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
spec...@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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and provide commented, minimal, self-contained, reproducible code.
