> install.packages("Matrix")
On Wed, Jun 6, 2018 at 2:24 PM, Bill Poling wrote:
> Good morning. In my continuing pursuit of self-taught R programming I am
> interested in following the tutorial provided by Bloggers.com "Beautiful
> and Powerful Correlation Tables in R"
>
>
> https://www.r-blogg
Hi Ray,
Have you done any search at all? I did a search on "R package ROC latent
variable" and got several hits.
In particular the package randomLCA seems relevant
https://cran.r-project.org/web/packages/randomLCA/vignettes/randomLCA-package.pdf
I glanced at the documentation which mentions 2 oth
Try this. Suppose your list of matrices is in the list locL.
nc <- 3
locL2 <- list()
for ( i in 1:length(locL )
locL2[[i]] <- as.numeric(t(locL[[i]]))
bigMat <- matrix(unlist(locL3), ncol=nc, byrow=TRUE)
HTH,
Eric
On Sun, Jun 10, 2018 at 6:33 PM, Christofer Bogaso <
bogaso.christo...@gmail.
Sorry typos
Try this. Suppose your list of matrices is in the list locL.
nc <- 3
locL2 <- list()
for ( i in 1:length(locL) )
locL2[[i]] <- as.numeric( t( locL[[i]] ) )
bigMat <- matrix(unlist(locL2), ncol=nc, byrow=TRUE)
HTH,
Eric
On Sun, Jun 10, 2018 at 7:10 PM, Eric B
Hi Sam,
I use the littler package for scripting. You may find it meets your needs.
https://github.com/eddelbuettel/littler
HTH,
Eric
On Wed, Jun 13, 2018 at 5:33 AM, Sam Tuck wrote:
> Hi All,
> I am new to R and am wondering if there is a way to pass
> arguments between rscripts. I
sapply( 1:length(YH), function(i) { YH[[i]][iuhV[i]]})
On Mon, Jun 18, 2018 at 1:55 PM, akshay kulkarni
wrote:
> dear members,
> I have list YH and index vector iuhV. I want
> to select iuhV[1] from YH[[1]], iuhv[2] from YH[[2]], iuhv[3] from
> YH[[3]]..iuhv[n] f
My response does not have an explicit for loop.
On Mon, Jun 18, 2018 at 2:15 PM, akshay kulkarni
wrote:
> correctionI want the method without a for loop
>
> From: akshay kulkarni
> Sent: Monday, June 18, 2018 4:25 PM
> To: R help Mailing list
> Subject: subs
The statement
dvec <- -hsmooth
looks like it might be the source of the problem, depending on what hsmooth
is.
On Tue, Jun 26, 2018 at 11:16 AM, Maija Sirkjärvi wrote:
> Thanks for the reply! I got that figured out, but still have some problems
> with the quadratic programming.
>
> It seems
If you want a "fresh" R session when you start to run the script you could
consider putting as the first line
rm(list=ls())
This will remove objects from your environment (variables, functions, ..)
HTH,
Eric
On Mon, Jul 2, 2018 at 5:34 PM, PIKAL Petr wrote:
> Hi
>
> Without code it is just f
Hi WHP,
It should be
.Tsp=c(2016,2018.417,12)
(off by one error)
HTH,
Eric
On Tue, Jul 3, 2018 at 6:23 PM, Bill Poling wrote:
> Hi, obviously missing something here? Getting error: Error in
> attributes(.Data) <- c(attributes(.Data), attrib) : invalid time series
> parameters specified
>
For what it's worth, for larger vectors, and following on from your
observation that the do.call() approach is faster, the following provides
some modest additional speedup:
cbind(x,t(do.call(cbind, lapply(x, function(y) vec
Rgds,
Eric
On Tue, Jul 3, 2018 at 8:12 PM, Viechtbauer, Wolfgang
Hi Catalin,
This should work. I set the number of repetitions and sample sizes as
variables so it would be clear how to modify for your actual case.
nreps<- 3
sampSize <- 2
w <- unlist( lapply(1:nreps, function(i) {
rep(paste("R",i,sep=""),sampSize) } ) )
aa2 <- cbind( as.data.frame(aa), w)
> If (identical(snlcqn, snlcqna)) snlcqn else snlcqna
??
