Perl regex question

figaro · #1 (**permalink**) 07-16-2008

I have the following code:

Code:

#!/usr/bin/perl -w

@files = <*.csv>;
foreach $file (@files) {
  open(FH, $file) || die("Error: Cannot open file $file for reading.");
  my @dt = ($file =~ /^(\w+).(\d{6})\.csv$/);
  while (<FH>) { 
    print "@dt[0] $_\n";
  }
  close(FH);
}

There is redundancy in this code as it first checks for all files ending in ".csv" (line 3) and subsequently parses the filename (line 6) looking for characters and digits. How do I change line 3 into a regular expression, such that line 6 can be removed and the array @dt be determined there?

KevinADC · #2 (**permalink**) 07-16-2008

You can't. And there really is no redundancy as the glob <> first finds all files with .csv extention so you can open them, the regexp then parses those strings (the filenames) to extract more specific information.

KevinADC · #3 (**permalink**) 07-16-2008

Well, I did come up with this, but it may not be any more efficient than what you had and might even be less efficient, you would have to benchmark both codes to know which is really better.

Code:

my %files = map {/^(\w+).\d{6}\.csv$/; $_ => $1} <*.csv>;
print Dumper \%files;
foreach my $file (keys %files) {
  open(FH, $file) || die("Error: Cannot open file $file for reading.");
  while (<FH>) { 
    print "$files{$file} $_\n";
  }
  close(FH);
}

this regexp probably needs refining:

/^(\w+).\d{6}\.csv$/

what is the dot in there for after (\w+)?

figaro · #4 (**permalink**) 07-17-2008

Thank you for your reply and have been experimenting with this a little. Performance gain (or loss) is minor. Am still working on a built-in timer, but the differential is mere seconds (if any) on a total body of about 200 files and combined requiring 40MB.

And the dot (.) is the part of the file name: w+ being the standard file name and d{6} being the 24hr time of the time of download. So a file would have a name such as: scores.234506.csv

KevinADC · #5 (**permalink**) 07-17-2008

the dot should be escaped then, like the other dot in the regexp:

my %files = map {/^(\w+)\.\d{6}\.csv$/; $_ => $1} <*.csv>;

ghostdog74 · #6 (**permalink**) 07-17-2008

if the file names are always named like scores.234506.csv, you can just use split on dots and then get array element 1. That should be your number. Easier than regexp.

KevinADC · #7 (**permalink**) 07-17-2008

split() is a regexp.

More UNIX and Linux Forum Topics You Might Find Helpful
Thread	Thread Starter	Forum	Replies	Last Post
regex question	xiamin	Shell Programming and Scripting	3	03-05-2009 02:53 AM
Perl regex help - matching parentheses	cvp	Shell Programming and Scripting	7	02-27-2009 05:38 PM
how do i strip this line using perl regex.	ramky79	Shell Programming and Scripting	1	03-18-2008 12:10 PM
regex question	arushunter	Shell Programming and Scripting	8	01-04-2007 05:49 PM
regex question	rocketkids	UNIX for Dummies Questions & Answers	6	02-12-2004 05:49 AM

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search
Display Modes	Rate This Thread
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode	Rate This Thread: