I'm trying to install FEAR package on my personal laptop with Windows 7
operating system. I've followed the instructions step by step but still
encountered this problem:I thought I installed FEAR in the default
location,C:Program FilesRR-3.0.1libraryFEARHowever, I couldn't find it in the
librar
I am still trying to get my head around biomod2. I have run through the
tutorial a few times, which works really well in a linear format.
But, I want to see the models and assess them at every part of the process. So,
I need to:
1: be able to re-access all the files from /.BIOMOD_DATA/ once R i
Hi,
I am a bit surprised that an example code from an R-package installed
from CRAN can fail:
> n<-20
> p<-10
> X<-matrix(rnorm(n*p),ncol=p)
> pc<-adalasso.net(X,k=5)
Performing local (adaptive) lasso regressions
Vertex no Error in glmnet(XXtrain, ytrain, type.gaussian = type,
standardize = FALS
On 15/08/2013 10:56, Witold E Wolski wrote:
Hi,
I am a bit surprised that an example code from an R-package installed
from CRAN can fail:
n<-20
p<-10
X<-matrix(rnorm(n*p),ncol=p)
pc<-adalasso.net(X,k=5)
Performing local (adaptive) lasso regressions
Vertex no Error in glmnet(XXtrain, ytrain,
One of those simple tasks, but I can't get to first base with it. I've got a
data set of observations of subjects over a 10 year period beginning on 1st
April 2001 and ending on 31st March 2011. One of may variables is a score
based on an intervention on a given date. Before the intervention t
Sorry, I can't generate an error when running those commands in R on Linux
64-bit. But if I move to Windows (R version 3.0.1, XML_3.98-1.1), I get a
different error ...
> require(XML)
Loading required package: XML
> doc <- htmlTreeParse("
http://www.sec.gov/cgi-bin/browse-edgar?CIK=MSFT&Find=Searc
G'day,
I wonder if there exists a package that deals with the prediction of
time series based on a method proposed by Farmer & Sidorowich (1987) and
further developments by others. This method relies on the state space
representation of a time series and uses the nearest neighbor of a state
fo
Hi,One way would be:
df<-
data.frame(case,obsdate=as.Date(obsdate,format="%d/%m/%Y"),score,stringsAsFactors=FALSE)
#using as.data.frame(cbind(... should be avoided
df$date_end<-as.Date(unlist(lapply(with(df,tapply(obsdate,case,FUN=function(x)
x-1)),function(x)
c(x[-1],as.Date("31/03/2011",fo
Dear list,
I have 50 sites where information was recorded over a 45 year time period.
The recorded data could take one of four forms: Fishing effort,
Environmental, Both or Inconclusive.
What i am aiming to do is cluster sites based on their similarity through
time, essentially i view this
Dear list,
I have 50 sites where information was recorded over a 45 year time period. The
recorded data could take one of four forms: Fishing effort, Environmental, Both
or Inconclusive.
What i am aiming to do is cluster sites based on their similarity through time,
essentially i view this as
On Wed, 14 Aug 2013, Hervé Pagès wrote:
Hi Zhang,
First note that a list is a vector (try is.vector(list())).
The documentation for sapply() and Vectorize() should say *atomic*
vector instead of vector in the desccription of the 'simplify' and
'SIMPLIFY' arguments.
So in order for sapply() t
When you select a single row from a matrix, the dimsion is lost:
n <- matrix(nrow = 3, ncol = 5)
dim(n)
[1] 3 5
dim(n[1,])
NULL
dim(n[2,])
NULL
This doesn't happen if you select more than one row:
dim(n[1:2,])
[1] 2 5
This is causing trouble. SCENARIO: when I filter out unqualified sam
Hello all,
I’ve fitted a bivariate smoothing model (with GAM) to some data, using two
explanatory variables, x and y. Now I’d like to add the surface corresponding
to my fit to a 3D scatterplot generated using plot3d().