Why not just always return snicqna ?
On Mon, Jul 9, 2018 at 9:43 AM, PIKAL Petr wrote:
> Hi
>
> You definitely should not use HTML formated mail. This is plain text
> mailing list for reason.
>
> If you experience space between "NSE/"
I found the following at
https://stackoverflow.com/questions/25269425/merge-zoo-removes-time-zone
library(xts)
merge2=function(x,y) {
as.zoo(merge(as.xts(x), as.xts(y)))}
If you define the function merge2() as above then
merge2(Dat1,Dat2)
should be ok
HTH,
Eric
On Mon, Jul 9, 2018 at 1:22
Hi Kevin,
It's good that you provided the background to the problem.
Rather than asking this list to "debug" your proposed solution, I think you
would be better off showing some of the "corrupted" data frame and ask for
suggestions how to deal with it.
(Suggestions may or may not match your initial
Hi Sumit,
I was not able to reproduce this problem.
I tried it in both R 3.5.1 and R 3.4.4.
Both gave the expected output (which differs from yours.)
Eric
On Tue, Jul 10, 2018 at 1:32 AM, Bert Gunter wrote:
> Dunno.
>
> But if I understand correctly, here's a base R way to do it:
>
> (## using
> Yours sincerely,
> AKSHAY M KULKARNI
>
> --
> *From:* Eric Berger
> *Sent:* Monday, July 9, 2018 12:15 PM
> *To:* PIKAL Petr
> *Cc:* akshay kulkarni; R help Mailing list
> *Subject:* Re: [R] inconsistency in display of character vector
>
> > If (identical(
Hi Rich,
This may not be the most efficient but it will identify the offenders.
> foo <- paste(wy2016$date, wy2016$time))
> uu <- sapply(1:length(foo),
function(i) { a <- try(as.POSIXct(foo[i]),silent=TRUE)
"POSIXct" %in% class(a) })
> which(!uu)
HTH,
Eric
On Fri, J
Some additional comments that might be relevant to people interested in
these topics.
1. For R scripts you should also consider the package littler developed by
Dirk Eddelbuettel, Highly recommended.
For info http://dirk.eddelbuettel.com/code/littler.html or the github
repository.
2. Scripts can
You should be able to do all this within R.
>From what you have written I don't see a compelling reason to use scripts
at the shell level.
Best,
Eric
On Wed, Jul 25, 2018 at 9:41 PM, Rich Shepard
wrote:
> On Wed, 25 Jul 2018, Eric Berger wrote:
>
> 1. For R scripts you sho
Hi Rich,
Thanks for posting this question.
I also use emacs with ESS for editing R files and I have been living with
the comment indentation problem you described.
Based on the comments in this thread I did a search and found a posted
solution that works for me. See
https://stat.ethz.ch/pipermail/e
I have been successfully using RODBC for a long time (years) to connect to
MS SQL Server from R.
This week I wanted to try using odbc but I am seeing some problems which
may be related to how I set up my driver and/or connection.
The dbWriteTable manual page gives as an example command:
dbWriteTab
use my brevity.
>
> On August 29, 2017 2:21:44 AM PDT, Eric Berger
> wrote:
> >I have been successfully using RODBC for a long time (years) to connect
> >to
> >MS SQL Server from R.
> >This week I wanted to try using odbc but I am seeing some problems
> >which
Hi Lars,
Two comments:
1. You can achieve what you want with a slight modification of your
definition of s(), using the hint from the error message that you need an
argument '.':
s <- function(.) {
dplyr::summarise(., x1m = mean(X1),
x2m = mean(X2),
x3m = mea
Hi Eric,
Bert's solution is very elegant. His final comment prompted me to check out
the aperm() function which I have never used.
