Shi, Tao shidaxia at yahoo.com writes:
Could someone explain to me why the following result is not a integer?
difftime(strptime(24NOV2004, format=%d%b%Y), strptime(13MAY2004,
format=%d%b%Y), units=days)
Time difference of 195.0417 days
Presumably because this goes across a
Hi all,
I am really new to dealing with Excel date and time formats to be used in R.
I have an Excel exported data matrix in TXT format.
One column, labeled DATE, is of the following format examples:
1/8/98 1:00
1/8/98 23:00
to mean that the observation was made on 1st of August 1998 at 1AM
If those are the only formats and the dates are in the current
timezone ...
dates - c(1/8/98 1:00, 1/8/98 23:00)
as.POSIXct(dates, format = %d/%m/%y %H:%M)
[1] 1998-08-01 01:00:00 1998-08-01 23:00:00
As for how to handle such a time series in R, it depends if it is a
regular or irregular
On Thu, Oct 21, 2010 at 1:31 PM, Amy Young-King
amy.young.k...@gmail.com wrote:
Hi all,
I am really new to dealing with Excel date and time formats to be used in R.
I have an Excel exported data matrix in TXT format.
One column, labeled DATE, is of the following format examples:
1/8/98
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
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 d.pen...@civil.gla.ac.uk wrote:
I am trying to convert an array from numeric values back to date and time
format. The code I have
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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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 d.pen...@civil.gla.ac.uk wrote:
Thanks Henrique,
Do you have any idea why the first entry doesn't have the time as the
Perfect.
Thanks
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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
__
On Sun, Sep 19, 2010 at 1:31 AM, Santosh Srinivas
santosh.srini...@gmail.com wrote:
I tried this and it works too (For most part) strangely for certain
dates (20090831) it is giving NA ...
FnO_Data$Date[m:l]
[1] 20090828 20090828 20090828 20090828 20090828 20090828 20090828 20090828
Strangely this is not working ... what am I doing wrong here?
tDate - FnO_Data$Date[1]
tDate
[1] 20090101
as.Date(c(tDate),format=%Y%m%d)
[1] NA
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PLEASE do read the
On Sep 18, 2010, at 11:25 AM, Santosh Srinivas wrote:
Strangely this is not working ... what am I doing wrong here?
tDate - FnO_Data$Date[1]
tDate
[1] 20090101
as.Date(c(tDate),format=%Y%m%d)
[1] NA
?sasDate
as.Date does not take numeric arguments. Try:
as.Date(as.character(tDate),
On Sep 18, 2010, at 10:25 AM, Santosh Srinivas wrote:
Strangely this is not working ... what am I doing wrong here?
tDate - FnO_Data$Date[1]
tDate
[1] 20090101
as.Date(c(tDate),format=%Y%m%d)
[1] NA
What version of R are you running?
What is the output of:
str(FnO_Data$Date)
and
On Sat, Sep 18, 2010 at 11:25 AM, Santosh Srinivas
santosh.srini...@gmail.com wrote:
Strangely this is not working ... what am I doing wrong here?
tDate - FnO_Data$Date[1]
tDate
[1] 20090101
as.Date(c(tDate),format=%Y%m%d)
[1] NA
Do you have zoo loaded? If you do then a minimal
Thanks.
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: 18 September 2010 21:07
To: Santosh Srinivas
Cc: r-help@r-project.org
Subject: Re: [R] Date issues
On Sep 18, 2010, at 11:25 AM, Santosh Srinivas wrote:
Strangely this is not working ... what am I
On Sep 18, 2010, at 11:36 AM, David Winsemius wrote:
On Sep 18, 2010, at 11:25 AM, Santosh Srinivas wrote:
Strangely this is not working ... what am I doing wrong here?
tDate - FnO_Data$Date[1]
tDate
[1] 20090101
as.Date(c(tDate),format=%Y%m%d)
[1] NA
?sasDate
as.Date does not take
-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: 18 September 2010 21:27
To: Santosh Srinivas
Cc: r-help@r-project.org
Subject: Re: [R] Date issues
On Sat, Sep 18, 2010 at 11:25 AM, Santosh Srinivas
santosh.srini...@gmail.com wrote:
Strangely this is not working ... what am I
Subject: Re: [R] Date issues
On Sat, Sep 18, 2010 at 11:25 AM, Santosh Srinivas
santosh.srini...@gmail.com wrote:
Strangely this is not working ... what am I doing wrong here?
tDate - FnO_Data$Date[1]
tDate
[1] 20090101
as.Date(c(tDate),format=%Y%m%d)
[1] NA
Do you have zoo loaded? If you do
Hello,
I have a dataframe with data such as:
dat$BEGINDATUM[3]
[1] 13-09-2007
dat$BEGINDATUM[4]
[1] 01-11-2007
class(dat$BEGINDATUM[3])
[1] factor
Now I need to make calculation with these dates.
