I would like to run a GLMM on categorical data. In this case the response
data are abundance of insects of 4 trophic groups.
I am unsure whether I need to cbind the categories and run with the model
with family=multinomial4 or family=gaussian.
Also I'm unsure whether I should use lme4 package or
On Nov 1, 2012, at 7:18 PM, kebrab67 wrote:
> Yes thanks guys the pmin did the trick. Primitif maybe but effective.
>
> is.primitive(pmin)
[1] FALSE
David Winsemius, MD
Alameda, CA, USA
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mail
Yes thanks guys the pmin did the trick. Primitif maybe but effective.
On Fri, Nov 2, 2012 at 3:12 AM, David Winsemius [via R] <
ml-node+s789695n4648205...@n4.nabble.com> wrote:
>
> On Nov 1, 2012, at 6:38 PM, arun wrote:
>
> > Hi,
> >
> > You can try this also :
> > set.seed(132)
> >
>
> dat1<-
On Nov 1, 2012, at 6:38 PM, arun wrote:
> Hi,
>
> You can try this also :
> set.seed(132)
>
> dat1<-data.frame(Company=rep(LETTERS[1:4],c(3,4,5,6)),Last_year=sample(2005:2012,18,replace=TRUE))
> dat2<-dat1[order(dat1$Company,dat1$Last_year),]
> dat2$Last<-ave(dat2$Last_year,dat2$Company,FUN=fu
Hi,
You can try this also :
set.seed(132)
dat1<-data.frame(Company=rep(LETTERS[1:4],c(3,4,5,6)),Last_year=sample(2005:2012,18,replace=TRUE))
dat2<-dat1[order(dat1$Company,dat1$Last_year),]
dat2$Last<-ave(dat2$Last_year,dat2$Company,FUN=function(x)
ifelse(x>=2009,2009,x))
head(dat2)
# Company
It is much more efficient to calculate the value of the log-likelihood and
it's gradient at the same time rather than calculating both by separate
functions. Is there a way to use optim so that there is one call to a
function that calculates both.
--
View this message in context:
http://r.
On Nov 1, 2012, at 5:41 PM, kebrab67 wrote:
> I have a set of data with 205 988 observation sand 10 variables , three of
> which are Legal_status, Date_of_incorporation and . I set my time
> horizon from 1989 to 2009. Now I want to know when a company is dead. If
> Last_year is bigger or equal t
I have a set of data with 205 988 observation sand 10 variables , three of
which are Legal_status, Date_of_incorporation and Last_year. I set my time
horizon from 1989 to 2009. Now I want to know when a company is dead. If
Last_year is bigger or equal to 2009 then I say that a new "variable" last
... or let R do all the setup:
lmModels <- lapply(seq_len(nrow(dat2)) , FUN = etc. ## the last result
is returned as the list component
I never declare lists and use for() loops to fill them any more.
-- Bert
On Thu, Nov 1, 2012 at 2:51 PM, Peter Alspach
wrote:
> Tena koe Toby
>
> Try creating
Did you ever show us the call to your function that made it fail
on one machine and succeed on two others? In addition, seeing
the output of traceback() after the failure and the output of
sessionInfo() and conflicts() would help track down the problem.
Bill Dunlap
Spotfire, TIBCO Software
wdun
On 11/01/2012 01:23 PM, risclab wrote:
Hello everyone,
Im working with graphNEL object and want to extract all the nodes which
have adjacent nodes with at least 20 nodes in between them.
acc(graph, graphnodes) obviously provides a list for the accessable
nodes of every node from a node and a nu
Thanks, Mick and Greg.
Greg, I have one doubt: taking into account that scope is the cause and that I
need to pass a data.frame in the data of the lm, why does it run in some
machines?. That point is the one I do not understand.
In any case, I made the change (data.frame) and it will be tes
HI,
Instead of splitting, you could also do this:
dat1<-read.table(text="
Subj Trial tt
1 1 1
1 1 2
1 1 3
1 1 4
1 1 5
1 2 1
1 2 2
1 2 3
1
Hi All,
I am having some real difficulty in trying to carry out constrained
optimization. I have had no problems with the Optim() function but when I
try to constrain the problem I am getting all sorts of error messages.
