Using gsub to extract character string before white space in R

Question

I have a list of birthdays that look something like this:

dob <- c("9/9/43 12:00 AM/PM", "9/17/88 12:00 AM/PM", "11/21/48 12:00 AM/PM") 

I want to just grab the calendar date from this variable (ie drop everything after the first occurrence of white-space).

Here's what I have tried so far:

dob.abridged <- substring(dob,1,8) dob [1] "9/9/43 1" "9/17/88 " "11/21/48" dob.abridged <- gsub(" $","", dob.abridged, perl=T) > dob.abridged [1] "9/9/43 1" "9/17/88"  "11/21/48" 

So my code works for calendar dates of length 6 or 7, but not length 8. Any pointers on a more effective regex to use with gsub that can handle calendar dates of length 6, 7 or 8?

Thank you.

Answer 1

No need for substring, just use gsub:

gsub( " .*$", "", dob ) # [1] "9/9/43"   "9/17/88"  "11/21/48" 

A space (), then any character (.) any number of times (*) until the end of the string ($). See ?regex to learn regular expressions.

Answer 2

I often use strsplit for these sorts of problems but liked how simple Romain's answer was. I thought it would be interesting to compare Romain's solution to a strsplit answer:

Here's a strsplit solution:

sapply(strsplit(dob, "\\s+"), "[", 1) 

Using the microbenchmark package and dob <- rep(dob, 1000) with the original data:

Unit: milliseconds                                     expr       min        lq    median                    gsub(" .*$", "", dob)  4.228843  4.247969  4.258232  sapply(strsplit(dob, "\\\\s+"), "[", 1) 14.438241 14.558832 14.634638         uq       max neval   4.268029  5.081608  1000  14.756628 53.344984  1000 

The clear winner on a Win 7 machine is the gsub regex from Romain. Thanks for the answer and explanation Romain.