Is there a shorter way to extract a date from a string?

Question

I wrote code to extract the date from a given string. Given

  > \"Date: 2012-07-29, 12:59AM PDT\"

it extracts

  > \"         


        

Answer 1

Regex with backreferencing works:

> sub("^.+([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]).+$","\\1","Date: 2012-07-29, 12:59AM PDT")
[1] "2012-07-29"

But @Dirk is right that parsing it as a date is the right way to go.

Answer 2

As (pretty much) always, you've got multiple options here. Though none of them really frees you from getting used to some basic regular expression syntax (or its close friends).

raw_date <- "Date: 2012-07-29, 12:59AM PDT"

Alternative 1

> gsub(",", "", unlist(strsplit(raw_date, split=" "))[2])
[1] "2012-07-29"

Alternative 2

> temp <- gsub(".*: (?=\\d?)", "", raw_date, perl=TRUE)
> out <- gsub("(?<=\\d),.*", "", temp, perl=TRUE)
> out
[1] "2012-07-29"

Alternative 3

> require("stringr")
> str_extract(raw_date, "\\d{4}-\\d{2}-\\d{2}")
[1] "2012-07-29"

Answer 3

Something along the lines of this should work:

x <- "Date: 2012-07-29, 12:59AM PDT"
as.Date(substr(x, 7, 16), format="%Y-%m-%d")

Answer 4

You can use strptime() to parse time objects:

R> strptime("Date: 2012-07-29, 11:59AM PDT", "Date: %Y-%m-%d, %I:%M%p", tz="PDT")
[1] "2012-07-29 11:59:00 PDT"
R> 

Note that I shifted your input string as I am unsure that 12:59AM exists... Just to prove the point, shifted by three hours (expressed in seconds, the base units):

R> strptime("Date: 2012-07-29, 11:59AM PDT", 
+>          "Date: %Y-%m-%d, %I:%M%p", tz="PDT") + 60*60*3
[1] "2012-07-29 14:59:00 PDT"
R> 

Oh, and if you just want the date, it is of course even simpler:

R> as.Date(strptime("Date: 2012-07-29, 11:59AM PDT", "Date: %Y-%m-%d"))
[1] "2012-07-29"
R>