I have the following data set, in a csv file, looks like the following:

Jan 27, 2010  16:01:24,000 125 - - -
Jan 27, 2010  16:06:24,000 125 - - -
......
The first few columns are month, day, year, time with OS3 accuracy. And the
last number is the measurement I need to extract.
I wonder if there is a easy way to just take out the measurements only from
a specific day and hour
-- 
Xin Zhang
Ph.D Candidate
Department of Statistics
University of California, Riverside
-------------------------------------------------------------------

I use strptime to configure the date format in my times series dataset.
First check to see how the dates are read.
For example:
# check the structure
str(your_file)
'data.frame' ...etc
This tells me that my original date is a factor but not in POSIXlt format.

#check your column dates
head(your_file)
[1] "1984-01-26" "1984-02-09" "1984-03-01" "1984-03-15" "1984-03-29"
"1984-04-12"
These are discrete column dates.

#convert your date format
your_file$date<- strptime(your_file$date,"%m/%d/%Y")
call ?strptime for options

Example:
For a specific day or hour, strptime would utilize:
strptime(your_file$date,"%d/%I") for day and hour.

Once you extract the type of date format you want, run str(your_file) again
to confirm the format change.
Does this answer your question?
Best,



-----
-------------------------------
Heather A. Wright, PhD candidate
Ecology and Evolution of Plankton
Stazione Zoologica Anton Dohrn
Villa Comunale
80121 - Napoli, Italy
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