On 08/12/2023 4:30 p.m., David Winsemius wrote:
On 12/7/23 08:21, Sorkin, John wrote:
Colleagues,
I have a matrix of character data that represents date and time. The format of
each element of the matrix is
"2020-09-17_00:00:00"
How can I convert the elements into a valid R
On 12/7/23 08:21, Sorkin, John wrote:
Colleagues,
I have a matrix of character data that represents date and time. The format of
each element of the matrix is
"2020-09-17_00:00:00"
How can I convert the elements into a valid R date-time constant?
You will not be able to s
17_00:00:00", format = "%Y-%m-%d_%H:%M:%S")
[1] "2020-09-17 CEST"
>
(in my time zone).
> -Original Message- From: R-help
> On Behalf Of Sorkin, John
> Sent: Thursday, December 7, 2023 11:22 AM To:
> r-help@r-project.org (r-help@r-
`anytime` was written for this:
> anytime::anytime("2020-09-17_00:00:00")
[1] "2020-09-17 CDT"
> class(anytime::anytime("2020-09-17_00:00:00"))
[1] "POSIXct" "POSIXt"
>
Dirk
--
dirk.eddelbuettel.com | @eddelbuettel | e...@debian.org
__
Às 16:30 de 07/12/2023, Rui Barradas escreveu:
Às 16:21 de 07/12/2023, Sorkin, John escreveu:
Colleagues,
I have a matrix of character data that represents date and time. The
format of each element of the matrix is
"2020-09-17_00:00:00"
How can I convert the elements into a valid R
Às 16:21 de 07/12/2023, Sorkin, John escreveu:
Colleagues,
I have a matrix of character data that represents date and time. The format of
each element of the matrix is
"2020-09-17_00:00:00"
How can I convert the elements into a valid R date-time constant?
Thank you,
John
John Da
Look at the lubridate package in R.
Regards,
Tim
-Original Message-
From: R-help On Behalf Of Sorkin, John
Sent: Thursday, December 7, 2023 11:22 AM
To: r-help@r-project.org (r-help@r-project.org)
Subject: [R] Convert character date time to R date-time variable.
[External Email
Colleagues,
I have a matrix of character data that represents date and time. The format of
each element of the matrix is
"2020-09-17_00:00:00"
How can I convert the elements into a valid R date-time constant?
Thank you,
John
John David Sorkin M.D., Ph.D.
Professor of Medicine, Uni
Dear Enrico,
Thanks a lot, that clarifies the topic for me.
Checking the numeric representation i was not aware of.
Best Regards
Tilmann
On 30.04.21 11:17, Enrico Schumann wrote:
> On Fri, 30 Apr 2021, Tilmann Faul writes:
>
>> Dear Jeff,
>>
>> Thanks for your answer.
>> Sys.timezone() gives
>
On Fri, 30 Apr 2021, Tilmann Faul writes:
> Dear Jeff,
>
> Thanks for your answer.
> Sys.timezone() gives
> [1] "Europe/Berlin"
> I tried "Europe/Berlin" as tz argument, giving the same result als using
> "CEST" (Central European Summer Time).
> It seems to me, that using as.POSIXct without tz ar
Dear Jeff,
Thanks for your answer.
Sys.timezone() gives
[1] "Europe/Berlin"
I tried "Europe/Berlin" as tz argument, giving the same result als using
"CEST" (Central European Summer Time).
It seems to me, that using as.POSIXct without tz argument defaults to tz
UTC and with tz argument, either "CE
="CET"),
0.3,
as.POSIXct("2021-04-21 00:00:00", tz="CET"),
0.2,
length=0.07, angle=15)
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Jeff Newmiller
> Sent: Thursday, April 29, 2021 11:20 PM
> To: r-hel
What is your TZ environment variable set to? That's what time conversion
defaults to ?DateTimeClasses
Also, I am not sure CEST is a valid timezone designation... it can be system
dependent, but using one of the elements listed in ?OlsonNames.
