How to subtract/add days from/to a date?

r date r-faq
Dan picture Dan · Feb 12, 2010 · Viewed 131.9k times · Source

I'm trying to build folders to store data pulls. I want to label the folders with the day of that data in the pull.

Ex. I pull 5 days ago data from mysql i want to name the folder the date from 5 days ago.

MySQL can easily handle date arithmetic. I'm not sure exactly how R does it. Should i just subtract the appropriate number of seconds in POSIXct and then convert to POSIXlt to name the folder MM_DD_YYYY?

Or is there a better way?

Answer

Shane picture Shane · Feb 12, 2010

Just subtract a number:

> as.Date("2009-10-01")
[1] "2009-10-01"
> as.Date("2009-10-01")-5
[1] "2009-09-26"

Since the Date class only has days, you can just do basic arithmetic on it.

If you want to use POSIXlt for some reason, then you can use it's slots:

> a <- as.POSIXlt("2009-10-04")
> names(unclass(as.POSIXlt("2009-10-04")))
[1] "sec"   "min"   "hour"  "mday"  "mon"   "year"  "wday"  "yday"  "isdst"
> a$mday <- a$mday - 6
> a
[1] "2009-09-28 EDT"

Related questions

Find the day of a week

Let's say that I have a date in R and it's formatted as follows. date 2012-02-01 2012-02-01 2012-02-02 Is there any way in R to add another column with the day of the week associated with the …

Read More
Convert character to class Date

I have a data frame with a character column of dates. When I use as.Date, most of my dates are parsed correctly, except for a few instances. The example below will hopefully show you what is going on. # my …

Read More
Changing date format in R

I have some very simple data in R that needs to have its date format changed: date midpoint 1 31/08/2011 0.8378 2 31/07/2011 0.8457 3 30/06/2011 0.8147 4 31/05/2011 0.7970 5 30/04/2011 0.7877 6 31/03/2011 0.7411 7 28/02/2011 0.7624 8 31/01/2011 0.7665 9 31/12/2010 0.7500 10 30/11/2010 0.7734 11 31/10/2010 0.7511 12 30/09/2010 0.7263 13 31/08/2010 0.7158 14 31/07/2010 0.7110 15 30/06/2010 0.6921 16 31/05/2010 0.7005 17 30/04/2010 0.7113 18 31/03/2010 0.7027 19 28/02/2010 0.6973 20 31/01/2010 0.7260 21 31/12/2009 0.7154 22 30/11/2009 0.7287 23 31/10/2009 0.7375 Rather than %d/%m/%Y, I would like it in the standard R format of %Y-%m-%d How can I make …

Read More