[REBOL] Re: Tip for splitting very long string ?
From: carl:cybercraft at: 17-Apr-2002 19:36
On 17-Apr-02, Ingo Hohmann wrote:
> Hi Carl,
> Am Die, 2002-04-16 um 13.16 schrieb Carl Read:
>> I tried to cut out the need for the charset in the following
>> function but it ends up in an infinate loop when it starts
>> comparing an empty string with an empty string. Can anyone think of
>> a rule that would override that? Be interesting to know if this
>> would be faster than the above. (If it worked...)
>> split2: func [
>> "This don't work..."
>> str [string!]
>> num [integer!]
>> /local blk s
>> blk: copy 
>> parse/all str [some[
>> s: (s: copy/part s num) s (insert tail blk s)
> you could do it like this:
> split2: func [
> "This don't work... Now it does ;-)"
> str [string!]
> num [integer!]
> /local blk s
> blk: copy 
> parse/all str [some[
> s: (s: copy/part s num insert tail blk s) num skip
> just 'skip the right number of characters.
> The 'insert is in the first paren!, because otherwise the last
> charecters would be added if the string does not contain a multiple
> of num characters.
Very nice Ingo! And I now realise skip needs its numbers preceeding
it, not following it. No wonder I'd never been able to get it to
work properly before. (:
So, here's the final version of the function (perhaps:)...
split: func [
"Copy sub-strings of a set length from a string."
str [string!] "String to be split."
num [integer!] "Length of sub-strings."
/local blk s
blk: copy 
parse/all str [some[
s: 1 num skip (insert tail blk copy/part s num)
Note I moved the skip to before the paren and gave it a range with a
low value of 1. This was because your version would add an empty
string to the block when the string could be divided evenly. ie...
>> split2 "abcde" 5
== ["abcde" ""]
I reduced the contents of the paren a bit too, there being some
On my system this function is a bit faster than my charset version and
about 30% faster than the looping version posted by Anton, so thumbs
up for parsing in this case.