Dear Ivan,
Thanks a lot.!
Thanking you,
Yours sincerely,
AKSHAY M KULKARNI
From: Ivan Krylov
Sent: Saturday, December 24, 2022 9:27 PM
To: akshay kulkarni
Cc: R help Mailing list
Subject: Re: [R] difference between script and a function
On Sat, 24 Dec 2022 15:47:14 +
akshay kulkarni wrote:
> How do you debug if there is an error, particularly if I run the
> script from the BASH prompt?
Post-mortem debugging for non-interactive R scripts can be enabled by
setting options(error = quote(dump.frames("A_SUITABLE_FILE_NAME",
...but from the
Linux prompt?
Thanking you,
Yours sincerely,
AKSHAY M KULKARNI
From: Ivan Krylov
Sent: Saturday, December 24, 2022 8:52 PM
To: akshay kulkarni
Cc: R help Mailing list
Subject: Re: [R] difference between script and a function
On Sat, 24 Dec 2022 14:54
On Sat, 24 Dec 2022 14:54:52 +
akshay kulkarni wrote:
> If there is some error in the script, the error will be output to
> the stdout? Or to the file that it creates for saving the output of
> the script?
When using Rscript: to stderr, to be precise.
When using R CMD BATCH: to the Rout
From: Andrew Simmons
Sent: Saturday, December 24, 2022 8:18 PM
To: akshay kulkarni
Cc: R help Mailing list
Subject: Re: [R] difference between script and a function
1. The execution environment for a script is the global environment. Each R
script run from a shell
1. The execution environment for a script is the global environment. Each R
script run from a shell will be given its own global environment. Each R
session has exactly one global environment, but you can have several active
R sessions.
2. Using return in a script instead of a function will throw
Dear members,
I will be running scripts automatically in RHEL
with crontab. I want to know the differences between running a script and a
function. in particular:
1. An execution environment will be created for the function. what about a
script? Is the
Also, is the default base for "log" the same in both programs?
On Tue, Apr 14, 2020 at 3:36 PM Sorkin, John wrote:
>
> Your question is unlikely to be answered unless you post code demonstrating
> the problem
> J
>
> John David Sorkin M.D., Ph.D.
> Professor of Medicine
> Chief, Biostatistics
Your question is unlikely to be answered unless you post code demonstrating the
problem
J
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and Geriatric
Medicine
Baltimore VA Medical Center
Hae guys,
When performing a poisson regression sometimes one has to input the
offset/exposure variable to account for individual time spent in a certain
therapy before acquiring a certain condition of interest, whereby in r
offset(log(months)) and in STATA offset(log(months)) results differ.
Hello,
I would like to know the diffrence between 2 commands : cor_auto (from qgraph
package) and lavCor (from lavaan package) to compute a polychoric correlation
matrix in order to do a network analysis.
I have the responses to the SF-36 questionnaire (36 items with ordered
responses) and I
Because ifelse is not intended to be an alternative to if ... else. They exist
for different purposes.
(besides the other replies, a careful reading of their help pages, and trying
the examples, should explain the different purposes).
--
Don MacQueen
Lawrence Livermore National Laboratory
7000
On 13/12/2017 10:31 AM, Jinsong Zhao wrote:
Hi there,
I don't know why the following codes are return different results.
> ifelse(3 > 2, 1:3, length(1:3))
[1] 1
> if (3 > 2) 1:3 else length(1:3)
[1] 1 2 3
Any hints?
The documentation in the help page ?ifelse and ?"if" explains it pretty
ifelse returns the "shape" of the first argument
In your ifelse the shape of "3 > 2" is a vector of length one, so it will
return a vector length one.
Avoid "ifelse" until you are very comfortable with it. It can often burn
you.
On Wed, Dec 13, 2017 at 5:33 PM, jeremiah rounds
ifelse is vectorized.
On Wed, Dec 13, 2017 at 7:31 AM, Jinsong Zhao wrote:
> Hi there,
>
> I don't know why the following codes are return different results.
