R subtracting 1 month from today's date gives NA

Question

I have a script in which I subset my data according to some set time periods and wanted to subset all the records that had occurred in the last month.

However if I try to subtract one month from today's date it yields an NA:

> today <- Sys.Date()
> today
[1] "2017-03-29"
> today - months(1)
[1] NA

I do have lubridate loaded but I think this calculation is being performed with base R. If I subtract 2 or more months it works fine:

> today - months(2)
[1] "2017-01-29"
> today - months(3)
[1] "2016-12-29"

Does anyone have any ideas about what might be going on?

UPDATE: I think this is something to do with simple date subtraction not handling leap year cases (2017 is not a leap year so "2017-02-29" does not exist).

Are there other packages / functions that take into account leap years? For the above example I would expect the answer to revert to the last day of the previous month, i.e.:

today - months(1)
# Should yield:
"2017-02-28"

Would it make sense for this calculation to give the same results for both today and yesterday (or what is the ISO convention for this)?

> sessionInfo()
R version 3.3.2 (2016-10-31)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 7 x64 (build 7601) Service Pack 1

locale:
[1] LC_COLLATE=English_United Kingdom.1252  LC_CTYPE=English_United Kingdom.1252   
[3] LC_MONETARY=English_United Kingdom.1252 LC_NUMERIC=C                           
[5] LC_TIME=English_United Kingdom.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
 [1] xlsx_0.5.7         xlsxjars_0.6.1     rJava_0.9-8        MRAtools_0.6.8     stringdist_0.9.4.4 stringr_1.2.0     
 [7] stringi_1.1.3      lubridate_1.6.0    data.table_1.10.4  PKI_0.1-3          base64enc_0.1-3    digest_0.6.12     
[13] getPass_0.1-1      RPostgreSQL_0.5-1  DBI_0.5-1         

loaded via a namespace (and not attached):
[1] magrittr_1.5   rstudioapi_0.6 tools_3.3.2    parallel_3.3.2

Pood

Answer 1

The calculation of months is indeed perfomed by base R but not the way your think. Months is used to get the month of a date object.

#Example
today <- Sys.Date()
months(today)
[1] "March"

To add or substract months, you should use %m+% from lubridate:

today <- Sys.Date()
today %m+% months(-1)
[1] "2017-02-28"

Kay

When using lubridate, the months function is overwritten and does not function like base R. – Bohemia

@Bohemia No, the lubridate function is month with no "s". Base R has months – Kay

Base R's months function won't take a number, but when I load lubridate, I can give months a number. Why is that? See also in the documentation: rdocumentation.org/packages/lubridate/versions/1.6.0/topics/… which seems to suggest it is overriding the base R function. – Bohemia

I think it's the %m+% function that accepts more inputs because lubridate::months(Sys.Date()) gives an error saying that months is not part of lubridate. – Kay

That's genuinely weird, because when lubridate is loaded, months(1) returns "1m 0d 0H 0M 0S". I'm not a big user of lubridate, so really just more for my own information than anything. :) – Bohemia

@Bohemia I agree. Not sure why months(1) works after lubridate is loaded. – Kay

Answer 2

One month ago is non-defined in this context. February 29th only exists in leap years.

See the lubridate documentation:

Note: Arithmetic with periods can results in undefined behavior when non-existent dates are involved (such as February 29th in non-leap years). Please see Period-class for more details and %m+% and add_with_rollback for alternative operations.

The lubridate package can handle what you are doing, but you need to perform the operaton using %m+%.

Bohemia

Have accepted @P Lapointe's worked answer but +1 for pointing me towards the specific help on this topic in lubridate. – Pood

Updated link - lubridate.tidyverse.org/reference/mplus.html – Sirenasirenic