The model is a simple 2 area, 2 fleet, 1 stock model of a fishery. The code
i
?tryCatch
As in:
> tryCatch(uniroot( function(x) x^2+1, lower=1, upper=2 )$root,
> error=function(e){warning(conditionMessage(e)); NA})
Warning in value[[3L]](cond) :
f() values at end points not of opposite sign
[1] NA
> tryCatch(uniroot( function(x) x^2-1.5, lower=1, upper=2 )$root,
> error=
Hi Matthew,
The problem is that the underlying JVM is running with a fixed maximum heap
size. When you load a workbook in XLConnect, an in-memory representation of
that workbook is created (in Java by Apache POI). Reading and writing data
additionally requires memory due to "communication" of your
Hi,
I'm using the uniroot function, and would like to detect an error which
occurs, for instance, when the values at endpoints are not of opposite
signs. For example:
uniroot( function(x) x^2+1, lower=1, upper=2 ).
I want to say something like:
if "error in uniroot(...)" return NA else return u
HI,
Try this:
I am not sure about the use of paste() here.
To get the output from x:
sapply(x,function(x) x)
[,1] [,2]
[1,] "A" "K"
[2,] "B" "L"
[3,] "C" "M"
[4,] "D" "N"
[5,] "E" "O"
If you want to get the result from:
res<-Reduce(paste, x)
do.call(rbind,strsplit(res," "))
#
I am so sorry for the lack of clarity.
I have a few subjects, every subject has done a number of trials and per
trial there is a measurement per millisecond (represented in the variable
tt), so every unique trial is represented in around 2500 rows.
An example of data would be:
Subj | Trial | t
HI,
Saw your response in Nabble.
Try this:
dat1<-read.table(text="
Subj Trial tt
1 1 1
1 1 2
1 1 3
1 1 4
1 1 5
1 2 1
1 2 2
1 2 3
1 3
I should have been more specific:
y <- list()
a <- c("A", "K")
b <- c("B", "L")
c <- c("C", "M")
d <- c("D", "N")
e <- c("E", "O")
y[[1]] <- a
y[[2]] <- b
y[[3]] <- c
y[[4]] <- d
y[[5]] <- e
y
[[1]]
[1] "A" "K"
[[2]]
[1] "B" "L"
[[3]]
[1] "C" "M"
[[4]]
[1] "D" "N"
[[5]]
[1] "E" "O"
How do I
Hi,
This is still a combination.
split(Reduce(cbind,x),1:nrow(Reduce(cbind,x)))
#$`1`
#[1] "A" "K"
#
#$`2`
#[1] "B" "L"
#
#$`3`
#[1] "C" "M"
#
#$`4`
#[1] "D" "N"
#
#$`5`
#[1] "E" "O"
A.K.
From: Mathijs de Vaan
To: arun
Cc: R help
Sent: Thursday, Novembe
HI,
May be this also helps:res<- subset(df,type=="Expresssion"|type=="DNase HS")
res$indx<-c(0,cumsum(abs(diff(as.integer(res$type=="Expresssion")
res1<-with(res,aggregate(dpeak,by=list(indx=indx),tail,1,type=="DNase HS"))
res1<-res1[!is.na(res1[,2]),]
res2<-res[res$type=="Expresssion"|res[,7]%
Tena koe Toby
Try creating a list and storing each model as an element of the list. For
example,
lmModels <- vector('list', nrow(dat2))
for (i in 1:nrow(dat2))
{
lmModels[[i]] <- etc
}
Generally, I add a line after creating the list to name the elements and then
address the list using these
Thanks very much for identifying this error Michael- of course I don't want to
exponentiate an exponential! I had copied the provided formula a little too
literally and hadn't done my QA/QCs
John
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Thu
Hi Marius,
I'm not sure if this helps:
1. knitr::spin() can convert a roxygen script to a report, although it
has no support to any roxygen @tags, e.g.
https://github.com/yihui/knitr/blob/master/inst/examples/knitr-spin.R
In your case, you can spin('code.R', format = 'Rnw')
2. knitr does have co
On Nov 1, 2012, at 2:16 PM, nrm2010 wrote:
>
>
> Dear helpeRs-
>
> I'm using a for loop to create a series of models.
> I'm trying to assign a name to each model created,
> using the loop index. The loop gets stuck at the name
> of the model, giving the error "target of assignment
> expan
Dear helpeRs-
I'm using a for loop to create a series of models.
I'm trying to assign a name to each model created,
using the loop index. The loop gets stuck at the name
of the model, giving the error "target of assignment
expands to non-language object". The linear model runs
without er
Hello everyone,
Im working with graphNEL object and want to extract all the nodes which
have adjacent nodes with at least 20 nodes in between them.
acc(graph, graphnodes) obviously provides a list for the accessable
nodes of every node from a node and a number of the edges between them.