>
> > ifelse(3 > 2, 1:3, length(1:3))
> [1] 1
> > if (3 > 2) 1:3 else length(1:3)
> [1] 1 2 3
>
> Any hints?
>
> Best,
>
Hi there,
I don't know why the following codes are return different results.
> ifelse(3 > 2, 1:3, length(1:3))
[1] 1
> if (3 > 2) 1:3 else length(1:3)
[1] 1 2 3
Any hints?
Best,
Jinsong
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and
Hello,
i want to estimate the causal impact on a scale variable, using the
difference-in-difference-method and the following 4 groups
- control- and treatment group (counterfactual analysis)
- two periods, measurement before and after treatment.
After discovering and estimating the
Well,...
cat() is as Jeff describes.
However, print() is a generic function (see ?UseMethod) for which
there are literally hundreds of different methods that may do far
more/different than merely output character strings. For example, the
print method for trellis objects, print.trellis, draws a
Please stop posting html email per the Posting Guide. You are only going to
reduce the chance of successfully communicating your questions to experienced
users on this list.
Re cat vs print: the purpose of print is to show values much as they are
entered in source code, so quotes and escaped
Since both *cat * as well as * print * create a character vector for
outputing on the screen. Still both give different results as apparant
below. My query is why so ?
> cat(10)
10
> print(10)
[1] 10
Why is the [1] of index number missing in case of *cat *?
Thanks
Ramnik
[[alternative
On 2017/4/7 23:13, Jeff Newmiller wrote:
I think it is a fundamental characteristic of graphics drivers that output will
look different in the details... you are on a wild goose chase. Postscript in
particular has a huge advantage in font presentation over other graphics output
mechanisms.
On 2017/4/7 23:13, Jeff Newmiller wrote:
I think it is a fundamental characteristic of graphics drivers that output will
look different in the details... you are on a wild goose chase. Postscript in
particular has a huge advantage in font presentation over other graphics output
mechanisms.
I think it is a fundamental characteristic of graphics drivers that output will
look different in the details... you are on a wild goose chase. Postscript in
particular has a huge advantage in font presentation over other graphics output
mechanisms.
--
Sent from my phone. Please excuse my
Hi there,
I try to plot with custom fonts, which have good shape Latin and CJK
characters. I set up all the fonts correctly. However, when I plot the
same code on png() and postscript(), I get different result. The main
problem is the space between characters is narrower in postscript() than
[This is drifting somewhat awy from the original intention of the topic, I
think].
This looks like a build dependency. I get
3.3.2 (yeah, I know, should upgrade):
> (1+2i)/0
[1] NaN+NaNi
R-devel, march 24:
> (1+2i)/0
[1] Inf+Infi
on the *same* machine. The difference is that one is stock
> On 31 Mar 2017, at 19:28, John McKown wrote:
>
> On Fri, Mar 31, 2017 at 12:15 PM, Berend Hasselman wrote:
>
> I have noted a difference between R on macOS en on Kubuntu Trusty (64bits)
> with complex division.
> I don't know what would happen
On Fri, Mar 31, 2017 at 12:15 PM, Berend Hasselman wrote:
>
> I have noted a difference between R on macOS en on Kubuntu Trusty (64bits)
> with complex division.
> I don't know what would happen R on Windows.
>
> R.3.3.3:
>
> macOS (10.11.6)
> -
> > (1+2i)/0
> [1]
On 31.03.2017 19:15, Berend Hasselman wrote:
I have noted a difference between R on macOS en on Kubuntu Trusty (64bits) with
complex division.
I don't know what would happen R on Windows.
R.3.3.3:
macOS (10.11.6)
-
(1+2i)/0
[1] NaN+NaNi
(-1+2i)/0
[1] NaN+NaNi
1i/0
I have noted a difference between R on macOS en on Kubuntu Trusty (64bits) with
complex division.
I don't know what would happen R on Windows.
