43

I use lubridate and figured that this would be so easy

ymd("2010-01-31")+months(0:23)

But look what one gets. It is all messed up!

 [1] "2010-01-31 UTC" "2010-03-03 UTC" "2010-03-31 UTC" "2010-05-01 UTC" "2010-05-31 UTC" "2010-07-01 UTC" "2010-07-31 UTC" "2010-08-31 UTC" "2010-10-01 UTC"
[10] "2010-10-31 UTC" "2010-12-01 UTC" "2010-12-31 UTC" "2011-01-31 UTC" "2011-03-03 UTC" "2011-03-31 UTC" "2011-05-01 UTC" "2011-05-31 UTC" "2011-07-01 UTC"
[19] "2011-07-31 UTC" "2011-08-31 UTC" "2011-10-01 UTC" "2011-10-31 UTC" "2011-12-01 UTC" "2011-12-31 UTC"

Then I read how lubridate caters to phenomenon such as interval, duration and period. So, OK I realize that a month is actually the number of days defined by (365*4+1)/48 = 30.438 days. So I tried to get smart and rewrite it as 

ymd("2010-01-31")+ as.period(months(0:23))

But that just gave an error. 

Error in as.period.default(months(0:23)) : 
  (list) object cannot be coerced to type 'double'
zx8754
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Farrel
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2 Answers2

95

Yes, you found the correct trick: going back a day from the first of the next month.

Here is as a one-liner in base R:

R> seq(as.Date("2010-02-01"), length=24, by="1 month") - 1
 [1] "2010-01-31" "2010-02-28" "2010-03-31" "2010-04-30" "2010-05-31"
 [6] "2010-06-30" "2010-07-31" "2010-08-31" "2010-09-30" "2010-10-31"
[11] "2010-11-30" "2010-12-31" "2011-01-31" "2011-02-28" "2011-03-31"
[16] "2011-04-30" "2011-05-31" "2011-06-30" "2011-07-31" "2011-08-31"
[21] "2011-09-30" "2011-10-31" "2011-11-30" "2011-12-31"
R>

So no need for lubridate which (while being a fine package) isn't needed for simple task like this. Plus, its overloading of existing base functions still strikes me as somewhat dangerous...

Dirk Eddelbuettel
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20

It is amazing how typing out a question focuses creative energy. I think I worked out the answer. I may as well post it here for the next poor soul who finds themselves wasting time.

ymd("2010-02-01")+ months(0:23)-days(1)

Simply specify the first day of the next month and generate a sequence from that but subtract 1 days from it to get the last day of the preceding month.

[1] "2010-01-31 UTC" "2010-02-28 UTC" "2010-03-31 UTC" "2010-04-30 UTC" "2010-05-31 UTC" "2010-06-30 UTC" "2010-07-31 UTC" "2010-08-31 UTC" "2010-09-30 UTC"
[10] "2010-10-31 UTC" "2010-11-30 UTC" "2010-12-31 UTC" "2011-01-31 UTC" "2011-02-28 UTC" "2011-03-31 UTC" "2011-04-30 UTC" "2011-05-31 UTC" "2011-06-30 UTC"
[19] "2011-07-31 UTC" "2011-08-31 UTC" "2011-09-30 UTC" "2011-10-31 UTC" "2011-11-30 UTC" "2011-12-31 UTC"

By the way, how do I get rid of the pesky "UTC" designations. Time zones are a life saver when they are needed. The rest of the time they are a nuisance.

Farrel
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    use `strftime(date)` to get rid of the timezone. so `strftime('2010-10-31 UTC')` would give you `2010-10-31`. – Ramnath Nov 30 '11 at 22:35
  • @SachaEpskamp: Stackoverflow will only allow me to accept my answer after two days. I guess that is quite clever. There may be another workaround that is more elegant. – Farrel Nov 30 '11 at 23:22
  • @Ramnath strftime gives me the date of the day before because where I live midnight UTC is actually 19:00 of the day before. I am in the eastern time zone. strftime(ymd("2010-02-01")+ months(0:23)-days(1)) [1] "2010-01-30 19:00:00" "2010-02-27 19:00:00" "2010-03-30 20:00:00" etc. – Farrel Nov 30 '11 at 23:30