[R] Keep date, strip time?

2011-02-08 Thread Mark Knecht
I have hundreds of CSV files coming in from another program that have a text field representing the date time combined together. I need to strip the time and keep the date. How could I do that? In the example below, on the first line I need to keep the 6/15/2009, turning it into a date that R

Re: [R] Keep date, strip time?

2011-02-08 Thread Phil Spector
Mark - Here's a few possibilites: dts = c('6/10/2009 10:04:00 AM','6/15/2009 9:47:00 AM','6/15/2009 9:47:00 AM') as.Date(sapply(strsplit(dts,' '),'[',1),'%m/%d/%Y') [1] 2009-06-10 2009-06-15 2009-06-15 as.Date(sub('(\\d+/\\d+/\\d+) .*','\\1',dts),'%m/%d/%Y') [1] 2009-06-10 2009-06-15

Re: [R] Keep date, strip time?

2011-02-08 Thread Prof Brian Ripley
Try as.Date() with a suitable format (it only knows about internationally standard formats), e.g. maybe you mean as.Date(6/10/2009 10:04:00 AM, format=%m/%d/%Y) [1] 2009-06-10 On Tue, 8 Feb 2011, Mark Knecht wrote: I have hundreds of CSV files coming in from another program that have a

Re: [R] Keep date, strip time?

2011-02-08 Thread Mark Knecht
On Tue, Feb 8, 2011 at 11:30 AM, Prof Brian Ripley rip...@stats.ox.ac.uk wrote: Try as.Date() with a suitable format (it only knows about internationally standard formats), e.g. maybe you mean as.Date(6/10/2009 10:04:00 AM, format=%m/%d/%Y) [1] 2009-06-10 Thank you! - Mark

Re: [R] Keep date, strip time?

2011-02-08 Thread Mark Knecht
On Tue, Feb 8, 2011 at 11:29 AM, Phil Spector spec...@stat.berkeley.edu wrote: Mark -   Here's a few possibilites: dts = c('6/10/2009 10:04:00 AM','6/15/2009 9:47:00 AM','6/15/2009 9:47:00 AM') as.Date(sapply(strsplit(dts,' '),'[',1),'%m/%d/%Y') [1] 2009-06-10 2009-06-15 2009-06-15