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

Enhanced PARSE version, as suggested by BrianH:

i: 1 ; add one as TextMate
parse text [any [thru newline (++ i)]]
print i

Answer 2

Here is the best for me:

temp: read/lines %mytext.txt
length? temp

Answer 3

Not the most efficient, but probably one of the fastest solution (anyway if a benchmark is run, I would like to see how this solution performs):

>> s: "1^/2^/ ^/^/3"
>> (length? s) - length? trim/with copy s newline
== 4

Answer 4

hehehe the read/lines length? temp is a great thing I though about read/lines -> foreach lines temps [ count: count + 1]

another way to do it would be to do

temp: "line 1 ^M line2 ^M  line3 ^M "
length? parse temp newline ; that cuts the strings into a block 
;of multiple strings that represent each a line [ "line 1" "line2" "line3" ] 
:then you count how much  strings you have in the block with length? 

I like to code in rebol it is so funny

Edit I didnt read the whole post so my solution already waas proposed in a different way...

ok to amend for my sin of posting a already posted solution I will bring insight comment of a unexpected behavior of that solution. Multiple chained carriage returns are not counted (using rebol3 linux ...)

>> a: "line1 ^M line2 ^M line3 ^M^M"
== "line1 ^M line2 ^M line3 ^M^M"

>> length? parse a newline 
== 3

Answer 5

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

Answer 6

count-lines: func [
    str
    /local sort-str ][
sort-str: sort join str "^/"
1 + subtract index? find/last sort-str "^/" index? find sort-str "^/"
]