Convert character month name to date time object - r

Question

I must be missing something simple.
I have a data.frame of various date formats and I'm using lubridate which works great with everything except month names by themselves. I can't get the month names to convert to date time objects.
> head(dates)
From To
1 June August
2 January December
3 05/01/2013 10/30/2013
4 July November
5 06/17/2013 10/14/2013
6 05/04/2013 11/23/2013
Trying to change June into date time object:
> as_date(dates[1,1])
Error in charToDate(x) :
character string is not in a standard unambiguous format
> as_date("June")
Error in charToDate(x) :
character string is not in a standard unambiguous format
The actual year and day do not matter. I only need the month. zx8754 suggested using dummy day and year.

Answer 1

lubridate can handle converting the name or abbreviation of a month to its number when it's paired with the rest of the information needed to make a proper date, i.e. a day and year. For example:
lubridate::mdy("August/01/2013", "08/01/2013", "Aug/01/2013")
#> [1] "2013-08-01" "2013-08-01" "2013-08-01"
You can utilize that to write a function that appends "/01/2013" to any month names (I threw in abbreviations as well to be safe). Then apply that to all your date columns (dplyr::mutate_all is just one way to do that).
name_to_date <- function(x) {
lubridate::mdy(ifelse(x %in% c(month.name, month.abb), paste0(x, "/01/2013"), x))
}
dplyr::mutate_all(dates, name_to_date)
#> From To
#> 1 2013-06-01 2013-08-01
#> 2 2013-01-01 2013-12-01
#> 3 2013-05-01 2013-10-30
#> 4 2013-07-01 2013-11-01
#> 5 2013-06-17 2013-10-14
#> 6 2013-05-04 2013-11-23

Answer 2

The following is a crude example of how you could achieve that.
Given that dummy values are fine:
match(dates[1, 1], month.abb)
The above would return you, given that we had Dec in dates[1. 1]:
12
To generate the returned value above along with dummy number in a date format, I tried:
tmp = paste(match(dates[1, 1], month.abb), "2013", sep="/")
which gives us:
12/2013
and then lastly:
result = paste("01", tmp, sep="/")
which returns:
01/12/2013
I am sure there are more flexible approaches than this; but this is just an idea, which I just tried.

Answer 3

Using a custom function:
# dummy data
df1 <- read.table(text = "
From To
1 June August
2 January December
3 05/01/2013 10/30/2013
4 July November
5 06/17/2013 10/14/2013
6 05/04/2013 11/23/2013", header = TRUE, as.is = TRUE)
# custom function
myFun <- function(x, dummyDay = "01", dummyYear = "2013"){
require(lubridate)
x <- ifelse(substr(x, 1, 3) %in% month.abb,
paste(match(substr(x, 1, 3), month.abb),
dummyDay,
dummyYear, sep = "/"), x)
#return date
mdy(x)
}
res <- data.frame(lapply(df1, myFun))
res
# From To
# 1 2013-06-01 2013-08-01
# 2 2013-01-01 2013-12-01
# 3 2013-05-01 2013-10-30
# 4 2013-07-01 2013-11-01
# 5 2013-06-17 2013-10-14
# 6 2013-05-04 2013-11-23