R.3.3.3:
macOS (10.11.6)
-
> (1+2i)/0
[1] NaN+NaNi
> (-1+2i)/0
[1] NaN+NaNi
>
> 1i/0
[1] NaN+NaNi
> 1i/(0+0i)
[1] NaN+NaNi
The only place I've noticed differences is in encoding and string sorting,
both of which are locale and library dependent.
Best,
Ista
On Mar 31, 2017 8:14 AM, "Neil Salkind" wrote:
> Can someone please direct me to an answer to the question as to how R
> differs for
File encodings differ when you move outside of standard ASCII code. Not really
R's problem, but it is a fly in the ointment when teaching classes with mixed
laptop armoury and there are also differences between classroom and desktop
computers. RStudio does have features to switch encodings, but
> On Mar 30, 2017, at 8:40 PM, Boris Steipe wrote:
>
> I can't remember having seen my students write code that runs correctly on
> one platform but not the other. Obviously under the hood there are
> significant differences, but as far as code goes, R seems quite
I can't remember having seen my students write code that runs correctly on one
platform but not the other. Obviously under the hood there are significant
differences, but as far as code goes, R seems quite foolproof. There are GUI
differences in base R - but AFAIK no such differences in the
Can someone please direct me to an answer to the question as to how R differs
for these two operating systems, if at all? Thanks - Neil
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
Dear R forum
I have following dataset
amounts =
Hi,
As Jeff said, more than one grouping variable can be supplied, and there
is an example at the bottom of the help page for ave(). The same goes
for by(), but the order that you supply the grouping variables becomes
important. Whichever grouping variable is supplied first to by() will
Now would be an excellent time to read the help page for ?ave. You can specify
multiple grouping variables.
--
Sent from my phone. Please excuse my brevity.
On October 28, 2016 7:28:44 PM PDT, Ashta wrote:
>Hi all thank you very much for your help. Worked very well for that
Hi all thank you very much for your help. Worked very well for that
data set. I just found out that one of the data sets have another
level and do the same thing, I want to calculate the difference
between successive row values (num) to the first row value within
city and year.
city, year, num
Haven't seen a solution using ave...
d$diff <- ave( d$num, d$year, FUN = function( v ) { v - v[ 1 ] } )
--
Sent from my phone. Please excuse my brevity.
On October 28, 2016 12:46:15 PM CDT, William Dunlap via R-help
wrote:
>You could use match() to find, for each row,
You could use match() to find, for each row, the index of the first
row with the give row's year:
> d <- data.frame(year=c(rep(2001, 3), rep(2002, 3)),
num=c(25,75,150,30,85,95))
> indexOfFirstOfYear <- with(d, match(year, year))
> indexOfFirstOfYear
[1] 1 1 1 4 4 4
> d$diff <-
I read the problem incorrectly; I did not see that you wanted the
difference from the first entry; trying again:
> require(dplyr)
> input <- read.table(text = "Year Num
+ 200125
+ 200175
+ 2001 150
+ 200230
+ 200285
+ 200295", header = TRUE)
>
> input %>%
+
Here are a couple of other ways using 'dplyr' and 'data.table'
> require(dplyr)
> input <- read.table(text = "Year Num
+ 200125
+ 200175
+ 2001 150
+ 200230
+ 200285
+ 200295", header = TRUE)
>
> input %>%
+ group_by(Year) %>%
+ mutate(diff = c(0, diff(Num)))
Hi,
You could use an anonymous function to operate on each `year-block' of
your dataset, then assign the result as a new column:
d <- data.frame(year=c(rep(2001, 3), rep(2002, 3)),
num=c(25,75,150,30,85,95))
d$diff <- unlist(by(d$num, d$year, function(x) x - x[1]))
d
year
Hi all,
I want to calculate the difference between successive row values to
the first row value within year.
How do I get that?
Here isthe sample of data
Year Num
200125
200175
2001 150
200230
200285
200295
Desired output
Year Num diff
200125 0
2001
You were clearly mistaken.
dataframe$column is almost the same as dataframe[["column"]], except that the $
does partial matching. Both of these "extract" a list element.
