5 ms·
The Panda Language: No Loops, No Ifs, Just Fun
- Gravityloss 6y agoReminds me of Ruby.
- boygobbo 6y agoIn case anyone's interested, here's a Ruby version: (1..10).select(&:odd?).map{|x| x*x }.select{|x| x < 50 }
- eesmith 6y agoFWIW, 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] And here's my interpretation of the C equivalent: #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; } Not quite a-la 1972, but then again the example 1972 code from that page wouldn't compile then either - variables had to be declared at the start of the function. I wonder what the APL looks like. I hacked this solution: (x<50)/x←(1=2|x)/x←((⍳10)*2) but I'm guessing the real solution would be 1/3 the size.
- gandercrews 6y agorewritten apl version w/ lambdas and commute (vs assignment & parens): {⍵/⍨⍵<50}{2*⍨⍵/⍨1=2|⍵}⍳10 1 9 25 49 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?).
- eesmith 6y agoThanks! 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: >>> 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] seems more difficult to pull off in a pipeline API.
- gandercrews 6y agonah 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 2209
- eesmith 6y agoOh, 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!