On April 29, 2021 12:22:44 PM PDT, Tilmann Faul
w
Hy,
stumbled over the following problem while plotting DateTime Objects.
plot(as.POSIXct(c("2021-04-21 00:00:00", "2021-04-21 23:59:59")), c(0,
1), type='l')
arrows(as.POSIXct("2021-04-21 00:00:00", tz="CEST"),
0.3,
as.POSIXct("2021-04-21 00:00:00", tz="CEST"),
0.2,
l
r-Help Community
Never mine figured it out just use the "as.POSIXct" function
Jeff
I need to convert a date-time field (column) but I'm losing the time when
I convert using ..
tsData <- myData[,10, drop=FALSE]
tsData$date_time <- as.Date(tsData$date_time, format="%m/%d/%y %H:%M"
Hello,
as.Date() outputs an object of class "Date", you want an object of class
c("POSIXt", "POSIXct"). Use as.POSIXct().
Hope this helps,
Rui Barradas
Às 16:04 de 08/05/19, reichm...@sbcglobal.net escreveu:
r-Help Community
I need to convert a date-time field (column) but I'm losing
r-Help Community
I need to convert a date-time field (column) but I'm losing the time when
I convert using ..
tsData <- myData[,10, drop=FALSE]
tsData$date_time <- as.Date(tsData$date_time, format="%m/%d/%y %H:%M")
head(tsData)
date_time
1
2013-06-20
On 25 April 2017 at 18:05, Duncan Murdoch wrote:
| On 25/04/2017 5:41 PM, Dirk Eddelbuettel wrote:
| >
| > On 25 April 2017 at 16:04, Jeff Reichman wrote:
| > | R Users
| > |
| > | Having problems converting the following DTG into an R recognized
date/time
| > | field
| > |
| > | 01-01-2016T14:02
On 25/04/2017 5:41 PM, Dirk Eddelbuettel wrote:
On 25 April 2017 at 16:04, Jeff Reichman wrote:
| R Users
|
| Having problems converting the following DTG into an R recognized date/time
| field
|
| 01-01-2016T14:02:23.325
|
| Would I separate it into a date field and time filed then put it back
On 25 April 2017 at 16:04, Jeff Reichman wrote:
| R Users
|
| Having problems converting the following DTG into an R recognized date/time
| field
|
| 01-01-2016T14:02:23.325
|
| Would I separate it into a date field and time filed then put it back
| together???
The anytime package (on CRAN) do
> z <- as.POSIXct("01-01-2016T14:02:23.325", format="%d-%m-%YT%H:%M:%OS")
> dput(z)
structure(1451685743.325, class = c("POSIXct", "POSIXt"), tzone = "")
> z
[1] "2016-01-01 14:02:23 PST"
> format(z, "%H:%M:%OS3 on %b %d, %Y")
[1] "14:02:23.325 on Jan 01, 2016"
(Don't separate the date and time pa
On 25/04/2017 5:04 PM, Jeff Reichman wrote:
R Users
Having problems converting the following DTG into an R recognized date/time
field
01-01-2016T14:02:23.325
Would I separate it into a date field and time filed then put it back
together???
This appears to work (though I'm not sure whe
R Users
Having problems converting the following DTG into an R recognized date/time
field
01-01-2016T14:02:23.325
Would I separate it into a date field and time filed then put it back
together???
Jeff
[[alternative HTML version deleted]]
__
2:16 PM
>>>>> > *To:* Tom Wright
>>>>> > *Cc:* David L Carlson ; r-help <
>>>>> r-help@r-project.org>
>>>>>
>>>>> >
>>>>> > *Subject:* Re: [R] Date Time in R
>>>>> >
>&
>
>>>> >
>>>> > Please find the details: (i have changed name from eir to a1, rest
>>>> all is
>>>> > same)
>>>> >
>>>> >
>>>> >
>>>> > str(a1$date)
>>>> >
>>&
;>> >
>>> > # extract the weekdays as number
>>> >
>>> > weekdays <- wday(a1$date)
>>> >
>>> >
>>> >
>>> > # note can’t have spaces in variable names
>>> >
>>> > week_names <- wda
by
>> default
>> >
>> > # so no need for the mdy() or ymd() function
>> >
>> >
>> >
>> > # extract the weekdays as number
>> >
>> > weekdays <- wday(a1$date)
>> >
>> >
>> >
>> >
names
> >
> > week_names <- wday(a1$date, label=TRUE)
> >
> >
> >
> >
> >
> > *From:* Shivi Bhatia [mailto:shivipm...@gmail.com]
> > *Sent:* July 26, 2016 12:16 PM
> > *To:* Tom Wright
> > *Cc:* David L Carlson ; r-help
>
gt;
>
>
> *From:* Shivi Bhatia [mailto:shivipm...@gmail.com]
> *Sent:* July 26, 2016 12:16 PM
> *To:* Tom Wright
> *Cc:* David L Carlson ; r-help
>
> *Subject:* Re: [R] Date Time in R
>
>
>
> Hello Tom,
>
>
>
> Please find the details: (i have change
- wday(a1$date)
# note can’t have spaces in variable names
week_names <- wday(a1$date, label=TRUE)
*From:* Shivi Bhatia [mailto:shivipm...@gmail.com]
*Sent:* July 26, 2016 12:16 PM
*To:* Tom Wright
*Cc:* David L Carlson ; r-help
*Subject:* Re: [R] Date Time in R
Hello Tom,
Please
rmation before doing any processing on the date
> (i.e. before the as.Date() function) we may be able to help.
