Re^3: build a distribution

OK, then the way I'd approach the task would be something like this:

my %bins;
open my $INF, '<', $FileName or die $!;
while (<$INF>) {
    chomp;
    $bins{get_bin($_)}++;
}
printf "%-6.6s %u items\n", $_, $bins{$_} for sort keys %bins;

sub get_bin {
   # Determine the name of the bin to put the value into
   my $val = shift;
   my $bin_min = int($val / 10);
   my $bin_max = $bin_min + 10;
   return "$bin_min-$bin_max";
}
[download]

You'll want to wrap in some error checking, testing, as well as any options you want...

...roboticus

Comment on Re^3: build a distribution Download Code

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Re^4: build a distribution by Grig (Novice) on Aug 07, 2010 at 18:29 UTC
Your script is very nice. Unfortunately, I am not quite familiar with all those specific perl cryptography yet, but if the following data are used `1 1 2 3 3 3 6 10 13 20 22 22 22 22 23 34 34 34 35 36 37 37 40 41 42 43` [download] your script gives the following output: `0-10 7 items 1-11 2 items 2-12 6 items 3-13 7 items 4-14 4 items` [download] How it can be changed to get the number of items for the intervals listed: `0-10 10-20 20-30 30-40` [download] ? Thank you for your help.	[reply] [d/l] [select]
Re^5: build a distribution by roboticus (Chancellor) on Aug 07, 2010 at 20:38 UTC
Grig: Ah, I made a mistake in the subroutine. It was: `sub get_bin { # Determine the name of the bin to put the value into my $val = shift; my $bin_min = int($val / 10); my $bin_max = $bin_min + 10; return "$bin_min-$bin_max"; }` [download] but it should have been (untested): `sub get_bin { # Determine the name of the bin to put the value into my $val = shift; my $bin_min = 10int($val / 10); my $bin_max = $bin_min + 10; return "$bin_min-$bin_max"; }` [download] I forgot the "10" when computing $bin_min, so instead of 0, 10, 20, etc., it was using 0, 1, 2, ... ...roboticus	[reply] [d/l] [select]