Re: How can I access a cross directory set of data files by most recently modified date?

You can use an array of all files (see File::Find or File::Find::Rule) and sort them by modification time:

@files = sort {(stat $b)[9] <=> (stat $a)[9])} @files;

Comment on Re: How can I access a cross directory set of data files by most recently modified date? Download Code

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Re^2: How can I access a cross directory set of data files by most recently modified date? by SkipHuffman (Monk) on Jun 07, 2007 at 16:31 UTC
I thought of that, but it seems kind of brute force for something someone probably has solved more elegantly.	[reply]
Re^3: How can I access a cross directory set of data files by most recently modified date? by Fletch (Bishop) on Jun 07, 2007 at 17:02 UTC
You can use a schwartzian transform to make it more efficient by only doing the stat call once per file, but other than that there's not that much more to it. Update: For those playing along at home that'd be `@files = map { $_->[0] } sort { $a->[1] <=> $b->[1] } map { [ $_, (stat $_)[9] ] } @files;`	[reply] [d/l]
Re^3: How can I access a cross directory set of data files by most recently modified date? by moritz (Cardinal) on Jun 07, 2007 at 17:00 UTC
Well, using a binary heap as a data structure might be more efficient, especially if you know you know that you need only the the knewest `$k` files, then all insertion and deletion operations will be in O(log `$k`) operations. But if you are talking about a more elegant interface: I don't know any :( Perl 6 in German	[reply] [d/l] [select]
Re^4: How can I access a cross directory set of data files by most recently modified date? by SkipHuffman (Monk) on Jun 07, 2007 at 17:05 UTC
Thanks, that may be the way I need to go, but I would rather not reinvent the square wheel if I don't have to. Skip	[reply]