Like th
Thanks to you all for your help with this code Basically, the purpose of this
was to create a column detailing whether it is the 1st, 2nd, 3rd, etc time that
the site was sampled. I need this for creating a graph which shows this new
variable on the x axis versus a variable of interest. The var
Does
?all.vars
##as in
> all.vars(y~x)
[1] "y" "x"
help?
-- Bert
On Thu, Nov 1, 2012 at 11:04 AM, Hafen, Ryan P wrote:
> I need to find all global variables being used in a function and
> findGlobals() in the codetools package works quite nicely. However, I am not
> able to find variables t
On Nov 1, 2012, at 12:57 PM, Marius Hofert wrote:
> So for putting in some lines of code in a
> paper, this is quite restrictive (that's why I was hoping for a latex-related
> solution -- still asking on r-help since I'd expect R users to have
> encountered
> this issue before).
>
> Cheers,
>
Dear Yihui, Dear Frank,
Thanks for helping.
I am aware of Sweave and knitr and this is an incredible development. Still,
this "machinery" is too heavy for my needs. I typically use pure R scripts for
my work (well documented with Roxygen). I can then easily debug and
are not disturbed by addition
It is not clear exactly what you want.
Do you want the final object to be a matrix with 2 columns (or the
number of columns equal to the number of items in the list? If so you
could just use do.call(cbind, x) and you would not need reduce or
paste (paste is the wrong tool unless you want a single
See comments inline
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Mick Cooney
> Sent: Thursday, November 01, 2012 9:23 AM
> To: Greg Snow
> Cc: r-help@r-project.org; Eva Prieto Castro
> Subject: Re: [R] Problem with lm
>
>
On Thu, Nov 1, 2012 at 6:47 PM, mdvaan wrote:
> I have a question about the Reduce function:
>
> x <- list()
> x[[1]] <- LETTERS[1:5]
> x[[2]] <- LETTERS[11:15]
> Reduce(paste, x)
> [1] "A K" "B L" "C M" "D N" "E O"
>
> How do I get this?:
> [1] "A" "K"
> [2] "B" "L"
> [3] "C" "M"
> [4] "D" "N"
>
What is "this"? A matrix with two columns? A data frame? You did an
excellent job of providing data to reproduce your attempt so far;
thank you. But I'm not sure what you want as result. But maybe one of
these two things?
> data.frame(x, stringsAsFactors=FALSE)
c..ABCDE.. c..K...
HI,
By splitting that code into smaller components:
res1<-strsplit(v2,"_") # will split the elements into substrings wherever it
matches substring "_". Result will be a list
head(res1)
#[[1]]
#[1] "sex" "1"
#[[2]]
#[1] "sex" "2"
#
[[3]]
#[1] "sex" "3"
#
#[[4]]
#[1] "age" "height" "
I need to find all global variables being used in a function and findGlobals()
in the codetools package works quite nicely. However, I am not able to find
variables that are used in formulae. Simply avoiding formulae in functions is
not an option because I do not have control over what functio
HI,
May be this is a bit closer:
res1<-unlist(lapply(strsplit(v2,"_"),function(x)
paste(x[1],x[3],x[2],x[4],sep="_")))
v1[1]
#[1] "age_1"
v2[grep(v1[1],res1)]
#[1] "age_height_1_1" "age_height_1_7"
v1[8]
#[1] "height_2"
v2[grep(v1[8],res1)]
#[1] "age_height_2_2"
A.K.
- Original Messa
Hi,
I am using premake4 generated make files to compile my projects, which
works fine if I am using mingw32-make.exe from the MinGW 32 bit
distribution, but doesn't work with the make.exe supplied by the latest
RTools 2.16.
That make.exe first complains it can't create directories and then that i
I am using the mle2 function to run a simple linear model with a normal
distribution. I have one continuous variable (X) and one factor (treatment)
with 6 levels. I have seen ANCOVA examples similar to this, but my goal is
to compare the differences in the slope parameters (the interaction term)
am
I have a question about the Reduce function:
x <- list()
x[[1]] <- LETTERS[1:5]
x[[2]] <- LETTERS[11:15]
Reduce(paste, x)
[1] "A K" "B L" "C M" "D N" "E O"
How do I get this?:
[1] "A" "K"
[2] "B" "L"
[3] "C" "M"
[4] "D" "N"
[5] "E" "O"
Thanks for your help!
--
View this message in context
But the 'y' is a parameter to the function, so does this mean that the
error is occurring when the function is invoked without that
parameter?
