3 ms·

I share your sentiments and it is incredible. To an outsider this might as well be a parody of programming blog posts. Such impenetrable syntax taken in stride:

I share your sentiments and it is incredible. To an outsider this might as well be a parody of programming blog posts. Such impenetrable syntax taken in stride: We can improve this approach by recognizing that we can handle the edge cases by using a catenation: ⍸0 1⍷0,⍵ ⍸1 0⍷⍵,0 Notice immediately how much better this feels. Ah yes, how immediate is the feeling, much better, indeed. APL is so ridiculous, you have to love it.

abrudz 3y ago
The syntax is actually quite simple, but you may be confused by the unfamiliar symbols and lack of syntax. 0 1 is simply [0,1] in JSON and ⍵ is the argument name. The other symbols cirrespond to prefix and infix operators. Compare the following JS expression which is syntactically (but not semantically) equivalent to the APL expression below it: - [0,1] * 0 ** w ⍸ 0 1 ⍷ 0 , ⍵
abrudz 3y ago
As for the semantics, here are the equivalent JS functions: I = y => [...y.keys()].filter((e,i) => y[i]) // ⍸y Indices of trues in y E = (x,y) => y.map((e,i) => x.every((e,j) => e == y[i+j])) // x⍷y mask indicating indices where x Exists as a sub-array in y C = (x,y) => [x,y].flat() // x,y Catenate x and y into a single array w = [1,1,1,1,0,1,1,1,0,1,0,1,1,0,0,0,1,1,1,1] I(E([0,1],C(0,w))) // [0,5,9,11,16] Now, if I, E, and C were prefix/infix JS operators, with x being the left argument (if any) and y being the right argument, we'd write: I([0,1] E (0 C w)) All APL operators have long right scope, so we don't need to parenthesise right arguments: I [0,1] E 0 C w // same syntax as APL's ⍸ 0 1 ⍷ 0 , ⍵ Alternatively, you can see the infix operators as being methods of all types: I([0,1].E(0.C(w))) Now we remove all the .() noise: I [0,1] E 0 C w // same syntax as APL's ⍸ 0 1 ⍷ 0 , ⍵ Wasn't that hard, was it?

abrudz 3y ago

The syntax is actually quite simple, but you may be confused by the unfamiliar symbols and lack of syntax. 0 1 is simply [0,1] in JSON and ⍵ is the argument name. The other symbols cirrespond to prefix and infix operators. Compare the following JS expression which is syntactically (but not semantically) equivalent to the APL expression below it: - [0,1] * 0 ** w ⍸ 0 1 ⍷ 0 , ⍵

abrudz 3y ago

As for the semantics, here are the equivalent JS functions: I = y => [...y.keys()].filter((e,i) => y[i]) // ⍸y Indices of trues in y E = (x,y) => y.map((e,i) => x.every((e,j) => e == y[i+j])) // x⍷y mask indicating indices where x Exists as a sub-array in y C = (x,y) => [x,y].flat() // x,y Catenate x and y into a single array w = [1,1,1,1,0,1,1,1,0,1,0,1,1,0,0,0,1,1,1,1] I(E([0,1],C(0,w))) // [0,5,9,11,16] Now, if I, E, and C were prefix/infix JS operators, with x being the left argument (if any) and y being the right argument, we'd write: I([0,1] E (0 C w)) All APL operators have long right scope, so we don't need to parenthesise right arguments: I [0,1] E 0 C w // same syntax as APL's ⍸ 0 1 ⍷ 0 , ⍵ Alternatively, you can see the infix operators as being methods of all types: I([0,1].E(0.C(w))) Now we remove all the .() noise: I [0,1] E 0 C w // same syntax as APL's ⍸ 0 1 ⍷ 0 , ⍵ Wasn't that hard, was it?