http://qs1969.pair.com?node_id=338745

Ryszard has asked for the wisdom of the Perl Monks concerning the following question:

I realise this is more than an algorithmic question, but figured someone here has a higher level of math geekyness than i.

What i want to do is round an integer to the closest unit.

In this specific case, i'm running a cronjob every five mins, and putting the data into a database for later processing.

What i'm after is rounding the timestamp from the database to the nearest 5 mins. I've come up with something, which seems to work, however i'm wondering if this is the most efficient way to do it.

#!/usr/bin/perl -w

use strict;

my @time = localtime(time);

my $prec   = 5;
my $number = 52;

$\="\n";

my $div = $number%$prec;
$div = 1 if ($div == 0);

my $num = sprintf("%.1f", $number/$div );
my $intnum = int sprintf("%.1f", $number/$div );

if ( ($num-$intnum) >= .5) {
    print $prec-($number%$prec)+$number;
} else {
    print $number-($number%$prec);
}

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Replies are listed 'Best First'.
Re: OT: Integer rounding?
by Roy Johnson (Monsignor) on Mar 22, 2004 at 20:21 UTC
    You might get some help from this Q&A.

    Update:
    Doing it by hand, I'd do: 

    sub round_to_nearest {
    my ($n, $scale) = @_;
    int(($n + $scale/2)/$scale) * $scale;
}

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    The PerlMonk tr/// Advocate
[reply]
[d/l]
      A few sources:

      perlfaq4 

      sub round {
    my($number) = shift;
    return int($number + .5);
}

      [download]

      perldoc -f int

      int Returns the integer portion of EXPR. If EXPR is omitted, uses "$_". You should not use this function for rounding: one because it truncates towards "0", and two because machine representations of floating point numbers can sometimes produce counterintuitive results. For example, "int(-6.725/0.025)" produces -268 rather than the correct -269; that's because it's really more like -268.99999999999994315658 instead. Usually, the "sprintf", "printf", or the "POSIX::floor" and "POSIX::ceil" functions will serve you better than will int().
[reply]
[d/l]
Re: OT: Integer rounding?
by Ryszard (Priest) on Mar 23, 2004 at 19:13 UTC
    After a bit more googling i found the answer: 
    dataless <dataless767@hotmail.com> wrote in article
<a8liar$m91$1@helle.btinternet.com>...
> Hi all,
> Can anyone recommend an algorithm to round an integer to the nearest
+ 5 or
> 10?


Let's round n to the nearest five first:

Divide n by 5
Add 0.5 to n
Discard the fractional part of n
Multiply n by 5

Now let's round to the nearest ten:

Divide n by 10
Add 0.5 to n
Discard the fractional part of n
Multiply n by 10


Now let's round to the nearest seven:

Divide n by 7
Add 0.5 to n
Discard the fractional part of n
Multiply n by 7


Do you see a pattern?

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    [download]
    or in perl 
    
#!/usr/bin/perl -w

use strict;

my $prec   = 5;
my $number = 32;

$\="\n";

print int(($number/$prec)+.5)*$prec;

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    smaller, anyone?
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[d/l]
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