Thanks to everyone for helping me overcome the problem faced .I have got
over with the problem .once again thanks for the immediate response of all
the members
--
View this message in context:
http://www.nabble.com/Datetime-conversion-tp25503138p25530176.html
Sent from the R help mailing list
On Sep 18, 2009, at 1:03 AM, premmad wrote:
I'm relatively new to R .I tried converting the datetime column with
values
like 01apr1985:00:00:00.000 using strptime(datetime,%d%b%Y).Could
anyone
help me in this regard .Please reply ASAP i need .
You will need to give us a more complete
Thanks for your reply
datetime
01OCT1987:00:00:00.000
12APR2004:00:00:00.000
01DEC1987:00:00:00.000
01OCT1975:00:00:00.000
01AUG1979:00:00:00.000
26JUN2003:00:00:00.000
01JAN1900:00:00:00.000
13MAY1998:00:00:00.000
30SEP1998:00:00:00.000
is in the file and i have imported it in to R and created
Seems to work alright for me.
datetime -c(
01OCT1987:00:00:00.000,
12APR2004:00:00:00.000,
01DEC1987:00:00:00.000,
01OCT1975:00:00:00.000,
01AUG1979:00:00:00.000,
26JUN2003:00:00:00.000,
01JAN1900:00:00:00.000,
13MAY1998:00:00:00.000,
30SEP1998:00:00:00.000)
date-strptime(datetime,%d%b%Y)
See if this is what you are looking for:
dt - as.data.frame(datetime)
date-strptime(as.character(dt$datetime),%d%b%Y)
date
[1] 1987-10-01 2004-04-12 1987-12-01 1975-10-01 1979-08-01
2003-06-26
[7] 1900-01-01 1998-05-13 1998-09-30
cheers,
-Girish
I'm not able to replicate your problem. Here's what I get. See if this is
what you want:
dt$date-strptime(as.character(dt$datetime),%d%b%Y)
dt
datetime date
1 01OCT1987:00:00:00.000 1987-10-01
2 12APR2004:00:00:00.000 2004-04-12
3 01DEC1987:00:00:00.000 1987-12-01
4
Girish it works for me also if its a vector.I have problem if the data is
stored as dataframe(rows and columns) please do help me in this
Girish A.R. wrote:
Seems to work alright for me.
datetime -c(
01OCT1987:00:00:00.000,
12APR2004:00:00:00.000,
01DEC1987:00:00:00.000,
It works but what i need is the result also as a column .
I tried using the following code .
dt$new-strptime(as.character(dt$datetime),%d%b%Y.
It shows the following error
Error in `$-.data.frame`(`*tmp*`, Sa_dt, value = list(sec = c(0, 0, :
replacement has 9 rows, data has 14.
Please help
The same what you have worked out is my need but i'm getting the following
error
Error in `$-.data.frame`(`*tmp*`, date, value = list(sec = c(0, 0, :
replacement has 9 rows, data has 14
Please help me in this
--
View this message in context:
The same what you have worked out is my need but i'm getting the following
error
Error in `$-.data.frame`(`*tmp*`, date, value = list(sec = c(0, 0, :
replacement has 9 rows, data has 14
Please give more detail about what you did. This error is certainly
not from the example used in
Can you post a reproducible code snippet, along with the output/error
messages, and the output of sessionInfo(). That way other folks on R-help
may be able to offer help.
Here's myl output of sessionInfo()
Your problem is that 'strptime' returns an object of POSIXlt type
which is 9 elements; what you what is: ( you need a POSIXct type)
dt$new-as.POSIXct(strptime(as.character(dt$datetime),%d%b%Y))
On Fri, Sep 18, 2009 at 6:26 AM, premmad mtechp...@gmail.com wrote:
It works but what i need is the
Thanks .I tried its working but when i tried to view the dataframe i got the
following error
Error in edit.data.frame(get(subx, envir = parent), title = subx, ...) :
can only handle vector and factor elements
--
View this message in context:
Sorry for confusing you all with my inexperienced posting .
I tried as u said if you have 9 rows in the data it is working fine but
please try out the same example as you have suggested earlier with morethan
9 rows.
I tried it as following
datetime -c(
+ 01OCT1987:00:00:00.000,
+
use View to view the dataframe.
On Fri, Sep 18, 2009 at 8:17 AM, premmad mtechp...@gmail.com wrote:
Thanks .I tried its working but when i tried to view the dataframe i got the
following error
Error in edit.data.frame(get(subx, envir = parent), title = subx, ...) :
can only handle vector
On Fri, Sep 18, 2009 at 04:32:27AM -0700, premmad wrote:
Sorry for confusing you all with my inexperienced posting .
I tried as u said if you have 9 rows in the data it is working fine but
please try out the same example as you have suggested earlier with morethan
9 rows.
I tried it as
I'm relatively new to R .I tried converting the datetime column with values
like 01apr1985:00:00:00.000 using strptime(datetime,%d%b%Y).Could anyone
help me in this regard .Please reply ASAP i need .
--
View this message in context:
17 matches
Mail list logo