On 1 November 2012 16:02, Greg Snow <538...@gmail.com> wrote:
> Yes, it is most likely due to scoping. It is safest to create a data
> frame with all the
On 1 November 2012 11:47, Hasan Diwan wrote:
> Sir,
>
> On 30 October 2012 04:32, aajit75 wrote:
>>
>> Please suggest where am i going wrong or alternate option to solve such
>> issues while reading large DB table.
>>
>
> You might consider setting the fetch size. I haven't used RJDBC in a few
Sir,
On 30 October 2012 04:32, aajit75 wrote:
>
> Please suggest where am i going wrong or alternate option to solve such
> issues while reading large DB table.
>
You might consider setting the fetch size. I haven't used RJDBC in a few
months, but hopefully, the hints on
http://www.rforge.net/
Or see https://github.com/yihui/knitr-examples/blob/master/015-listings.Rnw
for a minimal example.
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Thu, Nov 1, 2012 at 1:23 PM, Frank Harrell w
The knitr package makes this relatively easy to do. See for example
http://biostat.mc.vanderbilt.edu/KnitrHowto
Frank
Marius Hofert-3 wrote
> Dear expeRts,
>
> What's a 'good' (nice-looking, easy-to-read) setup for the LaTeX package
> 'listings' to display R code?
>
> The two versions below a
Hi Paul,
So you have described bootstrapping in SEM, but that does not address
the cross-validation. See inline.
On Thu, Nov 1, 2012 at 10:24 AM, Paul Miller wrote:
> Hello All,
>
> Recently, I was asked to help out with an SEM cross-validation analysis.
> Initially, the project was based on "
Hello,
A bit confusing:
" I would like to extract
all rows (so called* defined row*s) with type==Expression - subset (df,
type==Expression) - and the aforegoing type==DNase HS (which is not
necessarly row n-1 - assumung that the defined row is n"
In the dataset, there is "Expresssion" for column
Actually forget that loop, that was old code. Now remember i had to actually
split the one query down into several queries to get everything.
On 01.11.2012, at 18:38, Jessica Streicher wrote:
> I didn't have the same exception yet, but please make sure you have a stable
> connection (avoid wlan
On Thu, Nov 1, 2012 at 10:28 AM, Hermann Norpois
wrote:
> Hello,
>
> my data is sorted by start.ens (see below). And now I would like to extract
> all rows (so called* defined row*s) with type==Expression - subset (df,
> type==Expression) - and the aforegoing type==DNase HS (which is not
> necessa
I didn't have the same exception yet, but please make sure you have a stable
connection (avoid wlan).
I personally ran out of memory very often and had to make several queries,
combining the results in R instead.
This is probably not the best thing to do, but worked for me to fetch large
amount
Hello,
my data is sorted by start.ens (see below). And now I would like to extract
all rows (so called* defined row*s) with type==Expression - subset (df,
type==Expression) - and the aforegoing type==DNase HS (which is not
necessarly row n-1 - assumung that the defined row is n). I dont know how
t
Hello All,
Recently, I was asked to help out with an SEM cross-validation analysis.
Initially, the project was based on "sample-splitting" where half of cases were
randomly assigned to a training sample and half to a testing sample. Attempts
to replicate a model developed in the training sample
I really don't understand what you want to achieve and how your data presents
itself.
You have several combinations of trial/subject which are unique?
There is a variable for each such combination that you want to test against
(tt)?
In addition you have data on each such combination (several set
This error can still occur even if something is passed as the y
parameter to the function. There are several things that can
complicate the process (and I don't remember exactly which one causes
the problem in this case). Here are a couple:
In the original function, y is a parameter, but it is n
Dear Claudia:
Your question has many layers.
Technically speaking, when I do "2011-05-01 CEST" -"2011-04-01 CEST" I get:
> "2011-05-01 CEST" -"2011-04-01 CEST"
Error in "2011-05-01 CEST" - "2011-04-01 CEST" :
non-numeric argument to binary operator
As previous posters suggest, storing the dat
Yes, it is most likely due to scoping. It is safest to create a data
frame with all the data in it, then pass that to the data argument of
lm.
On Thu, Nov 1, 2012 at 2:25 AM, Eva Prieto Castro wrote:
> Hi,
>
> I have a problem in relation with a packahe I made. It runs on my machine
> (Windows,
Hello,
As I am new with R I am completely stuck in resolving a, no doubt, easy
problem.
