是否有更短的方法从字符串中提取日期？

Question

我编写了代码来从给定的字符串中提取日期.特定

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

它提取

  > "2012-07-29" 

问题是我的代码看起来很冗长而且很麻烦.我想知道这样做是否更优雅.

  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

您可以strptime()用来解析时间对象:

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> 

请注意,我移动了输入字符串,因为我不确定上午12:59存在...只是为了证明这一点,移动了三个小时(以秒为单位表示基本单位):

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> 

哦,如果你只是想要约会,那当然更简单:

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

Answer 2

有些事情应该有效:

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