My approach so far is to create a grid of x and y values and the correspo
Ah, yes. it was the
net$Time[net$From=="A" & net$To=="C"]),
notation I was missing. I suppose I'd think of that as similar to an array
slice Perl (no, I know it's not the same!)
I'll have to spend a little time looking at your sample data-generation code
too :)
Thanks very much,
jack
___
Dear Doran, Bert and Roger,
Thank you for attending my query and for your valuable responses.
The task is slightly more complex. Here's the real case... I have genetic
variation data (40,000 single nucleotide polymorphisms) from 90,000
individuals. This makes the 90,000 (samples) rows/columns o
You might also try (could be faster):
library(data.table)
dt1<- data.table(df,key=c("case","obsdate"))
dt1[,date_end:=obsdate-1]
dt1[,date_end:=c(date_end[-1],as.Date("2011-03-31")),by=case]
dt1<-subset(dt1,select=c(1,2,4,3))
dt1
# case obsdate date_end score
#1: a 2001-04-01 2007-05-1
On 15-08-2013, at 09:29, Zhang Weiwu wrote:
> When you select a single row from a matrix, the dimsion is lost:
>
>> n <- matrix(nrow = 3, ncol = 5)
>> dim(n)
> [1] 3 5
>> dim(n[1,])
> NULL
>> dim(n[2,])
> NULL
>
> This doesn't happen if you select more than one row:
>
>> dim(n[1:2,])
> [1] 2
Post to R-sig-ecology , not here.
-- Bert
On Thu, Aug 15, 2013 at 2:20 AM, Chris McOwen
wrote:
> Dear list,
>
> I have 50 sites where information was recorded over a 45 year time period.
> The recorded data could take one of four forms: Fishing effort,
> Environmental, Both or Inconclusive.
>
I think substitute() or bquote() will do a better job here than gsub() be
they work on the parsed formula rather than on the raw string. The
terms() function will interpret the formula-specific operators like "+"
and ":" to come up with a list of the 'variables' (or 'terms') in the formula
E.g.,
Hi,
You could try:
res<-data.frame(lapply(wkend,function(x) seq(x,as.Date("2013-12-26"),by=7)))
colnames(res)<- paste0("date",1:3)
head(res)
# date1 date2 date3
#1 2013-01-04 2013-01-05 2013-01-06
#2 2013-01-11 2013-01-12 2013-01-13
#3 2013-01-18 2013-01-19 2013-01-20
#4 2013-01-
Thanks very much A.K. I can see from answers to my previous posts, too, that I
will be using lapply() a lot in my R future. BNC
From: arun kirshna [via R] [mailto:ml-node+s789695n4673817...@n4.nabble.com]
Sent: Wednesday, August 14, 2013 11:06 PM
To: Crombie, Burnette N
Subject: Re: condense re
2013/8/15 Berend Hasselman :
> n[1,,drop=FALSE]
>
> Berend
Thanks a lot, and I wasn't aware I can do ?`[`
__
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/posting
Thanks very much for your contribution, Siraaj. I appreciate you taking the
time to help me learn loops, etc. BNC
-Original Message-
From: Siraaj Khandkar [mailto:sir...@khandkar.net]
Sent: Wednesday, August 14, 2013 9:08 PM
To: Crombie, Burnette N
Cc: r-help@r-project.org
Subject: Re:
On Thu, Aug 15, 2013 at 7:40 AM, Weiwu Zhang wrote:
> 2013/8/15 Berend Hasselman :
>> n[1,,drop=FALSE]
>>
>> Berend
>
> Thanks a lot, and I wasn't aware I can do ?`[`
See:
?"?"
;-) (Sorry, couldn't resist!)
