Converting date with additional characters in format

Question

I have a string variable that I want to parse to class Date. In addition to the day, year and month, the format has other characters like separators (,

Answer 1

Try this:

as.Date(gsub("[u',()]","",my_data$date), format = '%d %Y %m')

Example with a single string:

d <- "(u'9', u'2005', u'06')"
d <- gsub("[u',()]","",d)
d.date <- as.Date(d, "%d %Y %m")

Result:

d.date
[1] "2005-06-09"

Answer 2

You don't need to strip the characters not used in the conversion specification. In the Details section of ?strptime we find that:

"[a]ny character in the format string not part of a conversion specification is interpreted literally"

That is, in the format argument of as.Date, you may include not only the conversion specification (introduced by %) but also the "other characters":

Furthermore, from ?as.Date:

Character strings are processed as far as necessary for the format specified: any trailing characters are ignored

Thus, this works:

as.Date("(u'9', u'2005', u'06')", format = "(u'%d', u'%Y', u'%m")
# [1] "2005-06-09"

Answer 3

If it is character class, you can try:

library(lubridate)

test <- c("u'9'", "u'2005'", "u'06'")

dym(paste(gsub("u|'", "", test), collapse = "/"))
[1] "2005-06-09 UTC"

Here I use lubridate to convert the string where I removed "u" and the ' character into time format. The collapse character I used in paste is arbitrary, lubridate can handle pretty much anything as a separator between date parts.