But I get these result:
as.date(as.character(dat$BEGINDATUM[3]))
[1] NA
Hi,
I think you just need to add the format = argument. Does this help?
x - factor(01-11-2007)
as.character(x)
[1] 01-11-2007
as.Date(as.character(x), format = %d-%m-%Y)
[1] 2007-11-01
Cheers,
Josh
On Fri, Sep 3, 2010 at 2:11 PM, André de Boer rnie...@gmail.com wrote:
Hello,
I have a
Hello all,
I've 2 strings that representing the start and end values of a date and
time.
For example,
time1 - c(21/04/2005,23/05/2005,11/04/2005)
time2 - c(15/07/2009, 03/06/2008, 15/10/2005)
as.difftime(time1,time2)
Time differences in secs
[1] NA NA NA
attr(,tzone)
[1]
How can i calculate the
Hi Dunia,
You need to convert the character strings to Dates.
time1 - as.Date(c(21/04/2005,23/05/2005,11/04/2005), %d/%m/%Y)
time2 - as.Date(c(15/07/2009, 03/06/2008, 15/10/2005), %d/%m/%Y)
time2-time1
Best,
Ista
On Thu, Sep 2, 2010 at 10:32 AM, Dunia Scheid dunia.sch...@gmail.com wrote:
try to use difftime() instead of as.difftime().
On Thu, Sep 2, 2010 at 10:32 PM, Dunia Scheid dunia.sch...@gmail.com wrote:
Hello all,
I've 2 strings that representing the start and end values of a date and
time.
For example,
time1 - c(21/04/2005,23/05/2005,11/04/2005)
time2 -
On 08/21/2010 12:35 AM, ivo welch wrote:
...
(PS: Is there an easier way to tell R that I want a whole lot more
tick marks and/or labels than what it gives me by default?)
Hi Ivo,
I haven't been able to find a way to make the axis function print more
tick mark labels than it wants to. That's
This would make a nice feature for the next R release---a parameter
that overrides the default choice for the number of ticks or labels on
the axes. since it has to be calculated already, this should not be
hard---spoken by someone without knowledge of the innards, of course.
/iaw
On Sat, Aug
The treatment of dates seems to be a little inconsistent in R 2.11.1
(2010-05-31):
[1] The choice of origins?
as.integer(as.Date(1970-01-01))
works and assumes as origin 1970-01-01. However,
as.Date(1)
does not work. It requires an origin (as.Date(1,
origin=1970-01-01)). If we set a
On Fri, Aug 20, 2010 at 10:35 AM, ivo welch ivo.we...@gmail.com wrote:
The treatment of dates seems to be a little inconsistent in R 2.11.1
(2010-05-31):
[1] The choice of origins?
as.integer(as.Date(1970-01-01))
works and assumes as origin 1970-01-01. However,
as.Date(1)
The zoo
On Fri, 2010-08-20 at 10:35 -0400, ivo welch wrote:
The treatment of dates seems to be a little inconsistent in R 2.11.1
(2010-05-31):
snip /
[2] How do dates on axes work?
plot( c(as.Date(1:20, origin=1970-01-01)), 1:20 )
axis( side=3, c(as.Date(1:20, origin=1970-01-01)))
The
Hi all,
I am trying to convert all the dates (all days that are not Friday) in data
frame into dates to next Friday.
The following works but the result is returned as vector rather than the
original class.
It would be greatly apprecited if you could provide any solution to this
problem.
Many
On Wed, Aug 4, 2010 at 8:33 PM, Steven Kang stochastick...@gmail.com wrote:
Hi all,
I am trying to convert all the dates (all days that are not Friday) in data
frame into dates to next Friday.
The following works but the result is returned as vector rather than the
original class.
It
Hi,
Is there a function to get the last(or first) day of the week, given the
week number of the year?