>
>
>
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Shivi
> Bhatia
> Sent: July 26, 2016 11:46 AM
> To: David L Ca
; relevant in this context.
>
> On Tue, Jul 26, 2016 at 8:55 PM, David L Carlson wrote:
>
>> Show us the output, don’t just tell us what you are seeing. If the dates
>> are correct in the csv file, show us the structure of the data frame you
>> created with read.csv() an
-help-boun...@r-project.org] On Behalf Of Shivi Bhatia
Sent: July 26, 2016 11:46 AM
To: David L Carlson
Cc: r-help
Subject: Re: [R] Date Time in R
Hi David please see the code and some reproducible data:
eir$date<- as.Date(eir$date,format = "%m-%d-%y") then i had used the
lubridate li
n this context.
>
> On Tue, Jul 26, 2016 at 8:55 PM, David L Carlson
> wrote:
>
>> Show us the output, don’t just tell us what you are seeing. If the dates
>> are correct in the csv file, show us the structure of the data frame you
>> created with read.csv() and show
the command(s) you used to convert the
> character data to date format. The solution is likely to be simple if you
> will cut/paste the R console and not just describe what is happening.
>
>
>
> David C
>
>
>
> *From:* Shivi Bhatia [mailto:shivipm...@gmail.com]
>
om: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Shivi
> Bhatia
> Sent: Tuesday, July 26, 2016 4:43 AM
> To: Duncan Murdoch
> Cc: r-help@r-project.org
> Subject: Re: [R] Date Time in R
>
> Thanks Duncan for the quick response. I will check again as you suggested
will cut/paste the R
console and not just describe what is happening.
David C
From: Shivi Bhatia [mailto:shivipm...@gmail.com]
Sent: Tuesday, July 26, 2016 10:08 AM
To: David L Carlson
Subject: Re: [R] Date Time in R
Hi David,
This gives the results accurately. The first line shows all the variable
Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Shivi Bhatia
Sent: Tuesday, July 26, 2016 7:42 AM
To: Marc Schwartz
Cc: R-help
Subject: Re: [R] Date Time in R
Thanks Marc for the help. thi
Thanks Marc for the help. this really helps.
I think there is some issue with the data saved in csv format for this
variable as when i checked:
str(eir$date)- this results in :-
Date[1:5327], format: NA NA NA NA NA.
Thanks again.
On Tue, Jul 26, 2016 at 5:58 PM, Marc Schwartz wrote:
> Hi,
>
> T
Hi,
That eir$date might be a factor is irrelevant. There is an as.Date() method for
factors, which does the factor to character coercion internally and then calls
as.Date.character() on the result.
Using the example data below:
eir <- data.frame(date = c("05-30-16", "05-30-16", "05-30-16",
Hello Again,
While i tried your solution as you suggested above it seems to be working.
Here is the output
temp<- dput(head(eir$date))
c("05-30-16", "05-30-16", "05-30-16", "05-30-16", "05-30-16", "05-30-16")
however it still shows class(eir$date) as character and hence i cannot find
weekdays from
Thanks Duncan for the quick response. I will check again as you suggested.
If that doesn't work i will share a reproducible example.
Thanks again
On Tue, Jul 26, 2016 at 4:43 PM, Duncan Murdoch
wrote:
> On 26/07/2016 7:05 AM, Shivi Bhatia wrote:
>
>> Hi Team,
>>
>> This scenario may have co
On 26/07/2016 7:05 AM, Shivi Bhatia wrote:
Hi Team,
This scenario may have come across a number of times however i checked
nabble & SO and couldn't find a solution hence request assistance.
