Using Dyalog APL I would like to take a Mask Vector where '-' means use the original character and any other character is to replace that position in the Data Vector. So in the example below, Mask 'a' is used on Vector 'b' to create Vector 'c'.
┌──────┬────────┬────┐
│X--Y-Z│--34--91│-+-+│ Mask V...
@RubenVerg No. Its author doesn't really believe in such.
@RubenVerg ⍐ is "Set how much the file may hold". ⍺ doesn't need further description, as it would just be a subset of the docs — afaict, there's no concise summary that would work.
@Adám ^ is already a special case because there's a dialect that doesn't support the standard. including all dialects in the special case would probably mean that all other Unicode-equivalent glyphs supported in each dialect should also appear in the list?
@Adám hm, that's an unexpected answer, not believing in documentation? weird, I guess I'll use the language bar plus some testing
> A rational number is a number that can be expressed as the quotient of 2 integers p÷q — a numerator p and a denominator q. For example, for 1.5, p and q would be 3 and 2, respectively.
> Write a function that: • takes a single non-zero positive number right argument.
> • returns a 2-element "integer" result representing the smallest non-zero positive values for p and q respectively
can't make it to the quest, here's my solution: (,÷∨)∘1
… which looks deceptibly simply, but actually isn't.
We can use tacit.help to unpack it, giving: {(⍵,1)÷(⍵∨1)}
Here, ∨ is the Greatest Common Divisor. But what does that even mean for a non-integer?
The Greatest number you can divide e.g. 3.5 with to get an integer. But dividing 3.5 with anything, surely makes it even "less" whole‽ No, not if we divide it by a number between 0 and 1.
Indeed, it gives 0.5 and 3.5÷0.5 is 7, the first part of the desired result.
If 0.5 is one factor of 7 then the other factor has to be 1÷0.5 i.e. 2.
Maybe an easier to follow solution is the (mathematically equivalent) 1∘(∧÷,)
Here, ∧ is the Lowest Common Multiple. So we're asking for the smallest positive integer that we can get to by multiplying 3.5 with an integer. That would be 7, which is 3.5×2
The rest follows as before; divide 7 with 1 to get 7, and with 3.5 to get the denominator, 2.
If we don't like this whole business with extra divisions, would could write the longer 1∘∧(⊣,÷)⊢ or {n,⍵÷⍨n←1∧⍵}
It seems the test framework doesn't properly check that the numbers are as small as possible, so a few people submitted solutions that just multiply by a huge power of 10.
(Don't worry, they were well marked down for doing so.)
No, everyone either "cheated" or used ∧/∨
I do want to mention this stranded formulation, which reads nicely: {⍵ 1÷1∨⍵}
OK, here's a bonus quest or two…
What if I want this to work on non-scalar arrays of numbers?
So, two options: an array of the same shape, with each element becoming a 2-element vector , and yes, increasing the rank by adding a trailling (or leading!) axis.