The final step to complete his response is
prec_daily2 <- aperm(prec_daily, c(3,1,2))
Regards
On Wed, Sep 13, 2017 at 9:06 PM, Bert Gunter wrote:
> Thanks for the
You did not provide the data frame so I will first create one and then use
it to create an xts
library(xts)
df <- data.frame( year=1980:2009, cnsm=sample(170:180,30,replace=TRUE),
incm=rnorm(30,53,1), wlth=rnorm(30,60,1))
dates <- as.Date(paste(df$year,"-01-01",sep=""))
myXts <-
You can just use the same code that I provided before but now use your
dataset. Like this
df <- read.csv(file="data2.csv",header=TRUE)
dates <- as.Date(paste(df$year,"-01-01",sep=""))
myXts <- xts(df,order.by=dates)
head(myXts)
#The last command "head(myXts)" shows you the first few rows of the x
I am not familiar with the vegan package, so I am just making a guess here.
If 'na.action=na.omit' is part of the call to varpart, try removing it from
the function call and moving it above as follows:
options(na.action="na.omit")
RDA_Ger <- varpart(comm, x1, x2, x3, transfo="hellinger", scale = F
Hi Terry,
I take your question to mean how to label distinct rows of a data frame. If
that is not your question please clarify.
I found the row.match() function in the package prodlim that can be used to
solve this.
However since your request requires no additional dependencies I borrowed
the relev
What if you answer 'Yes' ?
On Wed, Sep 20, 2017 at 1:03 PM, AbouEl-Makarim Aboueissa <
abouelmakarim1...@gmail.com> wrote:
> Dear All: good morning
>
> I am trying to install the "" package, but I am getting this error message.
>
>
> *> utils:::menuInstallPkgs()*
> *Warning in install.packages(NU
Jim,
I don't see how that link could be related to John's issue. Symptoms
related to your link involve discrepancies of four years whereas John is
seeing discrepancies of one day.
John,
I do not see any attached files.
Regards
On Sat, Sep 23, 2017 at 1:30 PM, Jim Lemon wrote:
> Hi John,
> It c
e the shorter
date11<-as.Date(as.POSIXlt(a_col$date))
On Sun, Sep 24, 2017 at 8:51 AM, John wrote:
> Hi,
>
>Thank you for all your responses.
>For Eric, The files are attached. (I believe it was also attached in my
> first message)
>For David, Could you send me
Hi Ashraf,
It is not obvious to me what your structures are but one problem in your
function is the assignment tt1 <- SpatialLines(list(tt[[i]])).
This will set tt1 to just have one item.
Consider the following
test.func <- function(x) {
tt1 <- list()
for ( i in ... ) {
...
Hi David,
I was about to post a reply when Bert responded. His answer is good
and his comment to use the name 'dat' rather than 'data' is instructive.
I am providing my suggestion as well because I think it may address
what was causing you some confusion (mainly to use "which", but also
the missing
Hi Rich,
If I understand your comment about "uniformly distributed along the log=x
axis" then I think John's (second) set of commands needs a change to the
definition of xx, as follows:
xx <- exp(seq(from=log(min(x)),to=log(max(x)),length=50))
m <- lm(y ~ log(x))
yy <- predict(m, data.frame(x=xx))
Hi Chris,
Maybe the na.rm=TRUE is affecting things. Try this
apply(datTAF[,75:78],2,function(x){ sum(!is.na(x)) })
HTH,
Eric
On Tue, Sep 26, 2017 at 9:53 AM, Chris Evans wrote:
> I am hitting an odd message "Error in FUN(newX[, i], ...) : all arguments
> must have the same length". I can't su
d to convert each matrix to coordinates
> and then to a line and then to a spatial line as figured in the code.
>
> My data structure is a list of 141 matrices.Each matrix represents
> coordinates of the river lines position.
>
> Ashraf, cheers
>
>
> On Monday, 25 September
Hi Bayan,
In your code, 'a' is a vector and is.integer(a) is a logical of length 1 -
most likely FALSE if even one element of a is not an integer. (Since R will
coerce all the elements of a to the same type.)
