Re: Re: Efficiency Question

I'm sorry I forgot to address that. The dirs in %dirs will never exactly match the dirs listed in %config_dirs. The config file will only contain high level directories like /home /etc /usr etc. But the %dirs is going to contain things like /usr/local/bin etc. so an exact match won;t be possible.

Comment on Re: Re: Efficiency Question

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Re^3: Efficiency Question by tadman (Prior) on Mar 21, 2002 at 20:33 UTC
If you're brave, you can construct a regex which you can bang your various directories against. This can be done quite simply like so: `my $skip = join ('\|', map {quotemeta} keys %$config_dirs);` This will look something like "/usr/bin\|/home\|/var" or what have you. Now you can just go and check against this, like dragonchild suggested, but with a slight mod: `foreach (keys %$dirs) { next if /^($skip)/; # Do stuff }` [download] Presumably if the value of $skip does not change within your program, you can use the /o option to only "compile once" your regular expression.	[reply] [d/l] [select]
Re: Re: Re: Efficiency Question by CukiMnstr (Deacon) on Mar 21, 2002 at 21:00 UTC
But then if you have /home in the config file, you should not print /home/foo? (I did not understand that too well) If you are going to have only high level directories (and by that I mean subdirectories in /), and the directories you have to decide wether or not to print are subdirectories of these high level ones, then I guess you could strip everything after the second '/' and then make a direct comparaison... `sub print_report { foreach (keys %$dirs) { print BENCHMARK $_; m#^/[^/]+/#; next if defined $config_dirs->{$&}; print "$_\n"; } }` [download] (I hope I understood the problem correctly)	[reply] [d/l]


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