Question

I have some data that looks a bit like this:

require(zoo)

X <- rbind(c(date='20111001', fmt='%Y%m%d'),
            c('20111031', '%Y%m%d'),
            c('201110', '%Y%m'),
            c('102011', '%m%Y'),
            c('31/10/2011', '%d/%m/%Y'),
            c('20111000', '%Y%m%d'))
print(X)

#      date       fmt     
# [1,] "20111001" "%Y%m%d"
# [2,] "20111031" "%Y%m%d"
# [3,] "201110"   "%Y%m"  
# [4,] "102011"   "%m%Y"  
# [5,] "31/10/2011" "%d/%m/%Y"
# [6,] "20111000" "%Y%m%d"

I only want the year and month. I don’t need the day, so I’m not worried that the final day is invalid. R, unfortunately, is:

mapply(as.yearmon, X[, 'date'], X[, 'fmt'], SIMPLIFY=FALSE)

# $`20111001`
# [1] "Oct 2011"

# $`20111031`
# [1] "Oct 2011"

# $`201110`
# [1] "Oct 2011"

# $`102011`
# [1] "Oct 2011"

# $`31/10/2011`
# [1] "Oct 2011"

# $`20111000`
# Error in charToDate(x) : 
#   character string is not in a standard unambiguous format

I know that the usual answer is to fix the day part of the date, e.g. using paste(x, '01', sep=''). I don’t think that will work here, because I don’t know in advance what the date format will be, and therefore I cannot set the day without converting to some sort of date object first.

Answer 1

Assuming that I’m always given a date (and never a time), and that any illegal ‘day’ is less than 61, I can guarantee a legal date as follows, by treating the supplied day as ‘seconds’ and replacing the supplied day with the 1st.

require(stringr)

safe_date <- str_c('01', X[, 'date'])
safe_fmt <- str_c('%d', str_replace(X[, 'fmt'], '%d', '%S'))

mapply(as.yearmon, safe_date, safe_fmt, SIMPLIFY=FALSE)

# $`0120111001`
# [1] "Oct 2011"

# $`0120111031`
# [1] "Oct 2011"

# $`01201110`
# [1] "Oct 2011"

# $`01102011`
# [1] "Oct 2011"

# $`0131/10/2011`
# [1] "Oct 2011"

# $`0120111000`
# [1] "Oct 2011"