You need to decide whether something "close enough" to an integer is to be
considered an i
Hi John,
Thanks to Jim for pointing out the file.rename() function. You can try this:
# define the filename templates
strIn <- "XYZW--Genesis_ABC.mp3"
strOut <- "01Gen--.mp3"
# create the strings "01", "02", ..., "50"
v <- sapply(1:50, function(i) sprintf("%02d",i) )
# perform all the file rena
Hi Hollie,
I tried to reproduce your example but could not. Here is what I did. Please
explain if you did something different.
x <- c(9.358*10^-3, -8.165*10^-3, -1.689*10^-8,
-8.165*10^-3, 9.358*10^-3, 1.854*10^-8,
-1.689*10^-8, 1.854*10^-8, 9.358*10^-3)
A <- matrix(x, nrow=3
) should read B <- matrix(y, nrow=3,
> byrow = TRUE)
>
>
> Regards, Hollie
> ------
> *From:* R-help on behalf of Eric Berger <
> ericjber...@gmail.com>
> *Sent:* 02 October 2017 16:16:13
> *To:* Bert Gunter
> *Cc:* r-help@r-projec
3.855762e-6
You can get a positive answer with Miwa by cranking up the number of steps
it uses (to its maximum of 4098, for example)
pmvnorm( ..., algorithm=Miwa(steps=4098))
But this is still not a "correct" answer and the "error" attribute is set
to NA to flag it.
HTH,
Eric
I don't see any attached dataset.
On Wed, Oct 4, 2017 at 11:11 AM, Hemant Sain wrote:
> I'm trying to perform a RFM analysis on attached dataset,
> i'm able to get the results using the auto_rfm function but i want to
> define my own breaks for RFM,
> when i tried to define my own breaks i got t
Hi John,
Here's one way to do it:
vec <- c(2,4,5)
yrs <- seq(from=as.Date("1991-01-01"),by="1 year",length=length(vec))
a <- xts(x=vec, order.by=yrs)
HTH,
Eric
On Fri, Oct 6, 2017 at 9:56 AM, John wrote:
> Hi,
>
>I'd like to make a time series at an annual frequency.
>
> > a<-xts(x=c(2,4,
Hi Christofer,
The directory /srv/shiny-server would normally be owned by the root user.
Your options would seem to be to either (1) bring up the R session as root
(dangerous)
or (2) try copying the file to your local directory and read it from there
(if allowed).
e.g. from the Unix shell:
> cd ~
home/ubuntu/Dat.RData")
>
> shinyServer(function(input, output) {
> output$table = renderTable(head(data.frame(1:20, 1:20), 20))
> })
>
> with above setup when I deploy my shiny app I get below error :
>
> 18.221.184.94:3838 says
> The application unexpectedly exited
> Di
Hi Duncan,
You can try this:
library(readr)
f <- function(s) {
t <- unlist(readr::tokenize(paste0(gsub(" ",",",s),"\n",collapse="")))
i <- grep("[a-zA-Z0-9]*/[a-zA-Z0-9]*/",t)
u <- sub("[a-zA-Z0-9]*/[a-zA-Z0-9]*/","",t[i])
paste0(u,collapse=" ")
}
f("f 147/1315/587 2820/1320/587 3624/1321
; otherwise, locally my shiny app is working but not with AWS.
>
> On Mon, Oct 9, 2017 at 12:37 PM, Eric Berger
> wrote:
> > Hi Christofer,
> > The shiny code you have written does not depend on loading the Dat.RData
> > file.
> > I commented out that line and ran y
Along the lines of Petr's followup:
AXnew <- data.frame(lapply( AX, function(s) sub(",",".",s)))
On Wed, Oct 11, 2017 at 10:59 AM, PIKAL Petr wrote:
> And as follow up,
>
> fff<-function(x) gsub(",", ".", x)
>
> BX <- apply(apply(AX, 2, fff), 2, as.numeric)
>
> this seems to be easier.