A data frame is a list where all elements are vectors of the same length. A
list is a vector where each element can refer
Hi All,
I thought dataset$variable is the same as dataset["variable"]. I tried the
following:
> str(ZWW_Kunden$Branche)
chr [1:49673] "231" "151" "151" "231" "231" "111" "231" "111" "231" "231"
"151" "111" ...
> str(ZWW_Kunden["Branche"])
'data.frame':49673 obs. of 1 variable:
$
Dear Jean,
Have a look at
http://stackoverflow.com/questions/5595512/what-is-the-difference-between-require-and-library
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics &
Hi,
Is there any other difference between 'require' and 'library' than the error or
warning when the library is not found ?
Jean in France
Ce message et toutes les pièces jointes (ci-après le 'Message') sont établis à
l'intention exclusive des destinataires et les informations qui y
Hi catalin,
I think what you are trying to do is to retrieve the original
observations from the cumulated values. In that case Olivier's
suggestion will do what you want:
c(x[1],diff(x))
Jim
On Sat, Mar 5, 2016 at 1:59 AM, catalin roibu wrote:
> I mean the first row
I mean the first row value
În Vin, 4 mar. 2016, 16:15 Jeff Newmiller, a
scris:
> "Keep the first values" is imprecise, but mixing an absolute value with a
> bunch of differences doesn't usually work out well. I frequently choose
> among
>
> x <- sample( 10 )
> dxright
"Keep the first values" is imprecise, but mixing an absolute value with a bunch
of differences doesn't usually work out well. I frequently choose among
x <- sample( 10 )
dxright <- c( 0, diff(x) )
dxleft <- c( diff(x), 0 )
for calculation purposes depending on my needs.
--
Sent from my
Hi,
(1) You should provide a minimal working example;
(2) But anyway, does...
x = sample(10)
c(x[1],diff(x))
... do what you want?
Olivier.
On Fri, 4 Mar 2016
13:22:07 +0200 catalin roibu wrote:
> Dear all!
>
> I want to calculate difference between successive
Dear all!
I want to calculate difference between successive values (cumulative
values) with R. I used diff function, but I want to keep the first values.
Please help me to solve this problem!
Thank you!
Best regards!
CR
--
-
-
Catalin-Constantin ROIBU
Lecturer PhD, Forestry engineer
Hi,
The first thing to check is if both functions use the same 'base' option - they
usually don't.
On a more general note, the two approaches use a different data augmentation
procedure, but this should not have a very large impact on the results provided
enough draws and similar priors are
On 16-08-2015, at 16:38, Jinsong Zhao jsz...@yeah.net wrote:
Hi there,
I notice that write.csv is a wrap of write.table. However, I can't get the
same results using both functions. Here is a reproducible example:
x - matrix(1:6, nrow =2)
rownames(x) - letters[1:2]
colnames(x) -
On Aug 16, 2015, at 9:38 AM, Jinsong Zhao jsz...@yeah.net wrote:
Hi there,
I notice that write.csv is a wrap of write.table. However, I can't get the
same results using both functions. Here is a reproducible example:
x - matrix(1:6, nrow =2)
rownames(x) - letters[1:2]
colnames(x)
I think that if you do ?write.csv and then page down to the section
entitled CSV files the mystery will be solved for you in the first few
paragraphs.
On 16/08/2015 15:38, Jinsong Zhao wrote:
Hi there,
I notice that write.csv is a wrap of write.table. However, I can't get
the same results
Hi there,
I notice that write.csv is a wrap of write.table. However, I can't get
the same results using both functions. Here is a reproducible example:
x - matrix(1:6, nrow =2)
rownames(x) - letters[1:2]
colnames(x) - LETTERS[1:3]
write.csv(x, )
,A,B,C
a,1,3,5
b,2,4,6
write.table(x, ,
Dear list members,
I’m having some problems understanding why drop1() and anova() gives
different results for *Gaussian* glm models. Here’s a simple example:
d = data.frame(x=1:6,
group=factor(c(rep(A,2), rep(B, 4
l = glm(x~group, data=d)
Running the following code
I am somewhat surprised that _anything_ sensible comes out of anova.glm(l,
test=Chisq). I think it is mostly expected that you use F tests for that case.