I have a dataset with an enormous amount of rows and 17 columns. I need to
know per trial and subject number if another variable (tt) exceeds a
maximum. If this is true than the last 5000 rows of that specific
On 30.10.2012 15:59, tudor wrote:
Dear list:
I would like to recreate how the artificial data set BregFix was generated
in package flexmix (thanks Bettina and Friedrich). The data set is
thoroughly described in Grun and Leisch's Computational Statistics & Data
Analysis 51(11) :5247-5252 but r
Oh hey Jan, sorry, I just saw I did not read correctly and mistook your quote
of Don's mail as a signature. So thank's to you for the second posting
Bernhard
--
View this message in context:
http://r.789695.n4.nabble.com/Loop-over-several-variables-tp4648112p4648132.html
Sent from the R help
Hey Don,
I just saw your second post ... that's even better!
Thanks again
Bernhard
--
View this message in context:
http://r.789695.n4.nabble.com/Loop-over-several-variables-tp4648112p4648131.html
Sent from the R help mailing list archive at Nabble.com.
Hey Don,
many thanks, I tried that and it works just fine. I have two questions
though:
1. In Addition to generating the t1-t5 (I have five iterations, in fact),
the code will generate an additional ti(without subscript), which seems to
be a copy of ti5. Is that what it should do ( and if yes, wh
Or
ti <- aggregate(dataframename[paste0("y", 1:3)],
by=dataframename["aggregationvar"],
sum,na.rm=TRUE)
which gives you all results in one data.frame.
Jan
"MacQueen, Don" schreef:
Many ways. Here is one:
### supposing you have y1, y2, and y3 in your data frame
for (i in
Hi A.K.
Thank you so much for replying this could work but the problem you recreate the
newv1, to match v2 but I don't want to do that. I just want to check if v2 has
that component from v1
THanks
Date: Thu, 1 Nov 2012 06:32:03 -0700
From: ml-node+s789695n4648123...@n4.nabble.com
To: frespi
Thanks a lot!!!
--
View this message in context:
http://r.789695.n4.nabble.com/help-extract-data-using-column-names-tp4647869p4648125.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz
Hi,
Not sure whether I understand it correctly.
v1 <-
c("age_1","age_2","age_3","age_4","age_5","age_6","height_1","height_2","height_3","height_4","height_5","height_6","height_7","height_8")
v2 <-
c("sex_1","sex_2","sex_3","age_height_1_1","age_height_2_2","age_height_3_3","age_height_4_4"
Many ways. Here is one:
### supposing you have y1, y2, and y3 in your data frame
for (i in 1:3) {
yi <- paste('y',i,sep='')
ti <- aggregate(dataframename[[yi]],
by=data.frame(dataframename$aggregationvar),
sum,na.rm=TRUE)
assign( paste('ti',i,sep='') , ti, '.GlobalE
Hi,
I am new R user and I am still learn this fabulous software. I am stuck on
this part. I need to check 2 character vectors
v1 <-
c("age_1","age_2","age_3","age_4","age_5","age_6","height_1","height_2","height_3","height_4","height_5","height_6","height_7","height_8")
and
v2 <-
c("sex_1","s
Ah nice. I clearly misunderstood and thought that facet_wrap() made sense.
John Kane
Kingston ON Canada
> -Original Message-
> From: ruipbarra...@sapo.pt
> Sent: Wed, 31 Oct 2012 19:45:33 +
> To: dysonspher...@gmail.com
> Subject: Re: [R] boxplots of various levels
>
> Hello,
>
HI,
Without an example dataset, it is a bit difficult to test.
Just a comment about your code:
"
TableAP <- cbind(zoo(TableAP), Lagpolity4AP1, Lagpolity4AP2, Lagpolity4AP3,
Lagpolity4AP4, Lagpolity4AP5,
LaglrgdpchAP1, LaglrgdpchAP2, LaglrgdpchAP3, LaglrgdpchAP4, LaglrgdpchAP5)
TableAP <- data
Hey everybody,
I am looking for a way to loop commands over several variables in a
dataframe. Basically I want to do something like this:
ti1<-aggregate(dataframename$y1,
by=data.frame(dataframename$aggregationvar),
sum,na.rm=TRUE)
This works fine as it is but i wan
TableAPS1 <- subset(TableAP, year >= 1959) ??
John Kane
Kingston ON Canada
> -Original Message-
> From: iamsta...@gmail.com
> Sent: Wed, 31 Oct 2012 20:19:46 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] Subsetting year range
>
> Hi,
>
> I have a panel data set that I am tryin
Hi,
Just to add:
Infact, there is a small difference in both the approaches.