>
> Ch
Hi John,
You can try the following:
override.linetype=c("twodash","solid")
p <- ggplot(obs, aes(x = Timestamp))
p <- p + geom_line(aes(y = air_temp, colour = "Temperature", linetype
="Temperature"))
p <- p + geom_line(aes(y = rel_hum/5, colour = "Humidity",
linetype="Humidity"))
p <- p +
guides(co
Combining and completing the advice from Greg and Boris the complete
solution is two lines:
data_2 <- read.csv("excel_data.csv", stringsAsFactors = FALSE)
match_list <- match( data_2$data1, data_2$data2 )
The vector match_list will have the matching position when it exists and
NA's otherwise. Its
One additional comment. If you want 0 instead of NA when there is no match
then the match statement should read:
match_list <- match( data_2$data1, data_2$data2, nomatch=0)
On Fri, Oct 13, 2017 at 7:39 AM, Eric Berger wrote:
> Combining and completing the advice from Greg and Bor
Hi Kevin,
I think there are issues with Rui's proposed solution. For example, if
there are rows in myDF1 which have a studyno
which does not match any row in myDF2, then you will lose those rows. In
your original request you said that you wanted to keep those rows.
To demonstrate my point I need t
Hi John,
Why not just try both and see which one makes sense?
On Tue, Oct 17, 2017 at 12:24 PM, John wrote:
> Hi,
>
>I have a question on ggplot2 with the second axis, but I don't think one
> needs to know ggplot2 package in order to answer this question.
>
>In this example,
> https://r
I was able to reproduce the problem with this self-contained example. Maybe
it could be reproduced with an even smaller one ...
library(tidyquant) # Loads tidyverse, tidyquant, financial pkgs, xts/zoo
library(xts)
dtV <- as.Date("2017-01-01") + 1:100
locL <- list( foo=xts(rnorm(100), order.by=d
The ID matches in the first 16 characters.
How is your table declared?
On Mon, Jul 30, 2018 at 2:00 PM, Christofer Bogaso <
bogaso.christo...@gmail.com> wrote:
> Session Information for above error:
>
> > sessionInfo()
> R version 3.5.0 (2018-04-23)
> Platform: x86_64-w64-mingw32/x64 (64-bi
Hi Saptorshee,
Two comments:
1. no attachments made it through to the list. You probably need to include
the code directly in your email, and send your email as plain text
(otherwise information gets stripped)
2. for anyone interested in following up on Saptorshee's question, I
searched for the pap
upporting
the R community at large which greatly benefits from the work they do.
I hope that helps.
Regards,
Eric Berger
On Fri, Aug 3, 2018 at 11:25 AM, Rolf Turner
wrote:
> On 03/08/18 19:03, Rainer Krug wrote:
>
>> Let’s not alienate the business users!
>>
>> I agree tha
Hi Rolf,
A few edits because (i) nrow(a) should be nrow(A) and (ii) you have
calculated C[j,k,i] = A[i,j]*B[i,k], (iii) minor style change on lapply.
library(abind)
xxx <- lapply(1:nrow(A),function(i){A[i,]%o%B[i,]})
yyy <- do.call(abind,c(xxx,list(along=3)))
zzz <- aperm(yyy,c(3,1,2))
HTH,
Eric
Both loops are on 'i', which is a bad idea. :-)
Also myMatrix[i,i+1] will be out-of-bounds if i = ncol(myMatrix)
On Mon, Aug 6, 2018 at 12:02 PM, Maija Sirkjärvi
wrote:
> I have a basic for loop with a simple matrix. The code is doing what it is
> supposed to do, but I'm still wondering the err
You only need one "for loop"
for(i in 2:nrow(myMatrix)) {
myMatrix[i-1,i-1] = -1
myMatrix[i-1,i] = 1
}
HTH,
Eric
On Wed, Aug 8, 2018 at 12:40 PM, Maija Sirkjärvi
wrote:
> Thanks!
>
> If I do it like this:
>
> myMatrix <- matrix(0,5,5*2-3)
> print(myMatrix)
> for(i in 2:nrow(myMatrix))
>
Hi Shivi,
I have no experience with the rattle package but I just installed it with
no problem.