What does seem to come out is the same as for drop1(l, test=Rao), which gives
the scaled score test, which would seem to be equivalent to
Dear Duncan,
I had been thinking about FAQ 7.31. I tried to create a dummy dataset with
the same structure to replicate the problem with the need of sending my
dataset. However all of them gave identical() results between 32-bit and
64-bit. Note that coef()$fRow is a 1266 x 6 data.frame. Is it
low probability of occurring was just statisticians lingo for rare ;-)
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
To call in
On 04/06/2015 3:59 AM, Thierry Onkelinx wrote:
Dear Duncan,
I had been thinking about FAQ 7.31. I tried to create a dummy dataset
with the same structure to replicate the problem with the need of
sending my dataset. However all of them gave identical() results between
32-bit and 64-bit.
On 03/06/2015 11:56 AM, Thierry Onkelinx wrote:
Dear all,
I'm a bit puzzled by the difference in an object when created in R 32-bit
and R 64-bit.
Consider the code below. test.rda is available at
https://drive.google.com/file/d/0BzBrlGSuB9n-NFBWeC1TR093Sms/view?usp=sharing
# Run in R
Dear all,
I'm a bit puzzled by the difference in an object when created in R 32-bit
and R 64-bit.
Consider the code below. test.rda is available at
https://drive.google.com/file/d/0BzBrlGSuB9n-NFBWeC1TR093Sms/view?usp=sharing
# Run in R 3.2.0 Windows 32-bit, lme4 1.1-8
library(lme4)
]
Sent: zaterdag 14 februari 2015 8:16
To: 'Martyn Byng'; r-help@r-project.org
Cc: Franckx Laurent
Subject: RE: [R] difference between max in summary table and max function
I thought I'd chime in ... submitting the following:
?summary
Provides the following documentation for the default
HI Farnoosh,
Not sure I understand the expected output. The difference between the first 2
days is 136 days
May be this helps
library(data.table)
dcast.data.table(setDT(df)[, list(Visit=.N, Diff=
as.numeric(abs(diff(as.Date(Date, format='%d-%b-%y') ,
by = ID],
That's exactly what I was thinking. Thanks tons.
Sent from Yahoo Mail on Android
From:arun smartpink...@yahoo.com
Date:Sun, Feb 15, 2015 at 2:47 AM
Subject:Re: Difference in dates for unique ID
HI Farnoosh,
Not sure I understand the expected output.� The difference between the first 2
days
Dear all
I have found out that the max in the summary of an integer vector is not always
equal to the actual maximum of that vector. For example:
testrow - c(1:131509)
summary(testrow)
Min. 1st Qu. MedianMean 3rd Qu.Max.
1 32880 65760 65760 98630 131500
Its a formatting thing, try
summary(testrow,digits=20)
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Franckx Laurent
Sent: 13 February 2015 11:00
To: r-help@r-project.org
Subject: [R] difference between max in summary table and max function
Dear all
Byng
Sent: Friday, February 13, 2015 3:15 AM
To: Franckx Laurent; r-help@r-project.org
Subject: Re: [R] difference between max in summary table and max function
Its a formatting thing, try
summary(testrow,digits=20)
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org
Dear R users and contributors,
I recently observed a difference between the outputs of the classic
eigen() function, and the Arnoldi variant eigs() that extracts only
the few first eigenpairs. Here is some sample code illustrating the
problem:
library(rARPACK)
library(speccalt)
set.seed(1)
#
eigs() is from a contributed package. No idea what it is about, but my
guess is these are actually numerical differences coming from different
algorithms used to calculate the eigenvalues.
For details, please ask the author of the corresponding contributed package.
Best,
Uwe Ligges
On
Dear Rena,
As Peter points out, it is better to ask the maintainer of the program for
detailed questions.
