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I'm trying to subtract 2 character vectors containing date time information in the following format:
> dput(train2)
structure(list(time2 = c("2011-09-01 23:44:52.533", "2011-09-05 12:25:37.42",
"2011-08-24 12:56:58.91", "2011-10-25 07:18:14.722", "2011-10-25 07:19:51.697"
), time3 = c("2011-09-01 23:43:59.752", "2011-09-05 12:25:01.187",
"2011-08-24 12:55:13.012", "2011-10-25 07:16:51.759", "2011-10-25 07:16:51.759"
)), .Names = c("time2", "time3"), row.names = c(NA, 5L), class = "data.frame")
I've hunted around and played with zoo, as.Date, as.POSIXct, etc. to try and find the correct code to subtract 2 datetime objects and get an answer in seconds but without luck.
I'd appreciate any suggestions.
804
You can use the difftime() function to calculate the time difference between two dates or datetimes in R. where: time1, time2: The two dates or datetimes. units: The units to use for time difference (default is “days”, but other options include “secs”, “mins”, “hours”, and “weeks”)
Total seconds between times: To get the total seconds between two times, you multiply the time difference by 86400, which is the number of seconds in one day (24 hours * 60 minutes * 60 seconds = 86400).
The difftime() method in R is used to compute the difference in the timestamps given. It is used to return an object of the class difftime itself accompanied by units attribute.
To format = , provide a character string (in quotes) that represents the current date format using the special “strptime” abbreviations below. For example, if your character dates are currently in the format “DD/MM/YYYY”, like “24/04/1968”, then you would use format = "%d/%m/%Y" to convert the values into dates.
> x1<-"2013-03-03 23:26:46.315"
> x2<-"2013-03-03 23:31:53.091"
> x1 <- strptime(x1, "%Y-%m-%d %H:%M:%OS")
> x2 <- strptime(x2, "%Y-%m-%d %H:%M:%OS")
> x1
[1] "2013-03-03 23:26:46"
> x2
[1] "2013-03-03 23:31:53"
I followed the answer of @Dirk Eddelbuettel, but I am losing precision. How can I force R to not be cuting parts of second?
Thankfully (man of strptime) I answered my question myself:
op <- options(digits.secs = 3)
After applying this setting the precision will be used.
http://stat.ethz.ch/R-manual/R-devel/library/base/html/strptime.html
The belowe may be useful if you would like to get difference in seconds, but get in minutes:
> as.numeric(x2-x1,units="secs")
[1] 306.776
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