in win.print() : unable to start
device devWindows\n") {
plotFunctiond(screen = FALSE)
On 11/14/2013 09:28 PM, peron wrote:
Hello
I have data in following form : 100 ind describe by two variables x and y.
D<-data .frame( x=rnorm(3), y=rnorm(3), size=c(50,10,40))
I want data for individual, i.e, 100 observations for my 100 ind.
Hi Olivier,
From the above, it seems you hav
Hello
I have data in following form : 100 ind describe by two variables x and y.
D<-data .frame( x=rnorm(3), y=rnorm(3), size=c(50,10,40))
I want data for individual, i.e, 100 observations for my 100 ind.
Thank for your help
Olivier Peron
[[alternative HTML version de
Maybe you are after "weights" option given by 'lm' or 'glm'
See:
http://stackoverflow.com/questions/6375650/function-for-weighted-least-squares-estimates
On 14 November 2013 10:01, Erkcan Özcan wrote:
> Thanks, but if you have another closer look to my post, you will see that my
> question has
Thanks, but if you have another closer look to my post, you will see that my
question has nothing to do with drawing error bars on a plot.
What I want is to do a curve fit to a data with error bars.
Best,
e.
On 14 Nov 2013, at 04:21, Suzen, Mehmet wrote:
> If you are after adding error bars in
On 11/14/2013 04:02 PM, dila radi wrote:
Hi all,
I have a data set which treat missing value as NA and now I need to replace
all these NA's by using number in the same row but different column.
Here is the part of my data:
V1 V2 V3 V4 V5 V6 V7 0 0 0 1.2 0 0 0.259 0 0 12.8 0 23.7 0 8.495 6
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