I have a date variable in my data-set eir. The class of this var was
character while i had read the file
Hi Team,
This scenario may have come across a number of times however i checked
nabble & SO and couldn't find a solution hence request assistance.
I have a date variable in my data-set eir. The class of this var was
character while i had read the file in r studio. Example of date - 05-30-16
To c
You are still posting in HTML, and it is continuing to impede this
conversation. Learn how to post in plain text before posting again. Gmail
does have this option.
You are not using dput, as previously asked, either. Read the web page I
referenced to learn how to send R data unambiguously.
Y
Thank you very much Jeff. Below is the data I used:
> Corrected_data
SA_LST SA_GHI_mean
61759 3/11/2007 1:00 0.0
67517 3/11/2007 2:00 0.0
70017 3/11/2007 3:00 0.0
70524 3/11/2007 4:00 0.0
71061 3/11/2007 5:00 0.0
71638 3/11/2007 6:00 0.0
You have not provided a reproducible example, so anything I say could be
wrong just due to misinterpretation. Please read [1] for suggestions on
making your examples reproducible, particularly regarding the use of dput
to provide example data. You have also posted in HTML format,
which can caus
I would use the 'lubridate' package for this:
> z <- Sys.time()
> z
[1] "2014-12-07 15:43:50 EST"
> require(lubridate)
> with_tz(z, "UTC")
[1] "2014-12-07 20:43:50 UTC"
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want
Dear R users
I am puzzled by the following result from R script. I am trying to convert
local time to UTC time. Time zone is -5, therefore I used the following
approach.
Below is the script.
> Corrected_SA_data$date_time[k-1]
[1] "2007-03-11 01:00:00"
> Corrected_SA_data$TZ[k-1]
[1] -5
> Correcte
)
> date2<-"2013-01-26"
> wday(ymd(date2),label=TRUE)
> # 1 parsed with %Y-%m-%d
> #[1] Sat
>
>
> week(ymd(date2))
> # 1 parsed with %Y-%m-%d
> #[1] 4
> A.K.
>
> ----- Original Message -
> From: yash kajaria
> To: r-help@r-project.org
>
t.org
Cc:
Sent: Thursday, March 14, 2013 9:45 AM
Subject: [R] date & time manipulation- R 2.15.1 windows 7
Hi,
I wanted to learn how to solve a date and time manipulation where i can
do the following two
1. difference of two dates eg (differnce between 5th jan 2013 and 1st
jan 2013)
On Thu, Mar 14, 2013 at 9:45 AM, yash kajaria wrote:
> Hi,
> I wanted to learn how to solve a date and time manipulation where i can
> do the following two
> 1. difference of two dates eg (differnce between 5th jan 2013 and 1st
> jan 2013)
>
> 2.Suppose i have week number of the year,
-project.org
> Subject: [R] date & time manipulation- R 2.15.1 windows 7
>
> Hi,
> I wanted to learn how to solve a date and time manipulation where i
> can
> do the following two
> 1. difference of two dates eg (differnce between 5th jan 2013 and 1st
> jan 2013)
>
Hi,
I wanted to learn how to solve a date and time manipulation where i can
do the following two
1. difference of two dates eg (differnce between 5th jan 2013 and 1st
jan 2013)
2.Suppose i have week number of the year, i want to know if i can find
out the day it refers to eg( say week
Hi all,
i have a date-time series plot like this :
Datum-Ura;padavine;P-kum-15dni;BLPV-1N;T-BLPV-1N;BLPV-4N;T-BLPV-4N;BLPV-5N;T-BLPV-5N;pretok;N-min;N-max
1.1.2011
18:00;0.0;0.0;174.62;14.60;174.56;14.10;174.64;14.20;299.26;174.18;181.78
2.1.2011
0:00;0.0;0.0;174.60;14.60;174.60;14.10;174.64;14.2
On Sat, Jan 14, 2012 at 4:35 AM, claire5 wrote:
> I have been trying for some time to nicely plot some of my data, I have
> around 1800 values for some light intensity, taken every hour of the day
> over almost 2 months.
>
> My data file looks like:
>
> Date Time. GMT.02.00 Intensity
> 1
Well, I am not sure how to use the zoo package to be honest.
I am trying to plot all 1800 data in the same graph but of course it looks
super messy. And R does not really recognize the time data input.
so i just want to plot the time series kind of. The problem is that the x
value is date and time
What exactly is your problem? People quite like the "zoo" package for
handling and plotting time series: perhaps that will work for you?
Michael
On Sat, Jan 14, 2012 at 4:35 AM, claire5 wrote:
> Hey guys,
>
> I have been trying for some time to nicely plot some of my data, I have
> around 1800 v
Hey guys,
I have been trying for some time to nicely plot some of my data, I have
around 1800 values for some light intensity, taken every hour of the day
over almost 2 months.
My data file looks like:
DateTime. GMT.02.00 Intensity
106.10.11 11:00:00AM x
206.10.11
On 03.12.2010 16:31, Alexander Salim wrote:
Hi all,
I have a dataset called ,dataSet1'. The time column is given in a numeric
code beginning with the year and ending with the minutes. Frist I tried the
strptime() function to solve the problem. It gave me just the date back (and
not the date an
Hi all,
I have a dataset called ,dataSet1'. The time column is given in a numeric
code beginning with the year and ending with the minutes. Frist I tried the
strptime() function to solve the problem. It gave me just the date back (and
not the date and time). There is also the ISOdatetime function
Perfect.
Thanks
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Try this:
sapply(0::(length(DateTime3)-1), function(i)format(strptime(start,
"%m/%d/%Y %H:%M") + i * interval, "%Y-%m-%d %H:%M:%S"))
On Wed, Oct 13, 2010 at 10:16 AM, dpender wrote:
>
> Thanks Henrique,
>
> Do you have any idea why the first entry doesn't have the time as the start
> specifie
Thanks Henrique,
Do you have any idea why the first entry doesn't have the time as the start
specified is "1/1/1981 00:00"?
Is it something to do with being at midnight?
Doug
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Sent from the
Try this:
sapply(0:(length(DateTime3)-1), function(i)as.character(strptime(start,
"%m/%d/%Y %H:%M") + i * interval))
On Wed, Oct 13, 2010 at 8:51 AM, dpender wrote:
>
> I am trying to convert an array from numeric values back to date and time
> format. The code I have used is as follows;
>
>
I am trying to convert an array from numeric values back to date and time
format. The code I have used is as follows;
for (i in 0:(length(DateTime3)-1)) {
DateTime3[i] <- (strptime(start, "%m/%d/%Y %H:%M")+ i*interval)
where start <- [1] "1/1/1981 00:00"
However the created array
Hello,
Is it possible to instruct (permanently) R to write on csv (and read
from csv) time series, where the time stamp has a particular format:
say:
-mm-dd
i.e.,
as in
format(Sys.Date(), "%Y-%m-%d")
Many thanks in advance,
Costas
__
R-help
Another solution, as a fix to my original algorithm, was found by a colleague
(Matthew Roberts). While he claims not too much for its elegance, it does
seem to work. This fix is based on the use of the 'pmax' function. This
function is a variant of the 'max' (maximum) function to return a vecto
esp wrote:
>
> For the function as defined above using 'sapply'
>> spot[,1]
> 01/09/2009 01/09/2009 00:00:01 01/09/2009 00:00:02 01/09/2009
> 00:00:03
> 1251759600 1251759601 1251759602
> 1251759603
>
> This was unexpected - it seems to have display
See below.
On Mon, Oct 5, 2009 at 6:50 PM, Gabor Grothendieck
wrote:
> Try this. First we read a line at a time into L except for the
> header. Then we use strapply to match on the given pattern. It
> passes the backreferences (the portions within parentheses in the
> pattern) to the function
Thank you all who replied, I will try out these ideas later today.
David Esp
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Sent from the R help mailing list archive at Nabble.com.
___
Try this. First we read a line at a time into L except for the
header. Then we use strapply to match on the given pattern. It
passes the backreferences (the portions within parentheses in the
pattern) to the function (defined via a formula) whose implicit
arguments are x, y and z. That function
On Oct 5, 2009, at 5:14 PM, esp wrote:
Date-Time-Stamp input method to correctly interpret user-specific
formats:coding is 90% there - based on exmple at
http://tolstoy.newcastle.edu.au/R/help/05/02/12003.html
...anyone got the last 10% please?
CONTEXT:
Data is received where one of the col
Off the top of my head, I think you're working to hard at this.
I would read in the timestamp column as a character string. Then,
find those where the string length is too short [using nchar()],
append "00:00:00" to those [using paste()], and then convert to
POSIXt [using as.POSIXct()].