-- Bert
>
> __
> R-help@r-project.org
R 3.0.1
OS X 10.8.4
Colleagues,
I am using sendmailR to send myself notifications when a lengthy R process
completes. Sample code is:
require("sendmailR")
NOTIFY <- "Notification"
FROM<- ""
TO <- FROM
CONTROL <-
I'm trying to write a function to make gam (in mgcv) give me the
fixed-effects "within" estimator, which is equivalent to the OLS dummy
variable estimator. Basically this involves subtracting the
within-group means from the IVs and the DV, adding their overall means
back in, and fitting the mo
This is a level of complexity I've not before encountered and I have not
seen the solution in Deepayan Sarkar's book. Briefly, I want to plot values
for each of 5 factors over a range of dates to visualize whether the
relative values of each factor change over time.
The data frame structure i
Hi All,
I have a number of plots to run and was hoping to use a function instead of
bock copying my code and retyping my parameters.
I am using R 3.0.0 on a mac.
My code is as follows:
library (car)
cres<-read.csv("//users//smits//r_work//cres.csv", header = TRUE)
attach(cres);
run_plots <- f
Hi list,
I have encountered the "Cannot find xml2-config" problem too during XML package
installation on my 64-bit Redhat (v. 6.4) linux machine. After looking through
the old posts I checked all the necessary libraries and they all seem to be
properly installed (see below). I don't understan
On Aug 15, 2013, at 2:23 AM, Lucas Holland wrote:
> Hello all,
>
> I’ve fitted a bivariate smoothing model (with GAM) to some data, using two
> explanatory variables, x and y. Now I’d like to add the surface
> corresponding to my fit to a 3D scatterplot generated using plot3d().
>
> My appr
I'm not an expert useR but I asked a similar question to stack overflow
that might give you new ideas.
http://stackoverflow.com/questions/17458556/how-can-i-speed-up-this-sapply-for-cross-checking-samples
On Thu, Aug 15, 2013 at 2:30 AM, Praveen Surendran
wrote:
> Dear Doran, Bert and Roger,
>
R users,
Poster abstracts are now being accepted for the 2014 ASA Conference on
Statistical Practice, February 20-24, Tampa, Florida, USA. Presenters are
required to host their posters during their assigned 90-minute session.
Based on a survey of the 2013 attendees there was particular interest
Hi Tao
In the same R session as you call install.packages(),
what does
system("which xml2-config", intern = TRUE)
return?
Basically, the error message from the configuration script for the
XML package is complaining that it cannot find the executable xml2-config
in your PATH.
(You can also
Hi Duncan,
Thank you very much for your quick response!
It turns out I was typing the shell commands in another xterm which was ssh'ed
to another linux computer and I completely forgot about it. After finding that
out and checking with my own computer, the libxml2 is installed but
libxml2-
HI,
xyplot(pct~sampdate|func_feed_grp,data='burns.date.ffg')
#Error in eval(substitute(groups), data, environment(x)) :
# invalid 'envir' argument of type 'character'
xyplot(pct~sampdate|func_feed_grp,data=burns.date.ffg) #no error
#Not sure about your exact specification, so this would get yo
typo
run_plots <- functon()
On Thu, Aug 15, 2013 at 1:10 PM, Gerard Smits wrote:
> Hi All,
>
> I have a number of plots to run and was hoping to use a function instead
> of bock copying my code and retyping my parameters.
>
> I am using R 3.0.0 on a mac.
>
> My code is as follows:
>
> library
Dumb error. Thanks for letting me know. Gerard
On Aug 15, 2013, at 11:38 AM, "Richard M. Heiberger" wrote:
> typo
>
> run_plots <- functon()
>
>
> On Thu, Aug 15, 2013 at 1:10 PM, Gerard Smits wrote:
> Hi All,
>
> I have a number of plots to run and was hoping to use a function instead o
Speed comparison:
dat1<-structure(list(Date = c("06/01/2010", "06/01/2010", "06/01/2010",
"06/01/2010", "06/02/2010", "06/02/2010", "06/02/2010", "06/02/2010",
"06/02/2010", "06/02/2010", "06/02/2010"), Time = c(1358L, 1359L,
1400L, 1700L, 331L, 332L, 334L, 335L, 336L, 337L, 338L), O = c(136.4
Hi,
Looks like you have some free time on your hands :-)
Something looks a bit off here, though, I was surprised to see the
time you reported for the data.table option:
> #separate the data.table creation step:
> dt1 <- data.table(dat2, key=c('Date', 'Time'))
> system.time(ans <- dt1[, .SD[.N],
Hello everyone,
How can I create a data_table with 1000 variables (Var1:Var1000)?
