从R中的字符串中提取带有点子字符串的模式

Question

I have a set of characters like

data <- c("ABS Spring Meeting 5.14.15", "DEFG Sellors Tour 10.28.14", "DDCC Fun at the Museum 4.4.15", "GAME CS vs. Washington 11.01.14", "BSS Studio 54 5.13.15","Pas-12 3.5.15")

As you can notice, the last set of digit is date of event. I want to convert them into date

date <- c("2015-05-14","2014-10-28","2015-04-04","2014-11-01","2015-05-13","2015-03-05")

Feel like I have to substring this kind ("5.14.15", "10.28.14", "4.4.15", "11.01.14", "5.13.15", "3.5.15") of pattern, then do the date convert.

Can anyone help me with this? Thank you!

Answer 1

In base R, and provided the date is always at the end of the string, you can use

as.Date(sub(".*\\s", "", data), "%m.%d.%y")
# [1] "2015-05-14" "2014-10-28" "2015-04-04" "2014-11-01"

Here, the regular expression is simply

• .* everything
• \\\\s a space character

So this removes everything up to and including the final space character.

Answer 2

The quickest way is with lubridate . If you supply the general format, it will try to figure it out for you:

library(lubridate)
mdy(data)
[1] "2015-05-14 UTC" "2014-10-28 UTC" "2015-04-04 UTC" "2014-11-01 UTC"

If your data becomes more complicated with other numbers you can use a string extraction method. Like so:

mdy(sub(".*?([0-9.]+)$","\\1", data))

In the pattern ".*?([0-9.]+)$" :

• .*? matches all characters and spaces. The question mark allows the next part of pattern be matched fully.

• ([0-9.]+)$ searches for the longest stretch of digits and decimal points reaching the end of the string marked by the dollar sign. The parantheses create a group of the tokens inside of it. We will use that group for the next step.

• "\\\\1" returns the capture group from the pattern and discards the rest of the match.

There are many websites that will go much further into regular expressions than I can. Since it is used in nearly every programming language, it will be well worth your time to invest at least a few hours in its study.

I learned a lot with this free Perl book online. Check out Ch. 5 here:

https://www.perl.org/books/beginning-perl/

This site has a sub-section focusing on R

http://www.regular-expressions.info/rlanguage.html

Answer 3

data <- data <- c("ABS Spring Meeting 5.14.15", 
   "DEFG Sellors Tour 10.28.14", "DDCC Fun at the Museum 4.4.15", 
    "GAME CS vs. Washington 11.01.14", "BSS Studio 54 5.13.15",
    "Pas-12 3.5.15")
library("lubridate")
library("stringr")

mdy(str_extract(data,"[0-9]+(\\.[0-9]+){2}$"))
## [1] "2015-05-14 UTC" "2014-10-28 UTC" "2015-04-04 UTC" "2014-11-01 UTC"
## [5] "2015-05-13 UTC" "2015-03-05 UTC"

The regular expression "[0-9]+(\\\\.[0-9]+){2}$" means "more than one numeral ( [0-9]+ ), followed by two ( {2} ) instances of (one dot ( \\\\. ) followed by more than one numeral [0-9]+ ), followed by the end of the string ( $ )"