I am using a Windows 10 machine with R version 3.4.2.
I suggest you provide additional information so that others may have ideas.
e.g. your operating system version and output from sessionInfo() (in R)
Hi Rolf,
When faced with such a situation I take the following approach which often
helps.
Use the same setup that caused the seg fault (you need a reproducible
problem.)
Start your R session using valgrind. e.g. in linux I do:
$ valgrind R
Assuming that a seg fault still occurs then valgrind sho
Hi Claire,
In Unix (linux) the 'which' command is documented as searching for the
command according to the PATH environment variable.
The different results from
$ which
and
> Sys.which()
would point to the fact that the PATH variable is different in the two
cases.
Compare:
$ echo $PATH
versus
Sy
Li is defined as d1$a which is a vector. You should use
N <- length(Li)
HTH,
Eric
On Wed, Aug 22, 2018 at 6:02 PM, Ogbos Okike
wrote:
> Kind R-users,
> I run a simple regression. I am interested in using the Monte Carlo to test
> the slope parameter.
> Here is what I have done:
> d1<-read.tab
't go. I also tried replacing it
> with length(Li). The same error remains.
>
> Thank so much for looking at this again.
>
> Ogbos
>
>
> On Wed, Aug 22, 2018 at 5:06 PM Eric Berger wrote:
>
>> Li is defined as d1$a which is a vector. You should use
>>
R")
> Error in eval(predvars, data, env) :
> numeric 'envir' arg not of length one
> Thank you for additional assitance.
> Ogbos
>
>
>
> On Wed, Aug 22, 2018 at 5:23 PM Eric Berger wrote:
>
>> You have an extra comma ... it should be
>>
>&g
Your problem is that the command you entered
> the_data<-read.csv(file=“c:/file_name.csv,header=TRUE,sep=“,”)
is missing a double quote after the .csv. The statement should be
> the_data<-read.csv(file=“c:/file_name.csv",header=TRUE,sep=“,”)
The '+' sign is a prompt from R that indicates it has
Hi Ivan,
Unfortunately I cannot answer your question.
However, I do have quite a bit of experience using R's reference classes
and you might want to consider the more recent R6 package.
It provides R6 classes which have advantages over
reference classes. See for example:
1. Hadley Wickham on R6 (
Hi Andrew,
I don't have any experience in this area but I was intrigued by your
question. Here is what I learned.
1, A bit of poking around turned up a thread on stats.stackexchange that
mentions that "smallest space analysis" (SSA) is a special case of
"multidimensional scaling" (MDS).
See the th
Hi Kevin,
I did something along these lines using shiny and I had a good experience
with it.
You would require a server (virtual or physical) to run the shiny-server
program.
This approach is particularly suitable if your target users do not know (or
use) R.
If you go down this route I also suggest
See also this thread in stats.stackexchange
https://stats.stackexchange.com/questions/26176/removal-of-statistically-significant-intercept-term-increases-r2-in-linear-mo
On Thu, Sep 27, 2018 at 3:43 PM, J C Nash wrote:
> This issue that traces back to the very unfortunate use
> of R-squared a
Not exactly sure what you are looking for but here is my workflow which may
give you another perspective
1. my OS is linux
2. I edit my files in emacs (with ESS pulled in)
3. I use make (and Makefile) to compile .o's and .so's from .cpp where
necessary (i.e. I use Rcpp etc)
(and also to build s
Hi Knut,
You are almost done.
> v <- match(Mydata$DATA1, needles, nomatch=NA)
> found <- Mydata[ !is.na(v), ]
> missing <- Mdata[ is.na(v), ]
HTH,
Eric
On Mon, Oct 22, 2018 at 5:51 PM Bert Gunter wrote:
> Re-read ?match and note the examples for %in%
>
> -- Bert
> Bert Gunter
>
> "The trouble
Hi Hamed,
I disagree with your criticism.
