If I want to count from zero in steps I have used.

print $_,'' for map{$_*5} 0 .. 5
0 5 10 15 20 25
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You could also do

foreach ( ('a'..'z')[map{$_*5}0..5] ) { 
    print $_,''; 
}
a f k p u z
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or for an array

@array = (0..50)
foreach ( @array[map{$_*5} 0..@array/5] ) { 
    print $_,''; 
}
0 5 10 15 20 25 30 35 40 45 50
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Or for a hash

@hash{'a'..'z'} = (1..26);

foreach ( (keys %hash)[map{$_*5}0 .. (scalar keys %hash)/5] ) { 
print "hash{$_} = $hash{$_}\n"; 
}
hash{a} = 1
hash{f} = 6
hash{k} = 11
hash{p} = 16
hash{u} = 21
hash{z} = 26
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my @list=(a,b,c,...z); // or whatever is in your list

for (my $i=0;$i<$#list;$i+=5) {
  my $val = $list[$i];

  ...
}
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sub step {
    my $step = shift;
    return grep {($_ - $_[0]) % $step == 0} @_
}

foreach (step 5, 2 .. 100) {
  print "$_ "
}
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2 7 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
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Hi,

But the fastest solution is using a closure:

sub step {
   my $step=shift;
   my $count=0;
   my @array=@_;
   return sub {
      $return=$array[$count];
      $count+=$step;
      return $return
      };
   };
 
for ( my $s=step(5,2..100); my $res=&$s ; 1 ) {
   print "$res ";
   }
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My benchmark shows 6 times faster than the map, and 10 times faster than the step method.

The reason is that I transfer an array only once, and than do queries in the array, whilst with map or step, one iterates over more than one array (map/grep and foreach).
---------------------------
Dr. Mark Ceulemans
Senior Consultant
IT Masters, Belgium