What's the fastest/most efficient way to count lines in Rebol?

Question

Given a string string, what is the fastest/most-efficient way to count lines therein? Will accept best answers for any flavour of Rebol. I\'ve been working unde

Answer 1

remove-each can be fast as it is native

s: "1^/2^/3"
a: length? s
print a - length? remove-each v s [v = #"^/"]
; >> 2

or as a function

>> f: func [s] [print [(length? s) - (length? remove-each v s [v = #"^/"])]]
>> f "1^/2^/3"
== 2