For a random variable X
X: D - - - > R
its CDF F is defined by
F: R - - - > [0,1]
F(z) = Prob(X <= z)
The fact that you wrote a convenient formula for the CDF
F(z) = (z-a)/(b-a) a <= z <= b
in a particular range for z is your decision, and as you noted th
on.
>
> Please see page 115 in
>
> https://books.google.co.uk/books?id=FEE8D1tRl30C&printsec=frontcover&dq=statistical+distribution&hl=en&sa=X&ved=0ahUKEwjp3PGZmJzeAhUQqxoKHV7OBJgQ6AEIKTAA#v=onepage&q=uniform&f=false
> The
>
>
> Thanks.
> Hamed.
>
&
> first_row_dist <- as.numeric(gower_dist)[1:(attr(gower_dist,"Size")-1)]
will give you the distances of the first row from the subsequent rows.
HTH,
Eric
On Fri, Oct 26, 2018 at 4:07 PM Aerenbkts bkts wrote:
> I have a data-frame with 30k rows and 10 features. I would like to
> calculate
Hi Marc,
Normally updating external libraries would not require re-installing an R
package that uses it.
The R package will be able to "find" the new one, because the updated
library would be in the same place (sort of) as the previous library.
On a linux system this is handled by system soft links
Hi Sebastien,
I like Duncan's response. An alternative approach is to pass around
environments, as in the following:
bar1 <- function(e) {
e$x <- e$y <- e$z <- 1
cat(sprintf('bar1: x=%d, y=%d, z=%d\n', e$x, e$y, e$z))
}
bar2 <- function(e) {
e$x <- e$y <- e$z <- 2
cat(sprintf('bar2: x=%d,
You need to read the Help page for the function survSplit
?survSplit
states that the id argument is a
"character string with the name of new id variable to create (optional).
This can be useful if the data set does not already contain an identifier"
The addicts data.frame has as its first column
:-)
On Thu, Nov 1, 2018 at 9:06 AM Rui Barradas wrote:
> Hello,
>
> This has nothing to do with R-help and I apologize in advance but this
> is really, really strange.
>
> A SO user is still using R 1.0.1:
>
>
> https://stackoverflow.com/questions/53096176/how-do-i-change-the-str-of-my-diff-time
Hi,
You have some problems with your setup. You set N based on the number of
rows in ts, but then in the call to approx_entropy you write ts[,i].
Note that ts[,i] is the i'th column of ts, whereas your definition of i
implies it is based on row numbers.
Maybe this is leading you to see problems el
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> Sender
> notified by
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> <https://mailtrack.io?utm_source=gmail&utm_medium=signature&utm_campaign=signaturevirality5&;
> Sender
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> 11/02/18,
> 4:02:18 PM
>
> On Fri, Nov 2, 2018 at 3:59 PM Eric Berger wrote:
>
>> How about something like this:
&g
What do you see at the OS level?
i.e. on windows
DIR rawData.rds
on linux
ls -l rawData.rds
compare the file sizes on both.
On Wed, Nov 7, 2018 at 9:56 AM Patrick Connolly
wrote:
> From a Windows R session, I do
>
> > object.size(rawData)
> 31736 bytes # from scraping a non-reproducible web ad
>
> Have I misunderstood what the idea is? I thought I'd get an identical
> object, irrespective of how different the OS stores and zips it.
>
>
>
> |>
> |> Rgds,
> |>
> |> Robert
> |>
> |>
> |> On 07/11/18 08:13, Eric Berger wrote:
>
Try google'ing for 'variance of an AR(1) process'.
With the same seed, if you set n=100, you will get something that will
compare well with what you discover from your search.
On Tue, Nov 13, 2018 at 2:04 PM Ashim Kapoor wrote:
> Dear All,
>
> Here is a reprex:
>
> set.seed(123)
> b <- arima
Eigen shows that the matrix is not positive definite (it has a negative
eigenvalue).
And isSymmetric() also shows it is not symmetric - compare (3,4) and (4,3)
On Tue, Nov 13, 2018 at 5:39 PM Hoffman, Gabriel
wrote:
> My understanding is that a Cholesky decomposition should work on any
> square,
Hi Ashim,
Per the help page for arima(), it fits an ARIMA model to the specified time
series - but the caller has to specify the order - i.e. (p,d,q) - of the
model.