For e.g when tested with your data
sapply(l,`[`,1:4)
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1.0 3 4 2 3
#[2,] 2.0 4 2 4 5
#[3,] 3.7 5 5 6 7
#[4,] NA 6 7 3 2
sapply(l,hea
Hi,
You can also use:
sapply(l,`[`,1:4)
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 3 4 2 3
#[2,] 2 4 2 4 5
#[3,] 3 5 5 6 7
#[4,] 7 6 7 3 2
A.K.
- Original Message -
From: D. Rizopoulos
To: Al Ehan
Cc: "r-help@r-project.org"
Sen
Dear R community,
I have recently discovered the package oblique.tree and I must admit that
it was a nice surprise for me,
since I have actually made my own version of a kind of a classifier which
uses the idea of oblique splits (splits by means of hyperplanes).
So I am now interested in comparing
On 01.11.2012 06:29, Manish Gupta wrote:
> HI,
>
> I need to parse an xml where key name are same but values are different.
>
>
>
> Some dummy text
>
> one
> two
> three
>
>
> When i use xmlGetAttr() function i always get one as value. How can i g
Dear expeRts,
What's a 'good' (nice-looking, easy-to-read) setup for the LaTeX package
'listings' to display R code?
The two versions below are partly inspired by the settings of the package
SweaveListingUtils and
http://r.789695.n4.nabble.com/R-How-to-format-R-code-in-LaTex-documents-td816055.ht
Eva,
What do you get when you type ls() on the machine that the code does not
run? It could be that "y" is not in that particular workspace.
HTH,
Jorge.-
On Thu, Nov 1, 2012 at 7:25 PM, Eva Prieto Castro <> wrote:
> Hi,
>
> I have a problem in relation with a packahe I made. It runs on my machi
On 10/31/2012 11:47 PM, fillay89 wrote:
I am trying to run the R Script below, I have actually simplified it to just
this part that is causing issues. When I run this script I continue to get
an error that says "cannot rescale a constant/zero column to a unit
variance". I cannot figure out what
Thanks a lot!
best
On Thu, Nov 1, 2012 at 8:19 AM, D. Rizopoulos wrote:
> try this:
>
> l <- list(c(1,2,3,7), c(3,4,5,6,3), c(4,2,5,7), c(2,4,6,3,2), c(3,5,7,2))
>
> sapply(l, head, 4)
>
>
> I hope it helps.
>
> Best,
> Dimitris
>
>
> On 11/1/2012 9:11 AM, Al Ehan wrote:
> > Hi,
> >
> > I have t
Hi,
I have a problem in relation with a packahe I made. It runs on my machine
(Windows, where I made the package), and it runs in a Mac machine, but it does
not run in another Mac machine with the same R version.
The part of the code is giving problems:
singleCosinor <- function(t, y, period=2
try this:
l <- list(c(1,2,3,7), c(3,4,5,6,3), c(4,2,5,7), c(2,4,6,3,2), c(3,5,7,2))
sapply(l, head, 4)
I hope it helps.
Best,
Dimitris
On 11/1/2012 9:11 AM, Al Ehan wrote:
> Hi,
>
> I have this lame question. I want to convert a list (each with varies in
> length) to matrix with same row len
I have a r-script (rook.R) that starts a Rook server. To present users
from having to start R and type in source("rook.R"), I want to create
a bash script and bat file that starts R and sources the script.
However, to keep the Rook server running R should not close after
running the scrip
Hi,
I have this lame question. I want to convert a list (each with varies in
length) to matrix with same row length by eliminating vectors outside the
needed range.
For example:
l<-list(NULL)
l[[1]]=1,2,3.7
l[[2]]=3,4,5,6,3
l[[3]]=4,2,5,7
l[[4]]=2,4,6,3,2
l[[5]]=3,5,7,2
#so say I want to only h
On Wed, Oct 31, 2012 at 9:14 PM, Worik R wrote:
> Compiling this little function gets me some strange behaviour
>
> .initDataDir <- function(){
> if(file.exists(LOCATION)) {
> if(as.logical(file.info(LOCATION)["isdir"]))return
> stop(LOCATION, " exists but is not a directory")
> }
>
On Thu, Nov 1, 2012 at 2:44 AM, wrote:
> Hi
> I'd like to fit an asymmetrical curve function to some physiological data.
> I've been told a weibull curve is a good place to start, but I'm having
> trouble specifying and fitting the function with nls and was wondering if
> someone could help.
>
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