(jeffa) 3Re: Eliminating Trailing Zeros

Use a sexeger (a reverse regex):

$n = reverse scalar sprintf("%2.4f",$n);
$n =~ s/^0+//g;
$n = reverse scalar $n;
$n =~ s/\.$//g;
[download]

now if could just condense those two regexes into one ... but sometimes, brute force is better than cleverness.

UPDATE: aha!

$n = reverse scalar sprintf("%2.4f",$n);
$n =~ s/^0+\.?//g;
$n = reverse scalar $n;
[download]

that was easier than i thought ... unless i missed something [And i did ... glad i retested, because i had forgotten to espace the dot .. bad jeffa. Also, the scalar keywords are not necessary].

And jmcnamara++ ... that's the best solution posted (IMHO) since you already are using sprintf.

jeffa

L-LL-L--L-LL-L--L-LL-L--
-R--R-RR-R--R-RR-R--R-RR
B--B--B--B--B--B--B--B--
H---H---H---H---H---H---
(the triplet paradiddle with high-hat)

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Re: (jeffa) 3Re: Eliminating Trailing Zeros by pemungkah (Priest) on Jul 17, 2003 at 20:06 UTC
An even more fun version: `$n = reverse scalar sprintf("%2.4f",$n); $n =~ s/\G0//g; $n = reverse scalar $n;` [download] (Hats off to merlyn's version of this in Effective Perl Programming!)	[reply] [d/l]
Re: Re: (jeffa) 3Re: Eliminating Trailing Zeros by dragonchild (Archbishop) on Jul 17, 2003 at 20:14 UTC
This brings up a very good point - oftentimes it's better, both coding- and performance-wise, to use reverse when handling strings than it is to try and work from the end. For example, when trying to handle group separators locale-independently for formatting numbers, the easiest way is to: `my $g = ","; # Could be ' ' or '.' in other locales my $n = 1234567890; $n = reverse $n; $n =~ s/(\d{3})/\1$g/g; $n = reverse $n;` [download] Much better than the options without reverse. ------ We are the carpenters and bricklayers of the Information Age. Don't go borrowing trouble. For programmers, this means Worry only about what you need to implement. Please remember that I'm crufty and crochety. All opinions are purely mine and all code is untested, unless otherwise specified.	[reply] [d/l]