For eg, week number for 7/20 is 29 as obtained by format(Sys.Date(),%U),
is there a function which returns 7/25 - the last day of week # 29
TIA,
Rao.
[[alternative HTML version
On Jul 20, 2010, at 6:37 PM, H Rao wrote:
Hi,
Is there a function to get the last(or first) day of the week, given
the
week number of the year?
For eg, week number for 7/20 is 29 as obtained by
format(Sys.Date(),%U),
is there a function which returns 7/25 - the last day of week # 29
On Tue, Jul 20, 2010 at 6:37 PM, H Rao hydsd...@gmail.com wrote:
Hi,
Is there a function to get the last(or first) day of the week, given the
week number of the year?
For eg, week number for 7/20 is 29 as obtained by format(Sys.Date(),%U),
is there a function which returns 7/25 - the last
jwiley.ps...@gmail.com
To: Felipe Carrillo mazatlanmex...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Sent: Thu, June 10, 2010 1:18:27 PM
Subject: Re: [R] Date conversion
Hello Felipe,
Is this what you want?
format(as.Date(3/10/10,
format=%m/%d/%y), %B %d, %Y)
Josh
On Thu, Jun 10, 2010 at 8:29
Hi:
Can't find a way to convert from shortDate to LongDate format. I got:
3/10/10 that I want to convert to March 10, 2010. I am using:
\documentclass[11pt]{article}
\usepackage{longtable,verbatim}
\usepackage{ctable}
\usepackage{datetime}
\title{my title}
\begin{document}
% Convert date
Hello Felipe,
Is this what you want?
format(as.Date(3/10/10, format=%m/%d/%y), %B %d, %Y)
Josh
On Thu, Jun 10, 2010 at 8:29 AM, Felipe Carrillo
mazatlanmex...@yahoo.com wrote:
Hi:
Can't find a way to convert from shortDate to LongDate format. I got:
3/10/10 that I want to convert to March
Thanks - that works
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I am parsing dates as follows:
z[1:10,1:3]
V1 V2 V3
10 03/02/09 22:20:51.274
2 100 03/02/09 22:28:18.801
3 200 03/02/09 22:33:33.762
4 300 03/02/09 22:40:21.826
5 400 03/02/09 22:41:38.361
6 500 03/02/09 22:42:50.882
7 600 03/02/09 22:45:19.885
8 700 03/02/09
Sorry
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PLEASE
Use %y indeed of %Y.
On Wed, Mar 17, 2010 at 12:00 PM, ManInMoon xmoon2...@googlemail.com wrote:
I am parsing dates as follows:
z[1:10,1:3]
V1 V2 V3
1 0 03/02/09 22:20:51.274
2 100 03/02/09 22:28:18.801
3 200 03/02/09 22:33:33.762
4 300 03/02/09 22:40:21.826
5
Hi All,
I have a character data.frame that contains character columns and date
columns. I've manage to convert some of my character columns to a date
format using as.Date(x, format=%m/%d/%y).
An example of one of my dates is
PROCHIDtDeath icdcucd date_admission1 date_admission_2
as.Date('17/02/2005','%d/%m/%Y')
[1] 2005-02-17
(Read the documentation more carefully to distinguish between %y and
%Y; I guess you tried lots of combinations but never tried the
correct one, so just be more careful at matching what your data is
with the format string you create.)
Good morning,
I am currently displaying a time series with the time on the Y axis, and the
values on the X axis.
The problem is that I wanted to reverse the Y axis (that is, to have the
latest date the closest to the X axis).
So I looked for a way to do this on this mailing list and I found a
On Feb 19, 2010, at 2:20 AM, Jeremie Smaga wrote:
Good morning,
I am currently displaying a time series with the time on the Y axis,
and the
values on the X axis.
The problem is that I wanted to reverse the Y axis (that is, to have
the
latest date the closest to the X axis).
So I looked
Dear R community
I would like to perform some statistical analysis on a data set containing
the following items: date, time, index of observation and various
covariates.
The date and time are originally extracted in the following format:
dd.mm. and hh:mm:ss respectively. R and more
Read the article in R News 4/1.
2010/1/27 Robert Kalicki robert.kali...@mph.unibe.ch:
Dear R community
I would like to perform some statistical analysis on a data set containing
the following items: date, time, index of observation and various
covariates.