The default order is (0,0,0) (per the help page). Hence your two calls are
different. The first call is equivalent to order=c(0,0,0)
Hi Aveek,
1. This is an "all-text" mailing list. Your attachment did not come
through.
You can check out the posting guide (see the link at the bottom of your
email)
and/or
use dput(...) on your structures and paste them into your email so
that members of the list can try to reproduce
I often use sprintf() to control formatting of text strings. e.g.
sprintf("%s is not between 1 or 15\n",nums)
See ?sprintf for details
HTH,
Eric
On Tue, Dec 18, 2018 at 10:56 AM Andrew wrote:
> Hi all
>
> Playing around with some branching using if, if else, and else, I wrote
> the following
You can do a web search that focuses on R related hits via
step 1: go to rseek.org
step 2: do your search there - e.g. 3d krig
This returns a lot of "hits" that seem to address your question.
HTH,
Eric
On Thu, Dec 20, 2018 at 5:56 AM Francois Chartier
wrote:
> Hi,
>
> I would like to create a
Hi Steven,
Here's one way, using print
try5<-function(A,B){
C<-A+B
#cat("\nA =",A,"\nC = ",C)
cat("\nA = ")
print(A)
cat("\nC = ")
print(C)
structure(list(A=A,B=B,C=C))
}
HTH,
Eric
On Sat, Dec 22, 2018 at 4:32 PM Steven Yen wrote:
> How do I print a matrix running a procedure? I
Since you don't provide lambda, rh or qext it is impossible to reproduce
what you are seeing.
Also note that in this mailing list HTML formatted emails are not passed
along.
On Tue, Dec 25, 2018 at 4:13 AM M P wrote:
> Hello,
> I used commands below to obtain a surface, can plot it and all loo
5,] 0.6467642 0.6467642 0.6467642
>> [6,] 0.5597143 0.5597143 0.5597143
>> [7,] 0.4854133 0.4854133 0.4854133
>> [8,] 0.4278326 0.4278326 0.4278326
>> [9,] 0.3834149 0.3834149 0.3834149
>> [10,] 0.3433031 0.3433031 0.3433031
>>
>>
>>
>> On Tue, D
discussion
-- Forwarded message -
From: M P
Date: Wed, Dec 26, 2018 at 6:32 PM
Subject: Re: [R] fields package question
To: Eric Berger
Thanks, dput outputs below. I'd like to evaluate surface e.g. at (-10.5,
0.935)
> dput(x)
structure(c(-10.9251387646973, -10.89773
Shiny (from RStudio - and free)
A wonderful tool. And the app is accessed via the user's browser.
On Thu, Jan 10, 2019 at 4:18 PM Bernard McGarvey <
mcgarvey.bern...@comcast.net> wrote:
> I want to create an R application that includes a user interface where the
> user inputs values etc and then
9 = c(0, 0), X60 = c(0, 0), X61 = c(0,
>
> 0), X62 = c(0, 0), X63 = c(0, 0), X64 = c(0, 0), X65 = c(0,
>
> 0), X66 = c(0, 0), X67 = c(0, 0), X68 = c(0, 0), X69 = c(0,
>
> 0), X70 = c(0, 0), X71 = c(0, 0), X72 = c(0, 0), X73 = c(0,
>
> 0), X74 = c(0, 0), X75 =
Hi Lionel,
Your choice of variable names is a bit odd (the roles of x and y seem to be
reversed from the usual.)
Assuming that you are looking for linear interpolation (in spite of the
subject of your email),
does the following give you what you need?
u1 <- approx(x=y1,y=x1,xout=y)
u2 <- approx(x=
I have experience with R6 classes but I have not used operator overloading
with them.
Out of curiosity I did a quick search and found this link which provides a
step by step (I am not claiming it is the only way)
https://stackoverflow.com/questions/49463235/arithmetic-operators-overload-for-r6class
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