I always find regex puzzles amusing, so after changing the unicode
typo quotes and dashes to ascii, the following simple prescription,
similar to those proffered by others, seems to produce what you
requested with your example:
x <- c("leucocyten + gramnegatieve staven +++ grampositieve staven
Sometimes you need to NOT use a regular expression and do things simpler. You
have a fairly simple example that not only does not need great power but may be
a pain to do using a very powerful technique, especially if you want to play
with look-ahead and look behind.
Assuming you have a line
I thought replacing the spaces following instances of +++,++,+,- with "\n" and
then reading with scan should succeed. Like Ivan Krylov I was fairly sure that
you meant the minus sign to be "-" rather than "–", but perhaps your were using
MS Word as an editor which is inconsistent with effective
Dear Jeff,
Regrets for not trying that before...never knew that % was the
culprit...!
THanking you,
Yours sincerely,
AKSHAY M KULKARNI
From: Jeff Newmiller
Sent: Thursday, April 13, 2023 1:08 AM
To: r-help@r-project.org ; akshay kulkarni
; Rui
Isn't this like trying to tie up the horse after it has left the barn? Why not
figure all this out _before_ converting to xts?
On April 12, 2023 12:29:49 PM PDT, akshay kulkarni
wrote:
>Dear Rui,
> Not working. I have entirely removed the column containing %
> but am still
Dear Rui,
Not working. I have entirely removed the column containing %
but am still bootless:
> head(coredata(INFYTX))
INFY Historical Data INFY Historical Data INFY Historical Data INFY
Historical Data
[1,] "47.26" "44.28" "47.56"
Às 19:57 de 12/04/2023, akshay kulkarni escreveu:
Dear members,
I have an xts object:
head(INFYTX)
INFY Historical Data INFY Historical Data.1 INFY Historical Data.2
2003-04-16 "47.26" "44.28""47.56"
2003-04-17 "46.30"
Dear members,
I have an xts object:
> head(INFYTX)
INFY Historical Data INFY Historical Data.1 INFY Historical Data.2
2003-04-16 "47.26" "44.28""47.56"
2003-04-17 "46.30" "44.92""46.53"
2003-04-21
On Wed, 12 Apr 2023 16:04:55 +0800 (GMT+08:00)
"Dezhi Wang" wrote:
> I run R 4.2.2 on CentOS-7.6.
Thank you for this useful information. Could you tell us how this build
of R was installed on this machine?
> > install.packages('sf')
> --- Please select a CRAN mirror for use in this session
On Wed, 12 Apr 2023 08:29:50 +
Emily Bakker wrote:
> Some example data:
> “leucocyten + gramnegatieve staven +++ grampositieve staven ++”
> “leucocyten – grampositieve coccen +”
>
> I want to split the strings such that I get the following result:
> c(“leucocyten +”, “gramnegatieve staven
Hi Eric,
THanks a lot..
Thanking you,
Yours sincerely,
AKSHAY M KULKARNI
From: Eric Berger
Sent: Wednesday, April 12, 2023 8:20 PM
To: akshay kulkarni
Cc: R help Mailing list
Subject: Re: [R] converting to date object...
lubridate::dmy("12 APR
Hi Mark,
Thanks a lot...
Thanking you,
Yours sincerely,
AKSHAY M KULKARNI
From: Marc Schwartz
Sent: Wednesday, April 12, 2023 8:33 PM
To: akshay kulkarni ; R help Mailing list
Subject: Re: [R] converting to date object...
Hi,
You do not need
Dear Dirk,
Thaks a lot...
THanking you,
Yours sincerely,
AKSHAY M KULKARNI
From: Dirk Eddelbuettel
Sent: Wednesday, April 12, 2023 8:38 PM
To: akshay kulkarni
Cc: R help Mailing list
Subject: Re: [R] converting to date object...
