In Raku, how does one write the equivalent of Haskell's span function?

Question

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In Raku, how does one write the equivalent of Haskell's span function?

In Haskell, given a predicate and a list, one can split the list into two parts:

• the longest prefix of elements satisfying the predicate
• the remainder of the list

For example, the Haskell expression …

span (< 10) [2, 2, 2, 5, 5, 7, 13, 9, 6, 2, 20, 4]

… evaluates to …

([2,2,2,5,5,7],[13,9,6,2,20,4])

How does one write the Raku equivalent of Haskell's span function?

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Update 1

Based on the answer of @chenyf, I developed the following span subroutine (additional later update reflects negated predicate within span required to remain faithful to the positive logic of Haskell's span function) …

sub span( &predicate, @numberList )
  {
  my &negatedPredicate = { ! &predicate($^x) } ;
  my $idx = @numberList.first(&negatedPredicate):k ;
  my @lst is Array[List] = @numberList[0..$idx-1], @numberList[$idx..*] ;
  @lst ;
  } # end sub span

sub MAIN()
  {
  my &myPredicate = { $_ <= 10 } ;
  my @myNumberList is Array[Int] = [2, 2, 2, 5, 5, 7, 13, 9, 6, 2, 20, 4] ;
  my @result is Array[List] = span( &myPredicate, @myNumberList ) ;

  say '@result is ...' ;
  say @result ;
  say '@result[0] is ...' ;
  say @result[0] ;
  say @result[0].WHAT ;
  say '@result[1] is ...' ;
  say @result[1] ;
  say @result[1].WHAT ;
  } # end sub MAIN

Program output is …

@result is ...
[(2 2 2 5 5 7) (13 9 6 2 20 4)]
@result[0] is ...
(2 2 2 5 5 7)
(List)
@result[1] is ...
(13 9 6 2 20 4)
(List)

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Update 2

Utilizing information posted to StackOverflow concerning Raku's Nil, the following updated draft of subroutine span is …

sub span( &predicate, @numberList )
  {
  my &negatedPredicate = { ! &predicate($^x) } ;
  my $idx = @numberList.first( &negatedPredicate ):k ;
  if Nil ~~ any($idx) { $idx = @numberList.elems ; }
  my List $returnList = (@numberList[0..$idx-1], @numberList[$idx..*]) ;
  $returnList ;
  } # end sub span

sub MAIN()
  {
  say span( { $_ == 0 }, [2, 2, 5, 7, 4, 0] ) ;  #  (() (2 2 5 7 4 0))
  say span( { $_ <  6 }, (2, 2, 5, 7, 4, 0) ) ;  #  ((2 2 5) (7 4 0))
  say span( { $_ != 9 }, [2, 2, 5, 7, 4, 0] ) ;  #  ((2 2 5 7 4 0) ())
  } # end sub MAIN

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Answer 1

I use first method and :k adverb, like this:

my @num  = [2, 2, 2, 5, 5, 7, 13, 9, 6, 2, 20, 4];
my $idx = @num.first(* > 10):k;

@num[0..$idx-1], @num[$idx..*];