Convert months mmm to numeric

Question

I have been given a csv with a column called month as a char variable with the first three letters of the month. E.g.:

\"Jan\", \"Feb\",\"Mar\",...\"Dec\"


        

Answer 1

Use match and the predefined vector month.abb:

tst <- c("Jan","Mar","Dec")
match(tst,month.abb)
[1]  1  3 12

Answer 2

A couple of options using:

vec <- c("Jan","Dec","Jan","Apr")

are

> Months <- 1:12
> names(Months) <- month.abb
> unname(Months[vec])
[1]  1 12  1  4

and/or

> match(vec, month.abb)
[1]  1 12  1  4

Answer 3

Just adding to the existing answers and the comment in the question:

readr::parse_date("20/DEZEMBRO/18","%d/%B/%y",locale=locale("pt"))

Results date format "2018-12-20". locale("pt") is for Portuguese, which is used in Brazil, can do "es" for Spanish, "fr" for French etc.

Answer 4

You can use the built-in vector month.abb to check against when converting to a number, eg :

mm <- c("Jan","Dec","jan","Mar","Apr")

sapply(mm,function(x) grep(paste("(?i)",x,sep=""),month.abb))
Jan Dec jan Mar Apr 
  1  12   1   3   4 

The grep construct takes care of differences in capitalization. If that's not needed,

match(mm,month.abb) 

works just as fine.

If you also have a day and a year column, you can use any of the conversion functions, using the appropriate codes (see also ?strftime)

eg

mm <- c("Jan","Dec","jan","Mar","Apr")
year <- c(1998,1998,1999,1999,1999)
day <- c(4,10,3,16,25)

dates <- paste(year,mm,day,sep="-")

strptime(dates,format="%Y-%b-%d")
[1] "1998-01-04" "1998-12-10" "1999-01-03" "1999-03-16" "1999-04-25"