On 31 Jan 02 at 05:17:17PM, Pradeep Sethi wrote:
> Hi All,
>
> I want to change date 9/9/1987 to 09/09/1973
I assume the change in year was an accident.
> was wondering, what is the most efficient way ?
Well, this won't be the "most" efficient way, but it should be
reasonably efficient, and will work with dates that include varying
degrees of pre-existing zero-padding.
#!/usr/bin/perl -w
foreach $date (<DATA>)
{
chomp $date;
print "$date: ";
$date =~ s%^(\d)(?=/)%0$1%;
$date =~ s%(?<=^\d\d/)(\d)(?=/)%0$1%;
print "$date\n";
}
__DATA__
1/1/1999
03/04/2020
09/9/1987
9/09/1987
09/09/1987
12/12/2000
12/6/2000
6/12/2000
output:
1/1/1999: 01/01/1999
03/04/2020: 03/04/2020
09/9/1987: 09/09/1987
9/09/1987: 09/09/1987
09/09/1987: 09/09/1987
12/12/2000: 12/12/2000
12/6/2000: 12/06/2000
6/12/2000: 06/12/2000
The code assumes a well-formed date, but would leave the majority of
random strings unaffected.
I'm too lazy to generate other solutions and benchmark them, but you
could gather the solutions you receive and benchmark them against each
other.
Of course, if you mean efficiency in the golf sense, you could do much
better than this.
Regards,
Ian
