4 ms·

Would you like to post a representative snippet in your favourite lang? I'll try and rewrite in perl and we can compare. (All langs have particular sweet spots,

by jbert 14y ago

Would you like to post a representative snippet in your favourite lang? I'll try and rewrite in perl and we can compare. (All langs have particular sweet spots, if you do an R or APL oneliner or something it's going to be much more verbose in perl, obviously) It's not that I think perl will be significantly (or at all) nicer, but I think the readability difference is often overstated and I'd like to test that thought.

afc 14y ago
Lets see, what about factorial in constant memory? Exponentiation (a to the power of b, where they are positive integers) in logarithmic time (not using built-in exponentiation functions/syntax)? Anonymously add a constant to a number and return it (in a way that can be passed to a function like 'map' or some such)? def Factorial(x): output = 1 for i in xrange(x): output *= (i + 1) return output def Factorial2(x): return reduce(operator.mul, xrange(1, x + 1), 1) I provided two versions, both with constant memory, to show how it'd look with explicit iteration and without it. With this, I get: >>> Factorial(200) 788657867364790503552363213932185062295135977687173263294742533244359449963403342920304284011984623904177212138919638830257642790242637105061926624952829931113462857270763317237396988943922445621451664240254033291864131227428294853277524242407573903240321257405579568660226031904170324062351700858796178922222789623703897374720000000000000000000000000000000000000000000000000L Exponentiation: def Power(a, b): if not b: return 1 if b % 2: return Power(a, b - 1) * a x = Power(a, b/2) return x * x Add a constant: lambda x: x + 7 Or, if you want to name it globally: def TweakValue(x): return x + 7 How do they look in Perl?
- judofyr 14y ago
  use bignum; use v5.10; =cut def Factorial(x): output = 1 for i in xrange(x): output *= (i + 1) return output =cut # Direct translation sub factorial { my $x = shift; my $output = 1; for my $i (1 .. $x) { $output *= $i; } return $output; } say factorial(200); =cut def Factorial2(x): return reduce(operator.mul, xrange(1, x + 1), 1) =cut use List::Util "reduce"; sub factorial2 { my $x = shift; reduce { $a * $b } (1 .. $x); } say factorial(200); =cut def Power(a, b): if not b: return 1 if b % 2: return Power(a, b - 1) * a x = Power(a, b/2) return x * x =cut sub power { my $a = shift; my $b = shift // return 1; if ($b % 2) { return power($a, $b - 1) * $a } my $x = power($a, $b/2); return $x * $x; } # lambda x: x + 7 my $plus7 = sub { $_[0] + 7 }; # Or: my $plus7 = sub { my $x = shift; $x + 7 }; say $plus7->(14); =cut def TweakValue(x): return x + 7 =cut sub plus7 { $_[0] + 7 } say plus7(14);
  - colomon 14y ago
    Just in case anyone was wondering, in Perl 6: sub factorial($n) { my $output = 1; for 1..$n -> $i { $output *= $i; } $output; } or sub factorial2($n) { [*] 1..$n; } Pretty anonymous function to add a constant: -> $x { $x + 7 } Really short version: * + 7 Named: sub plus7($x) { $x + 7; }
- jbert 14y ago