No
Date-Time-Stamp input method to correctly interpret user-specific
formats:coding is 90% there - based on exmple at
http://tolstoy.newcastle.edu.au/R/help/05/02/12003.html
...anyone got the last 10% please?
CONTEXT:
Data is received where one of the columns is a datetimestamp. At midnight,
th
On Sun, Sep 20, 2009 at 1:55 PM, Gabor Grothendieck
wrote:
> On Sun, Sep 20, 2009 at 4:52 PM, Mark Knecht wrote:
>> Thanks Gabor,
>> I did try to use dput but it wasn't cooperating and wanted to send
>> FAR too much data.
>
> dput(head(x, 10))
>
As I said, I tried almost exactly that. I didn't
On Sun, Sep 20, 2009 at 4:52 PM, Mark Knecht wrote:
> Thanks Gabor,
> I did try to use dput but it wasn't cooperating and wanted to send
> FAR too much data.
dput(head(x, 10))
>
> Your method works well for me but as I look at it I don't
> understand the use of double brackets - DF[[1]] - wh
Thanks Gabor,
I did try to use dput but it wasn't cooperating and wanted to send
FAR too much data.
Your method works well for me but as I look at it I don't
understand the use of double brackets - DF[[1]] - why do you do that?
Anyway, thanks for the fast reponses from you and Jim. Both
Note that your explanation refers to strptime but the code uses
strftime which accounts for the error.
Try this:
> Lines <- "ENTRY DATE
+ 3/23/2009 6:30:00 AM
+ 3/23/2009 6:30:00 AM
+ 3/23/2009 6:39:00 AM
+ 3/23/2009 6:39:00 AM
+ 3/23/2009 6:48:00 AM
+ 3/23/2009 6:48:00 AM
+ 3/23/2009 7:00:00 AM"
Here is one way to do it. Not sure why you want columns with either
date or time since you already have them. This will create a POSIXct
object you can use for processing and then two character columns with
date and time. Exactly what are you going to do with the data.
> str(x)
'data.frame':
Hi,
Can strptime (or some other function) help me turn the following
column of a data.frame into two new columns, one as date and the other
as time, preserving the AM/PM value?
Thanks,
Mark
> B
ENTRY DATE
1 3/23/2009 6:30:00 AM
2 3/23/2009 6:30:00 AM
3 3/23/2009 6:39:00 AM
4
chron uses:
- separate arguments for date and time -- not a single one as you have
- a separate system for the codes, not the % codes of Date and POSIXt classes
If you want to specify the date and time all as one string
rather than two arguments, use as.chron instead of
chron -- as.chron does use
what am I doing wrong?
chron(as.character(f), format=c(dates="%m/%d/%y", times="%h:%m"))
f <- structure(c(51L, 60L, 66L, 87L, 90L, 115L, 23L, 35L, 37L, 6L,
12L, 55L, 84L, 96L, 109L, 17L, 29L, 41L, 3L, 74L, 94L, 102L,
30L, 8L, 46L, 69L, 107L, 15L, 25L, 39L, 1L, 71L, 95L, 19L, 56L,
62L, 76L, 85L, 9
Gabor this is better than using
x = seq(ISOdate(2006,1,1, 0, 0, 0), by = "15 min", length.out=35040)
this works but your code below is exactly what I needed instead of almost.
thankyou very much
Stephen
On Thu, Jun 26, 2008 at 6:06 PM, Gabor Grothendieck <[EMAIL PROTECTED]>
wrote:
> Try this:
Try this:
library(chron)
t1 <- chron("1/1/2006", "00:00:00")
t2 <- chron("12/31/2006", "23:45:00")
deltat <- times("00:15:00")
tt <- seq(t1, t2, by = times("00:15:00"))
Note that if you have some data and your intention is simply to
create the index for it so you can create a zoo or zooreg object
I would like a sequence of dates with a time step of 15 minutes
starting:
1/1/2006 00:00:00 - 12/31/2006 23:45:00
function(x) {
chron(sub(" .*", "", x), gsub(".* (.*)", "\\1:00", x))
}
this is the piece of code I use to read in zoo objects
for any help I would be grateful I have tried sequenc
Hi,
I have to connect to MSSQL server to get some TimeSeries value (datetime,
float) on output.
Datetime data type in R is a POSIXct date with information about timezone,
daylight or solar time.
In the DB I have not this kind of information and this cause some problem at
the change of time for e
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