The date and time are originally
Good day, i imported some data into R from Excel. By using the edit()
function, this is what one of the dates looks like in R:
x - structure(1254351600, class = c(POSIXt, POSIXct), tzone = )
[1] 2009-10-01 BST
However, when i do the following, the date changes:
as.Date(x, formate=%Y-%m-%d )
Hi
This is on WinXP with regional settings as EST (we are now on DST but I run
EST) R2.9.2
x - structure(1254351600, class = c(POSIXt, POSIXct), tzone = )
x
[1] 2009-10-01 09:00:00 EST
as.POSIXlt(x)
[1] 2009-10-01 09:00:00 EST
as.Date(x, formate=%Y-%m-%d )
[1] 2009-09-30
I had a similar
Thank you all who replied, I will try out these ideas later today.
David Esp
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See below.
On Mon, Oct 5, 2009 at 6:50 PM, Gabor Grothendieck
ggrothendi...@gmail.com 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
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 displayed the
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
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,
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
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
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
I have trouble with this:
as.Date(Sep-1981, format=%b-%Y)
Returns NA
From documentation for strftime
'%b' Abbreviated month name in the current locale. (Also matches
full name on input.)
'%Y' Year with century.
What am I doing wrong?
cheers
Worik
[[alternative HTML
Worik,
You need a day!
as in:
as.Date(1-Sep-1981, format=%d-%b-%Y) ## first of the month
HTH,
Jim Porzak
Ancestry.com
San Francisco, CA
www.linkedin.com/in/jimporzak
use R! Group SF: www.meetup.com/R-Users/
On Thu, Sep 24, 2009 at 4:15 PM, Worik R wor...@gmail.com wrote:
I have trouble
Try this:
library(zoo)
as.yearmon(Sep-1981, %b-%Y)
[1] Sep 1981
as.Date(as.yearmon(Sep-1981, %b-%Y))
[1] 1981-09-01
as.Date(paste(1, Sep-1981), %d %b-%Y)
[1] 1981-09-01
On Thu, Sep 24, 2009 at 7:15 PM, Worik R wor...@gmail.com wrote:
I have trouble with this:
as.Date(Sep-1981,
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
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':
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
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
On Sun, Sep 20, 2009 at 4:52 PM, Mark Knecht markkne...@gmail.com 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
On Sun, Sep 20, 2009 at 1:55 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Sun, Sep 20, 2009 at 4:52 PM, Mark Knecht markkne...@gmail.com 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
As suggested in the article R News 4/1, I used
as.Date(as.character(Phenology_VE$Date), %Y-%m-%d), however this function
returns me only NA values
as.Date(as.character(Phenology_VE$Date), %Y-%m-%d)
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[26] NA NA NA NA
A reproducible example would help. What is Phenology_VE$Date? This works
as.Date(2009-09-01, %Y-%m-%d)
[1] 2009-09-01
Is this the date you wanted:
as.Date(39936, origin='1900-2-1')
[1] 2009-06-05
On Wed, Sep 2, 2009 at 2:09 AM, swertiev_coudr...@voila.fr wrote:
As suggested in the
Hello, I plot the abundance of a species in relation to the date. To have the
date as a continous variable I put it in the format standard in excel
(f.ex. 39939 means 06.05.2009). R uses 39939 on the x axis, but I would like
to have 06.05. I tried to use as.Date as suggested in some discussion
To: r-help@r-project.org
Subject: [R] Date format in plot
Hello, I plot the abundance of a species in relation to the date. To have the
date as a continous variable I put it in the format standard in excel
(f.ex. 39939 means 06.05.2009). R uses 39939 on the x axis, but I would like
to have 06.05. I
On Sep 1, 2009, at 1:58 PM, swertie wrote:
Hello, I plot the abundance of a species in relation to the date. To
have the
date as a continous variable I put it in the format standard in
excel
(f.ex. 39939 means 06.05.2009). R uses 39939 on the x axis, but I
would like
to have 06.05. I
Subject: [R] Date format in plot
Hello, I plot the abundance of a species in relation to the date. To
have the
date as a continous variable I put it in the format standard in
excel
(f.ex. 39939 means 06.05.2009). R uses 39939 on the x axis, but I
would like
to have 06.05. I tried to use as.Date
See R News 4/1. The article on dates there discusses how they
work and discusses Excel's dates as well.
On Tue, Sep 1, 2009 at 1:58 PM, swertiev_coudr...@voila.fr wrote:
Hello, I plot the abundance of a species in relation to the date. To have the
date as a continous variable I put it in the
Hi all,
I'm having a little bit of trouble with some date conversions and
am hoping someone can help me out. Thanks in advance.
OK, I have two sources of data that provide date info in a csv file
differently. I've attached a small zipped file with two text files
that illustrate both. (Is it
On Sat, Jul 11, 2009 at 10:20 AM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
No attachment appeared. I don't think the list allows zip files
as attachments. Try plain text.
On Sat, Jul 11, 2009 at 1:10 PM, Mark Knechtmarkkne...@gmail.com wrote:
Hi all,
I'm having a little bit of
No attachment appeared. I don't think the list allows zip files
as attachments. Try plain text.
On Sat, Jul 11, 2009 at 1:10 PM, Mark Knechtmarkkne...@gmail.com wrote:
Hi all,
I'm having a little bit of trouble with some date conversions and
am hoping someone can help me out. Thanks in
You want %Y, not %y.
You might also want to look at the zoo package:
library(zoo)
z - read.zoo(Date1.txt, header = TRUE, sep = ,, format = %m/%d/%Y)
or using chron:
library(zoo)
library(chron)
z - read.zoo(Date1.txt, header = TRUE, sep = ,, FUN = as.chron)
There are three vignettes that come
On Sat, Jul 11, 2009 at 12:05 PM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
You want %Y, not %y.
You might also want to look at the zoo package:
library(zoo)
z - read.zoo(Date1.txt, header = TRUE, sep = ,, format = %m/%d/%Y)
or using chron:
library(zoo)
library(chron)
z -
Try:
format(d, %a) or format(d, %A) or as.POSIXlt(d)$wday
There is also day.of.week in chron.
On Sat, Jul 11, 2009 at 3:42 PM, Mark Knecht markkne...@gmail.com wrote:
On Sat, Jul 11, 2009 at 12:05 PM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
You want %Y, not %y.
You might also
Thanks. Will do.
Cheers,
Mark
On Sat, Jul 11, 2009 at 12:49 PM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
Try:
format(d, %a) or format(d, %A) or as.POSIXlt(d)$wday
There is also day.of.week in chron.
On Sat, Jul 11, 2009 at 3:42 PM, Mark Knecht markkne...@gmail.com wrote:
On Sat,
Hi,
Is there a function that will convert this sort of date code which
looks like years from 1900 + month_number + day_number with no
spaces? As an example Jan. 3rd 2008 would be written as 1080103.
Thanks,
Mark
__
R-help@r-project.org mailing list
Try help(strptime). Example
strptime(1080103L + 19e6L, %Y%m%d)
# [1] 2008-01-03
(This assumes your input is an integer but you can just drop the L if
you want)
On 04/07/09 19:37, Mark Knecht wrote:
Hi,
Is there a function that will convert this sort of date code which
looks like years
On Sat, Jul 4, 2009 at 12:10 PM, Allan Engelhardtall...@cybaea.com wrote:
Try help(strptime). Example
strptime(1080103L + 19e6L, %Y%m%d)
# [1] 2008-01-03
(This assumes your input is an integer but you can just drop the L if you
want)
On 04/07/09 19:37, Mark Knecht wrote:
Hi,
Is
On Sat, Jul 4, 2009 at 12:24 PM, Mark Knechtmarkkne...@gmail.com wrote:
On Sat, Jul 4, 2009 at 12:10 PM, Allan Engelhardtall...@cybaea.com wrote:
Try help(strptime). Example
strptime(1080103L + 19e6L, %Y%m%d)
# [1] 2008-01-03
(This assumes your input is an integer but you can just drop the
Hi, All,
How to make a data frame, each row of data frame store the different length of
vector?
Thanks.
Tammy
_
Show them the way! Add maps and directions to your party invites.
It is called a list.
On Wed, Mar 11, 2009 at 6:32 AM, Tammy Ma metal_lical...@live.com wrote:
Hi, All,
How to make a data frame, each row of data frame store the different length
of vector?
Thanks.
Tammy
_
Show them the
.
Tammy
Date: Wed, 11 Mar 2009 07:50:37 -0400
Subject: Re: [R] Date frame
From: jholt...@gmail.com
To: metal_lical...@live.com
CC: r-help@r-project.org
It is called a list.
