Re: How to list files in dir with respect to time ?

To crush it down to a one-liner, using glob seemed ~~fastest~~ the most apt:

perl -e 'print "$_\n" for ( sort { -M $a <=> -M $b } glob "/path/\*" )
+;'
[download]

Update: by fastest I was thinking about coding effort rather than meaning to suggest a specific focus on performance.

One world, one people

Comment on Re: How to list files in dir with respect to time ? Download Code

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Re^2: How to list files in dir with respect to time ? by Roger (Parson) on Aug 16, 2005 at 15:17 UTC
You could use the Orcish Maneuver to make your code go faster... `# Original code for ( sort { -M $a <=> -M $b } glob ... ) # Orcish Maneuver my %t; for ( sort { ($t{$a} \|\|= -M $a) <=> ($t{$b} \|\|= -M $b) } glob ... )` [download]	[reply] [d/l]
Re^3: How to list files in dir with respect to time ? by anonymized user 468275 (Curate) on Aug 17, 2005 at 08:36 UTC
I've looked into this now - I was expecting sort to be smart enough to do that manouvre itself. For those who don't understand what it's doing, it is forcing the sort routine to perform the -M operation only once per file, but at the overhead of a hash lookup per sort iteration (there are geometrically upon n sort iterations per n files). sort + block could be optimised for minimal calculation of each side of the <=> operator (or cmp as appropriate) -- after all it ought to know it's going to be doing that irrespective of what's in the block, so it's a safe optimisation for sort to be doing already. But the orcish manouevre appears to be necessary because no such optimisation has been implemented. Perhaps Sort::Key has the manoeuvre already, however. One world, one people	[reply]
Re^4: How to list files in dir with respect to time ? by salva (Canon) on Aug 17, 2005 at 08:44 UTC
Perhaps Sort::Key has the manoeuvre already, however well, it's not exactly the Orcish manouevre, but Sort::Key calculates every key only once.	[reply]


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