Thank you,
Pooya Lalehzari
THIS E-MAIL IS FOR THE SOLE USE OF THE INTENDED RECIPIENT(S) AND MAY CONTAIN
CONFIDENTIAL AND PRIVILEGED INFORMATION.ANY UNAUTHORIZED REVIEW, USE, DISCLOSURE
OR DISTRIBUTION IS PROHIBITE
Please:
1. Define "data_table" (??)
2. Follow the advice from the posting guide link below to post a
coherent question.
-- Bert
On Thu, Aug 15, 2013 at 1:16 PM, Pooya Lalehzari
wrote:
> Hello everyone,
> How can I create a data_table with 1000 variables (Var1:Var1000)?
>
> Thank you,
> Pooya
Sorry. data.table or even data.frame (as they are related). My question is, if
there is a short form to refer to all 1000 of them similar to what exists in
SAS.
Thank you,
Pooya Lalehzari.
-Original Message-
From: Bert Gunter [mailto:gunter.ber...@gene.com]
Sent: Thursday, August 15,
On Thu, 15 Aug 2013, arun wrote:
#Not sure about your exact specification, so this would get you started.
xyplot(pct~sampdate,data=burns.date.ffg,groups=func_feed_grp,pch=1:7,type="l")
A.K./Dennis;
Now I see the source of my error: I quoted the data file name! Removing
the quotation marks p
On 08/15/2013 04:16 PM, Pooya Lalehzari wrote:
Hello everyone,
How can I create a data_table with 1000 variables (Var1:Var1000)?
I'm not familiar with data_table, but here's an example to get you
started in figuring this out:
> n <- 3
> var.range <- 1:n
> prefix <- "Var"
> data <- as.data.f
I tried it again on a fresh start using the data.table alone:
Now.
dt1 <- data.table(dat2, key=c('Date', 'Time'))
system.time(ans <- dt1[, .SD[.N], by='Date'])
# user system elapsed
# 40.908 0.000 40.981
#Then tried:
system.time(res7<- dat2[cumsum(rle(dat2[,1])$lengths),])
# user syst
Hi Rich,
It's better to dput() your dataset. From what you showed:
burns.date.ffg<- read.table(text="
'sampdate','func_feed_grp','pct'
'2000-07-18','Filterer',0.0351
'2000-07-18','Gatherer',0.7054
'2000-07-18','Grazer',0.0442
'2000-07-18','Predator',0.1078
'2000-07-18','Shredder',0.1074
'2003-07-0
Hi,
On Thu, Aug 15, 2013 at 1:38 PM, arun wrote:
> I tried it again on a fresh start using the data.table alone:
> Now.
>
> dt1 <- data.table(dat2, key=c('Date', 'Time'))
> system.time(ans <- dt1[, .SD[.N], by='Date'])
> # user system elapsed
> # 40.908 0.000 40.981
> #Then tried:
> syste
Dr. Therneau,
Thank you as always for first writing, and second continuing the Cox model in R
(and earlier I believe in SAS).
While your comments concerning non-proportional hazards is helpful, it does not
fully address the question, "What alternatives do I have if I assume
proportional assump
I usually get better results with data.table except for this situation.
If I take another example unrelated to the current topic:
set.seed(1254)
name<- sample(letters,1e6,replace=TRUE)
number<- sample(1:10,1e6,replace=TRUE)
datTest<- data.frame(name,number,stringsAsFactors=FALSE)
system.time(r
Here is a first stab:
library(gsubfn)
test <- "y1 + y2 ~ a*(b + c) + d + f * (h == 3) + (sex == 'male')*i"
gsubfn( "([a-zA-Z][a-zA-Z0-9]*)((?=\\s*[-+~)*])|\\s*$)",
function(x,...) paste0(toupper(x),'z'), test, perl=TRUE )
On Wed, Aug 14, 2013 at 9:13 PM, Frank Harrell wrote:
> I would like t
Here is the sample code.
Large_Table = data.frame(
Var1=numeric(),
Var2=numeric(),
Var3=numeric(),
.
.
.
Var1000
On Thu, Aug 15, 2013 at 5:09 PM, Pooya Lalehzari
wrote:
>
> Here is the sample code.
>
> Large_Table = data.frame(
> Var1=numeric(),
> Var2=numeric(),
> Var3=numeric(),
> .
> .
>
Hi,
May be this helps:
Large_Table<- data.frame(lapply(paste0("Var",1:1000),function(x)
{x1<-data.frame(numeric());colnames(x1)<-x; x1}))
str(Large_Table)
'data.frame': 0 obs. of 1000 variables:
$ Var1 : num
$ Var2 : num
A.K.
- Original Message
... But of course R is not C++ or Java, so there is no need to
"declare" anything, and the OP's question strongly suggests to me that
he/she has made no effort to learn R. My advice would be:
1. Stop posting.
2. Read An Introduction to R or other R tutorial and learn how the
language works.
-- B
Just to add:
You could convert "logi" to "numeric" by:
Large_Table2<-data.frame(matrix(nrow = 0, ncol=1000, dimnames =
list(c(),paste0("Var", 1:1000
str(Large_Table2)
#'data.frame': 0 obs. of 1000 variables:
# $ Var1 : logi
# $ Var2 : logi
# $ Var3 : logi
---
Larg
That is great!
Thank you so much.
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Thursday, August 15, 2013 5:29 PM
To: Ista Zahn
Cc: Pooya Lalehzari; R help
Subject: Re: [R] How can I create a data.table with 1000 variables
(Var1:Var1000)
Just to add:
You could c
Dear Witold E Wolski,
Re:
> I am have a procedure which generates multidimensional arrays.
>
> To compute them is expensive so I want to store them in order to be
> able to analyse them later.
>
> I am using at the moment
>
> problem is that the array is always assigned to a variable ma (the
>
HI Steve,
Thanks for testing.
When I run a slightly bigger dataset:
set.seed(1254)
name<- sample(letters,1e7,replace=TRUE)
number<- sample(1:10,1e7,replace=TRUE)
datTest<- data.frame(name,number,stringsAsFactors=FALSE)
library(data.table)
dtTest<- data.table(datTest)
system.time(res3<- dtTest[
Slightly modified, also seems to work.
gsubfn( "([[:alpha:]][[:alnum:]]*)((?=\\s*[-+~*)])|$)",function(x,...)
paste0(toupper(x),'z'), test, perl=TRUE )
#[1] "Y1z + Y2z ~ Az*(Bz + Cz) + Dz + Fz * (h == 3) + (sex == 'male')*Iz"
A.K.
- Original Message -
From: Greg Snow <538...@gmail.com>
I really appreciate the excellent ideas from Bill Dunlap and Greg Snow.
Both suggestions almost work perfectly. Greg's recognizes expressions
such as sex=='female' but not ones such as age > 21, age < 21, a - b >
0, and possibly other legal R expressions. Bill's idea is similar to
what Dunca
On Thu, 15 Aug 2013, arun wrote:
It's better to dput() your dataset.
A.K.,
That's what I usually do; with such a small data set I thought the raw
data would be equally good.
