The Panda Language: No Loops, No Ifs, Just Fun
pandalang.orgFWIW, here's the modern Python equivalent to "1..10.odd.sqr.lt(50)"
>>> [s for i in range(50) if i&1 and (s:=i*i) < 50]
[1, 9, 25, 49]
#include <stdio.h>
int main() {
int i, s;
for(i = 1; i <= 10; i++) {
if (i & 1 && (s = i*i) < 50) {
printf("%d ", s);
}
}
return 0;
}
I wonder what the APL looks like. I hacked this solution:
(x<50)/x←(1=2|x)/x←((⍳10)*2)
rewritten apl version w/ lambdas and commute (vs assignment & parens):
About as short as I can make it, and while its longer than the pandas solution, the primitives are significantly more general (sqr, odd - why include these as language primitives?).{⍵/⍨⍵<50}{2*⍨⍵/⍨1=2|⍵}⍳10 1 9 25 49Thanks! It's been a long time since I tried APL. All I could remember was ⍳ and right-to-left evaluation. The rest was through monkeying around with Rosetta Code examples.
As to your "why" question - I'm assuming to make a neat demo. Something like:
seems more difficult to pull off in a pipeline API.>>> from math import cos, sin >>> [s for i in range(50) if i&1 and cos(s:=i*i) < sin(i)] [1, 9, 49, 225, 361, 441, 625, 1089, 1521, 1681, 2025, 2209]nah its pretty straightforward w/ higher order fns.
one of many possible solutions:
{(1○⍺)>2○⍵} {(⍹ ⍵)/⍨⍵ ⍶ ⍹ ⍵} {⍵*2} {⍵/⍨2|⍵}⍳50 1 9 49 225 361 441 625 1089 1521 1681 2025 2209Oh, sorry, I was too clever by half. I should have been more specific - hard to pull off in Panda.
Thanks for an APL solution in any case!
Reminds me of Ruby.
In case anyone's interested, here's a Ruby version:
(1..10).select(&:odd?).map{|x| x*x }.select{|x| x < 50 }