Re: placing values into bins

The numbers you give in your example suggest that you may be able to make some simplifying assumptions. In particular, the (one!) example of a bin range has ranges for x and y which are exactly one whole number wide. If you can say that that is true for all bins, then it is possible to calculate directly which bin a datum should go in: simply apply int() to x and to y. This assumption also makes it reasonable to use arrays rather than hashes to store the bins. So:

my @data = (
  [ 123.7, 456.7 ],
  [ 564.7, 234.9 ],
);

for my $datum ( @data ) {
  my( $x, $y ) = @$_;
  $bins[ int $x ][ int $y ]++;
}

for my $x ( 0 .. $#bins ) {
  defined $bins[$x] or next;
  for my $y ( 0 .. $#{$bins[$x]} ) {
    defined $bins[$x][$y] or next;
    print "$x $y $bins[$x][$y]\n";
  }
}
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Re^2: placing values into bins by Anonymous Monk on Apr 28, 2005 at 12:59 UTC
Hi there. Thanks for your replies so far. Much appreciated. But what if I wanted to choose 125.5-130.0 and 456.5-457.0 as a range for example? How simple would that be to do?	[reply]
Re^3: placing values into bins by tlm (Prior) on Apr 28, 2005 at 13:18 UTC
If you follow the approach I sketched out in my other reply, all you need to do is pick the "left ends" of the ranges (in this case `125.5` and `456.5`), and the desired bin widths; perl takes care of the right ends of the ranges depending on the actual data. the lowliest monk	[reply]
Re^3: placing values into bins by mrborisguy (Hermit) on Apr 28, 2005 at 14:05 UTC
if you want to use the int method, but decide you want to use different sized bins, my simple solution would to be to create a new sub that returned the left end of the range, and just replace 'int' with that sub.	[reply]