As Peter correctly surmised, print.psych (which is used to print the output
from the fa function), knows that you have an oblique solution and is reporting
the amount of variance
Firstly, there is no fa() function in base R. There is one in package psych(),
which has a maintainer, etc.
I guess that it is because fa() does a non-orthogonal factor rotation and its
print method knows about it, whereas the default print method for loadings
assumes that rotations are
Hello,
I am trying a factor analysis via R.
When running the pricipal axis analysis I do get different tables depending
on the print command.
This is my factor analysis:
fa.pa_cor_3_2- fa(ItemsCor_4, nfactors=3, fm=pa,rotate=oblimin)
To get the h2 I did the following print command:
print
This is pretty much standard. I'm quite sure that other stats packages do
likewise and I wouldn't know who everyone is. It is not unheard of that
textbook authors give suboptimal formulas in order not to confuse students,
though.
The basic point is that the t transformation gives the exact
$\rho= \rho_0 $ or $\rho_1 =
\rho_2$.
The two statistics will not be equivalent at $\rho=0$ because the
statistics are based on different assumptions.
Jeremy Miles jeremy.mi...@gmail.com
Sent by: r-help-boun...@r-project.org
10/16/2014 07:32 PM
To
r-help r-help@r-project.org,
cc
Subject
[R
I'm trying to understand how cor.test() is calculating the p-value of
a correlation. It gives a p-value based on t, but every text I've ever
seen gives the calculation based on z.
For example:
data(cars)
with(cars[1:10, ], cor.test(speed, dist))
Pearson's product-moment correlation
data:
Hi Jeremy,
I don't know about references, but this around. See for example:
http://afni.nimh.nih.gov/sscc/gangc/tr.html
the relevant line in cor.test is:
STATISTIC - c(t = sqrt(df) * r/sqrt(1 - r^2))
You can convert *t*s to *r*s and vice versa.
Best,
Josh
On Fri, Oct 17, 2014 at 10:32
Dear R users,
Hi, thank you so much for your help in advance.
I have been using Stata but new to R. For my paper revision using Aalen's
survival analysis, I need to use R, as the command including Aalen's survival
seems to be available in R (32-bit, version 3.1.0 (2014-04-10)) but less ready
In the Cox regression case, the probable explanation is that you have
ties in your data; Stata and coxph may have different defaults for
handling ties. Read the manuals!
The difference in sign in the other cases is simply due to different
definitions of the models. I am sure it is well
Hi,
It is better to show the example data using ?dput(). Here, it is not clear
whether the columns are character columns or lists.
##If it is the latter case
dat1 - data.frame(V1=I(list(1:3, c(1,2,4), c(2,3,4,5))), V2= I(list(c(3,6,5),
c(7,10,9), 2:5)))
dat1$V3 - mapply(`c`,mapply(`-`,
HI,
I guess this should be a bit faster.
#1st case
dat1$V3 - lapply(seq_along(dat1$V2),function(i) c(dat1$V2[[i]][-1] -
head(dat1$V1[[i]],-1), tail(dat1$V1[[i]],1)))
#2nd case
dat2$V3 - unlist(lapply(seq_along(lst1[,2]),function(i)
paste(c(lst1[,2][[i]][-1] - head(lst1[,1][[i]], -1),
On 18/04/2014 21:46, David Winsemius wrote:
On Apr 18, 2014, at 12:59 PM, Prof Brian Ripley wrote:
On 18/04/2014 19:46, Rui Barradas wrote:
Hello,
The reason why is that you've misspelled CET (not CEST)
Neither CET nor CEST are portable time-zone names. We have not been given the
'at a
On 20/04/2014 08:50, Prof Brian Ripley wrote:
On 18/04/2014 21:46, David Winsemius wrote:
On Apr 18, 2014, at 12:59 PM, Prof Brian Ripley wrote:
On 18/04/2014 19:46, Rui Barradas wrote:
Hello,
The reason why is that you've misspelled CET (not CEST)
Neither CET nor CEST are portable
On Apr 20, 2014, at 2:16 AM, Prof Brian Ripley wrote:
On 20/04/2014 08:50, Prof Brian Ripley wrote:
On 18/04/2014 21:46, David Winsemius wrote:
On Apr 18, 2014, at 12:59 PM, Prof Brian Ripley wrote:
On 18/04/2014 19:46, Rui Barradas wrote:
Hello,
The reason why is that you've
Thank you for your reply.