On Wed, Mar 11, 2009 at 6:32 AM, Tammy Ma metal_lical...@live.com wrote:
Hi, All,
How to make a data frame
to create a data frame with each row contains different
sequence and then put to 3rd column, which seems
couldn't be done.
thanks.
Tammy
Date: Wed, 11 Mar 2009 07:50:37 -0400
Subject: Re: [R] Date frame
From: jholt...@gmail.com
To: metal_lical...@live.com
CC: r-help@r-project.org
Hi R users,
I have a factor variable called date as shown below: Can anyone share the
best / most efficient way to extract year and week (e.g. year = 2006, week
= 52 for first record, etc..)? My data set has 1 million records.
DATE
11DEC2006
11SEP2006
01APR2007
02DEC2007
Thanks in
Pele wrote:
Hi R users,
I have a factor variable called date as shown below: Can anyone share the
best / most efficient way to extract year and week (e.g. year = 2006, week
= 52 for first record, etc..)? My data set has 1 million records.
DATE
11DEC2006
11SEP2006
01APR2007
Hi,
There are possibly several ways to do this. My approach would be:
dates - strptime(as.character(DATE), %d%b%Y)
year - dates$year + 1900
week - floor(dates$yday/365 * 52)
HTH,
--sundar
On Thu, Mar 5, 2009 at 8:58 AM, Pele drdi...@yahoo.com wrote:
Hi R users,
I have a factor variable
Hi Uwe,
You are correct - that was a type O (52) and thanks for you your suggestion
that works..
Pele wrote:
Hi R users,
I have a factor variable called date as shown below: Can anyone share the
best / most efficient way to extract year and week (e.g. year = 2006,
week = 52 for
Hi there,
I am completely new to R and would like to do two things with date
functions:
1. Compute any date from a specified starting point, e.g. x - 2 months
2. How do I determine the weekday of any given date?
Thanks in advance
--
View this message in context:
# 1
d - Sys.Date()
seq(d, len = 2, by = 2 months)[2]
# 2
as.numeric(format(d, %w)) # 0 = Sunday
# or
format(d, %a)
See R News 4/1 for more info and the table at the end of it in particular.
On Fri, Jan 30, 2009 at 9:54 AM, ehxpieterse
eduard.piete...@macquarie.com wrote:
Hi there,
I am
Thanks a lot
Envoyé de mon iPhone
Le 3 déc. 08 à 00:39, jim holtman [EMAIL PROTECTED] a écrit :
If you want to do the addition, 'unclass' the variable:
alpha2+4
[1] 2008-12-25
alpha2 + unclass(alpha1)
[1] 2009-02-15
On Tue, Dec 2, 2008 at 4:10 PM, Christophe Dutang
[EMAIL
What is odd is that it seems to run ok if we call +.Date directly:
+.Date(alpha1, alpha2)
[1] 2009-02-15
On Tue, Dec 2, 2008 at 4:10 PM, Christophe Dutang [EMAIL PROTECTED] wrote:
Hi all,
I'm dealing with dates in R (2.7.2), but some basic operations raise a
warning.
Incompatible methods
On Wed, 2008-12-03 at 13:18 -0500, Gabor Grothendieck wrote:
What is odd is that it seems to run ok if we call +.Date directly:
+.Date(alpha1, alpha2)
[1] 2009-02-15
It also works if you flip the ordering:
alpha2 + alpha1
[1] 2009-02-15
Warning message:
Incompatible methods (+.Date,
On Wed, Dec 3, 2008 at 1:57 PM, Gavin Simpson [EMAIL PROTECTED] wrote:
On Wed, 2008-12-03 at 13:18 -0500, Gabor Grothendieck wrote:
What is odd is that it seems to run ok if we call +.Date directly:
+.Date(alpha1, alpha2)
[1] 2009-02-15
It also works if you flip the ordering:
alpha2 +
On Wed, 2008-12-03 at 14:13 -0500, Gabor Grothendieck wrote:
snip /
Why it works is not odd if you look at the help for ?`+.Date`, which
shows that for this method (correct term?) we need 'date' + 'x', where
'date' is an object of class Date and 'x' is numeric.
In fact what it says is
Hi all,
I'm dealing with dates in R (2.7.2), but some basic operations raise a
warning.
Incompatible methods (+.Date, Ops.difftime) for +
I saw this topic in this mailing list, but I do not understand what to
do...
cf. https://stat.ethz.ch/pipermail/r-help/2008-June/165842.html
Do I have
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