The Cantor set without

Let's start working with numbers instead of text. Since we're replacing every one, we'll use a real looping construct:

      , ({⍵,0,⍵}⍤0) 1 0 1
1 0 1 0 0 0 1 0 1

In APL we usually prefer simple arithmetic over array manipulation:

      , ({⍵×1 0 1}⍤0) 1 0 1
1 0 1 0 0 0 1 0 1

      ,∘({⍵×1 0 1}⍤0)⍣3⊢ 1
1 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1

Well, it's shorter than regex…