  Thanks for that. First one (not 100% sure it's constant memory, I think perl optimises the for to avoid instantiating the (1..$n) list: #!/usr/bin/perl use Modern::Perl; use bigint; say fact(5000); sub fact { my ($n) = @_; my $output = 1; for my $i (1..$n) { $output *= $i; } return $output; } (Took the value up to 5000 to get something which ran long enough to get a measurement, on my laptop it runs (including startup time) in ~1.2s. (Can we compare runtime too? I know we're discussing readability, but I'm interested). The reduce version (again, I'm unsure of const mem req). Comes in at ~1.4s: #!/usr/bin/perl use Modern::Perl; use List::Util qw(reduce); use bigint; say fact(5000); sub fact { my ($n) = @_; return reduce(sub { $a * $b; }, 1, (1..$n)); } The add a constant (true lambda, closing over lexicals in scope, etc etc) is 'sub': sub { my ($x) = @_; $x + 7; } (doesn't have the python lambda limitations). (Be aware that perl GC is refcounted though, so ref cycles are possible). Edit: I think the languages are comparable in terms of features and also readability. People may dislike leading sigils and that's fair, but I dislike python's "no need to declare your vars, just hope you don't typo an assignment" approach to lexicals.
  - zem 14y ago
    that also highlights one of my perl peeves - having to scrape function args out of @_ rather than declaring them as part of the function definition.
    - jbert 14y ago
      Yes, that sucks. It's not much extra typing: def function(x, y, z): versus sub function { my ($x, $y, $x) = @_; } The overhead is only really the 'my = @_;' chars, but it is annoying. On the other hand, it does allow you to do things like partially unpack args and pass the rest as a bundle to a super class.
      - pyre 14y ago
        You can partially unpack in Python too: def method(self, *args): a, b = args[:2] super(self).method(*args[:2]) Perl's lack of using the definition for the variables doesn't necessarily make it 'more powerful.'
- irahul 14y ago
  > def Factorial(x): output = 1 for i in xrange(x): output = (i + 1) return output Umm, did you mean: def factorial(x): output = 1 for i in xrange(2, x): output *= i return output Here are the perl versions. Such trivial examples are going to look the same in languages which share the same paradigms. use List::Util qw(reduce); # This is how I would write it. sub fact1 { my $num = shift; my $output = 1; $output *= $_ for (2 .. $num); return $output; } # Most people don't prefer inline loops and $_ variable, though # I don't see why not. sub fact2 { my $num = shift; my $output = 1; for my $i (2 .. $num) { $output *= $i; } return $output; } sub fact3 { my $num = shift; return reduce { $a * $b } 2..$num }
  - colomon 14y ago
    Based on the examples, I presume xrange(x) is every integer up to but not including x. ie 1..^$x in Perl 6, don't recall if that's the Perl 5 syntax or not.
    - irahul 14y ago
      > Based on the examples, I presume xrange(x) is every integer up to but not including x Yes. That makes my Python example incorrect(should be xrange(2, num + 1)) > 1..^$x in Perl 6, don't recall if that's the Perl 5 syntax or not. Perl 5 doesn have 1..^$x. I have always used 1..($n - 1) where I needed it - don't know if there is a better alternative.
      - afc 14y ago
        > Yes. That makes my Python example incorrect(should be xrange(2, num + 1)) Or just xrange(num) and then, as in my original code, output *= num + 1. :-)
        irahul 14y ago
        > Or just xrange(num) and then, as in my original code, output *= num + 1. :-) HN ate your asterisk(it uses asterisk to mark italics), or something wrong with my chrome app for HN. In your post, I saw: def Factorial(x): output = 1 for i in xrange(x): output = (i + 1) return output