That is
This seems to do the job but there are probably more elegant solutions:
f <- function(s) { sub("^ ","",unlist(strsplit(gsub("\\+ ","+@ ",s),"@"))) }
g <- function(s) { sub("^ ","",unlist(strsplit(gsub("- ","-@ ",s),"@"))) }
h <- function(s) { g(f(s)) }
To try it out:
s <- “leucocyten +
Hello List,
I have a dataset consisting of strings that I want to split while saving the
delimiter.
Some example data:
“leucocyten + gramnegatieve staven +++ grampositieve staven ++”
“leucocyten – grampositieve coccen +”
I want to split the strings such that I get the following result:
Hello,
I run R 4.2.2 on CentOS-7.6. I install package through
install.packages("sf").It threw the above error:
> install.packages('sf')
--- Please select a CRAN mirror for use in this session ---
*** caught segfault ***
address 0x60, cause 'memory not mapped'
Traceback:
1: download.file(url,
That is what I wrote the anytime package for: effortless automatic
parsing. Also works for dates:
> library(anytime)
> anydate("12 APR 2023")
[1] "2023-04-12"
>
Dirk
--
dirk.eddelbuettel.com | @eddelbuettel | e...@debian.org
__
Hi,
You do not need to use third party packages for date or date/time objects in R.
If you review ?as.Date, you will see that there are two default formats for
Date objects that are specified in the 'tryFormats' argument:
tryFormats = c("%Y-%m-%d", "%Y/%m/%d")
If your date character vector
lubridate::dmy("12 APR 2023")
On Wed, Apr 12, 2023 at 5:34 PM akshay kulkarni
wrote:
> dear members,
> I want to convert "12 APR 2023" into a Date
> object. I tried as_Date() from lubridate, but it is not working:
>
> > as_date("12 APR 2023")
> [1] NA
> Warning
dear members,
I want to convert "12 APR 2023" into a Date object.
I tried as_Date() from lubridate, but it is not working:
> as_date("12 APR 2023")
[1] NA
Warning message:
All formats failed to parse. No formats found.
> as_date("12-APR-2023")
[1] NA
Warning message:
Dear Duncan,
But it is entirely different in my case...! As you
said, the problem may be with my version of RStudio. The main point is that it
doesn't pose any serious problems...
Thanks any way for your reply...
Thanking you,
Yours sincerely,
AKSHAY M KULKARNI
This is what I get:
> source("~/temp/test.R", echo = TRUE)
> print(1)
[1] 1
> stop("here")
Error in eval(ei, envir) : here
I get similar output in every variation I tried. It never prints the 2.
On 12/04/2023 8:13 a.m., akshay kulkarni wrote:
Dear Duncan,
What if
Dear Duncan,
What if I use source() with echo? I am using that in
RStudio.
THanking you,
Yours sincerely
AKSHAY M KULKARNI
From: Duncan Murdoch
Sent: Wednesday, April 12, 2023 5:35 PM
To: akshay kulkarni ; R help Mailing list
Subject:
On 12/04/2023 7:03 a.m., akshay kulkarni wrote:
Dear members,
I have a script which I source it interactively. I
have the following questions:
1. If there is an error in an expression, an error message is printed, but
the execution continues till the end of
Dear members,
I have a script which I source it interactively. I
have the following questions:
1. If there is an error in an expression, an error message is printed, but
the execution continues till the end of the script. I am sourcing with echo. Is
there any
Dear all,
We are pleased to announce that CRAN now hosts our packages
DNAmixturesLite and KinMixLite.
These are fully R-based versions of our packages DNAmixtures and KinMix,
which rely on a commercial software, HUGIN, for performing efficient
computations in Bayesian networks. The lite
You may take a look at the bigmemory package or other which deal with large
memory data in
https://cran.r-project.org/web/views/HighPerformanceComputing.html#large-memory-and-out-of-memory-data
Some extra explanation is in https://stackoverflow.com/a/11127229/997979
Iago
One possibility might be to use Rcpp.
An R matrix is stored in contiguous memory, which can be considered as a
vector.
Define a C++ function which operates on a vector in place, as in the
following:
library(Rcpp)
cppFunction(
'void subtractConst(NumericVector x, double c) {
for ( int i = 0; i
28 matches
Mail list logo
