[R] Keep date, strip time?

[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" "2009-06-15"
> as.Date(sub('\\d+:\\d+:\\d+ [AP]M','',dts),'%m/%d/%Y')
[1] "2009-06-10" "2009-06-15" "2009-06-15"
> as.Date(as.POSIXct(dts,format='%m/%d/%Y %H:%M:%S %p'))
[1] "2009-06-10" "2009-06-15" "2009-06-15"

 					- Phil Spector
 					 Statistical Computing Facility
 					 Department of Statistics
 					 UC Berkeley
 					 spector at stat.berkeley.edu



On Tue, 8 Feb 2011, Mark Knecht wrote:

> 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 recognizes, but I need to throw away the
> time portion completely.
>
>
>> read.csv("C:\\D1\\F1-V1.csv", header=FALSE)[,c(1,7)]
>                       V1    V7
> 1   6/10/2009 10:04:00 AM    91
> 2    6/15/2009 9:47:00 AM  -279
> 3    6/15/2009 9:47:00 AM   861
> 4    6/22/2009 9:47:00 AM   771
> 5    6/22/2009 4:01:00 PM  -179
> 6    6/24/2009 2:53:00 PM    61
> 7     7/2/2009 9:47:00 AM   491
> 8     7/6/2009 9:47:00 AM    81
> 9   7/13/2009 10:04:00 AM  1681
>
> Thanks,
> Mark
>
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