by jbert 14y ago

afc 14y ago

Lets see, what about factorial in constant memory? Exponentiation (a to the power of b, where they are positive integers) in logarithmic time (not using built-in exponentiation functions/syntax)? Anonymously add a constant to a number and return it (in a way that can be passed to a function like 'map' or some such)? def Factorial(x): output = 1 for i in xrange(x): output *= (i + 1) return output def Factorial2(x): return reduce(operator.mul, xrange(1, x + 1), 1) I provided two versions, both with constant memory, to show how it'd look with explicit iteration and without it. With this, I get: >>> Factorial(200) 788657867364790503552363213932185062295135977687173263294742533244359449963403342920304284011984623904177212138919638830257642790242637105061926624952829931113462857270763317237396988943922445621451664240254033291864131227428294853277524242407573903240321257405579568660226031904170324062351700858796178922222789623703897374720000000000000000000000000000000000000000000000000L Exponentiation: def Power(a, b): if not b: return 1 if b % 2: return Power(a, b - 1) * a x = Power(a, b/2) return x * x Add a constant: lambda x: x + 7 Or, if you want to name it globally: def TweakValue(x): return x + 7 How do they look in Perl?

judofyr 14y ago
use bignum; use v5.10; =cut def Factorial(x): output = 1 for i in xrange(x): output *= (i + 1) return output =cut # Direct translation sub factorial { my $x = shift; my $output = 1; for my $i (1 .. $x) { $output *= $i; } return $output; } say factorial(200); =cut def Factorial2(x): return reduce(operator.mul, xrange(1, x + 1), 1) =cut use List::Util "reduce"; sub factorial2 { my $x = shift; reduce { $a * $b } (1 .. $x); } say factorial(200); =cut def Power(a, b): if not b: return 1 if b % 2: return Power(a, b - 1) * a x = Power(a, b/2) return x * x =cut sub power { my $a = shift; my $b = shift // return 1; if ($b % 2) { return power($a, $b - 1) * $a } my $x = power($a, $b/2); return $x * $x; } # lambda x: x + 7 my $plus7 = sub { $_[0] + 7 }; # Or: my $plus7 = sub { my $x = shift; $x + 7 }; say $plus7->(14); =cut def TweakValue(x): return x + 7 =cut sub plus7 { $_[0] + 7 } say plus7(14);
- colomon 14y ago
  Just in case anyone was wondering, in Perl 6: sub factorial($n) { my $output = 1; for 1..$n -> $i { $output *= $i; } $output; } or sub factorial2($n) { [*] 1..$n; } Pretty anonymous function to add a constant: -> $x { $x + 7 } Really short version: * + 7 Named: sub plus7($x) { $x + 7; }
jbert 14y ago
Thanks for that. First one (not 100% sure it's constant memory, I think perl optimises the for to avoid instantiating the (1..$n) list: #!/usr/bin/perl use Modern::Perl; use bigint; say fact(5000); sub fact { my ($n) = @_; my $output = 1; for my $i (1..$n) { $output *= $i; } return $output; } (Took the value up to 5000 to get something which ran long enough to get a measurement, on my laptop it runs (including startup time) in ~1.2s. (Can we compare runtime too? I know we're discussing readability, but I'm interested). The reduce version (again, I'm unsure of const mem req). Comes in at ~1.4s: #!/usr/bin/perl use Modern::Perl; use List::Util qw(reduce); use bigint; say fact(5000); sub fact { my ($n) = @_; return reduce(sub { $a * $b; }, 1, (1..$n)); } The add a constant (true lambda, closing over lexicals in scope, etc etc) is 'sub': sub { my ($x) = @_; $x + 7; } (doesn't have the python lambda limitations). (Be aware that perl GC is refcounted though, so ref cycles are possible). Edit: I think the languages are comparable in terms of features and also readability. People may dislike leading sigils and that's fair, but I dislike python's "no need to declare your vars, just hope you don't typo an assignment" approach to lexicals.
- zem 14y ago
  that also highlights one of my perl peeves - having to scrape function args out of @_ rather than declaring them as part of the function definition.
  - jbert 14y ago
    Yes, that sucks. It's not much extra typing: def function(x, y, z): versus sub function { my ($x, $y, $x) = @_; } The overhead is only really the 'my = @_;' chars, but it is annoying. On the other hand, it does allow you to do things like partially unpack args and pass the rest as a bundle to a super class.
    - pyre 14y ago
      You can partially unpack in Python too: def method(self, *args): a, b = args[:2] super(self).method(*args[:2]) Perl's lack of using the definition for the variables doesn't necessarily make it 'more powerful.'
irahul 14y ago
> def Factorial(x): output = 1 for i in xrange(x): output = (i + 1) return output Umm, did you mean: def factorial(x): output = 1 for i in xrange(2, x): output *= i return output Here are the perl versions. Such trivial examples are going to look the same in languages which share the same paradigms. use List::Util qw(reduce); # This is how I would write it. sub fact1 { my $num = shift; my $output = 1; $output *= $_ for (2 .. $num); return $output; } # Most people don't prefer inline loops and $_ variable, though # I don't see why not. sub fact2 { my $num = shift; my $output = 1; for my $i (2 .. $num) { $output *= $i; } return $output; } sub fact3 { my $num = shift; return reduce { $a * $b } 2..$num }
- colomon 14y ago
  Based on the examples, I presume xrange(x) is every integer up to but not including x. ie 1..^$x in Perl 6, don't recall if that's the Perl 5 syntax or not.
  - irahul 14y ago
    > Based on the examples, I presume xrange(x) is every integer up to but not including x Yes. That makes my Python example incorrect(should be xrange(2, num + 1)) > 1..^$x in Perl 6, don't recall if that's the Perl 5 syntax or not. Perl 5 doesn have 1..^$x. I have always used 1..($n - 1) where I needed it - don't know if there is a better alternative.
    - afc 14y ago
      > Yes. That makes my Python example incorrect(should be xrange(2, num + 1)) Or just xrange(num) and then, as in my original code, output *= num + 1. :-)
      - irahul 14y ago
        > Or just xrange(num) and then, as in my original code, output *= num + 1. :-) HN ate your asterisk(it uses asterisk to mark italics), or something wrong with my chrome app for HN. In your post, I saw: def Factorial(x): output = 1 for i in xrange(x): output = (i + 1) return output