I discovered the OlsonNames() function to get the time-zone names in my
system. Rui get a warning message when using a not recognized tz. On my
system this doesn't succed.
I solved as follow:
dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
Europe/Rome)
I forgot:
sysname
release
Linux
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2014-04-19 14:03 GMT+02:00 Nicola Sturaro Sommacal
Hi.
I am new to POSIX and I'd like to understand the reason of this difference.
dt1 = as.POSIXct(2014-03-29 09.00, format=%Y-%m-%d %H.%M)
dt2 = as.POSIXct(2014-03-30 09.00, format=%Y-%m-%d %H.%M)
dt2-dt1
dt1[1] 2014-03-29 09:00:00 CET dt2[1] 2014-03-30 09:00:00 CEST dt2-dt1
Time difference of
Hello,
The reason why is that you've misspelled CET (not CEST)
dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz =
CEST)
Warning messages:
1: In strptime(x, format, tz = tz) : unknown timezone 'CEST'
2: In as.POSIXct.POSIXlt(as.POSIXlt(x, tz, ...), tz, ...) :
unknown timezone
On 18/04/2014 19:46, Rui Barradas wrote:
Hello,
The reason why is that you've misspelled CET (not CEST)
Neither CET nor CEST are portable time-zone names. We have not been
given the 'at a minimum' information required by the posting guide, so
please read ?Sys.timezone on your system.
On Apr 18, 2014, at 12:59 PM, Prof Brian Ripley wrote:
On 18/04/2014 19:46, Rui Barradas wrote:
Hello,
The reason why is that you've misspelled CET (not CEST)
Neither CET nor CEST are portable time-zone names. We have not been given
the 'at a minimum' information required by the
I have a data frame with variable datetime which is of class POSIXct.
Consecutive observations are separated by 30 minutes.
However, some of the differences reported by R give unexpected results.
For example consider the following two consecutive entries:
par.dat$datetime[5944]
[1] 2010-04-04
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of David Fox
Sent: Tuesday, January 28, 2014 5:15 PM
To: r-help@r-project.org
Subject: [R] Difference between two datetimes
I have a data frame with variable datetime which
Dear all,
When I run an arima(1,1,1) on an I(1) variable, y, I get different estimates to
when I first difference the variable myself, e.g y2-diff(y), and then run
arima(1,0,1) on y2. Shouldn't these two approaches give the same output?
Any help will be much appreciated.
george
Dear all,
When I run an arima(1,1,1) on an I(1) variable, e.g. y, I get different
estimates to when I first difference the variable myself, e.g y2-diff(y),
and then run arima(1,0,1) on y2. Shouldn't these two approaches give the
same output?
Any help will be much appreciated.
george
--
View
Hi George: Assuming it's still relevant, the link below will explain why.
http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm
On Thu, Jul 18, 2013 at 2:14 PM, George Milunovich
george.milunov...@mq.edu.au wrote:
Dear all,
When I run an arima(1,1,1) on an I(1) variable, y, I get different
Hi Mark,
This is very helpful!!
Much appreciated
Sent from my iPad
On Jul 19, 2013, at 3:51 AM, Mark Leeds marklee...@gmail.com wrote:
Hi George: Assuming it's still relevant, the link below will explain why.
http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm
On Thu, Jul 18, 2013 at
Hi,
Does anybody know the difference between the Lloyd and Forgy algorithms
specified for R's kmeans clustering options? I know how Lloyd works, but I
cannot access Forgy's paper and could not find any specific information on
the web about how it really differs from Lloyd's method.
I appreciate
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