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

  > "2012-07-29" 

The problem is my code looks lengthy and cumbersome to read. I was wondering if was a more elegant way of doing this.

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

  #extract the string from raw date
  index = regexpr("[0-9]{4}-[0-9]{2}-[0-9]{2}", raw_date) #returns 'start' and 'end' to be used in substring

  start = index #start represents the character position 's'. start+1 represents '='
  end = attr(index, "match.length")+start-1
  date = substr(raw_date,start,end); date

Answer 1

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 2

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")