#or
xyplot(pct~sampdate,data=burns.date.ffg,groups=func_feed_grp,pch=1:7,type="l")
What I see in the plots are 7
Try this one
ff <- function (expr)
{
if (is.call(expr) && is.name(expr[[1]]) &&
is.element(as.character(expr[[1]]), c("~", "+", "-", "*", "/", ":",
"%in%"))) {
# the above list should cover the standard formula operators.
for (i in seq_along(expr)[-1]) {
Oops, I left "(" out of the list of operators.
ff <- function(expr) {
if (is.call(expr) && is.name(expr[[1]]) &&
is.element(as.character(expr[[1]]),
c("~","+","-","*","/","%in%","("))) {
for(i in seq_along(expr)[-1]) {
expr[[i]] <- Recall(expr[[i]])
}
I have an 8 core machine and I wish to use 6 of them to run a task
much more complicated than this example.
multitest <- function(n = 1000, lam = 500)
{
### Purpose:- Simple parallel task to check why mclapply doesn't work
###
-
Bill that is very impresive. The only problem I'm having is that I want
the paste0(toupper(...)) to be a general function that returns a
character string that is a legal part of a formula object that can't be
converted to a 'name'.
Frank
---
Oops, I left "(" out
Hi,
On Thu, Aug 15, 2013 at 4:03 PM, arun wrote:
> HI Steve,
>
> Thanks for testing.
>
> When I run a slightly bigger dataset:
> set.seed(1254)
> name<- sample(letters,1e7,replace=TRUE)
> number<- sample(1:10,1e7,replace=TRUE)
>
> datTest<- data.frame(name,number,stringsAsFactors=FALSE)
> library
How about this:
df1000cols =
setNames(as.data.frame(matrix(numeric(0),ncol=1000)),paste0("V",1:1000))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Pooya Lalehzari
Sent: Friday, 16 August 2013 8:27a
To: Bert Gunter
Cc: r-help@r
Hi Steve,
Thanks. The error problem is solved by using quotes.
Will post at data-table mailing list regarding the issue of:
system.time(ans <- dt1[, .SD[.N], by='Date'])
# user system elapsed
# 39.284 0.000 39.352
A.K.
- Original Message -
From: Steve Lianoglou
To: arun
The function crossprod() might be useful?
crossprod(X) is a more efficient way of producing t(X) %*% X
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Praveen Surendran
Sent: Thursday, 15 August 2013 10:30p
To: r-help@r-project.
Hi, i was messing around with R and noticed that there is an object for
timeseries which can be initialized with the ts() function. however it seems
that it can only be initialized with a vector of data points and its assumed
that these data points occur at regular time periods, user to specify
Say I have a dataframe for plotting scatterplot. The dataframe would be
organized in the following fashion (in CSV format):
name ABC EFG132 45256 67
to, say 200 000 entries
I am going to first do a scatterplot, after which I am going to subset a
portion of the dataset into A using alpha
Perhaps this is simple and common, but it took me quite a while to admit I
cannot solve it in a simple way.
The data frame `df` has the following columns:
unixtime, value, factor
Now I need a matrix of:
unixtime, value-difference-between-factor1-and-factor2
The naive solution is:
Dear all,
I'm using package "ridge" to deal with multicollinearity. It's been convenient
to automatically choose lambda.
However, how do I tell whether the OLS results have been improved after
applying ridge regression?
I only notice that more variables become statistically significant and som
On 15/08/2013 21:11, Daren Zou wrote:
Hi, i was messing around with R and noticed that there is an object for
timeseries which can be initialized with the ts() function. however it seems
that it can only be initialized with a vector of data points and its assumed
that these data points occur a
Thanks, the second approach worked fine on Windows.
--JJS
On Thu, August 15, 2013 8:38 am, Jeffrey Dick wrote:
> Sorry, I can't generate an error when running those commands in R on Linux
> 64-bit. But if I move to Windows (R version 3.0.1, XML_3.98-1.1), I get a
> different error ...
>
>> requir
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