[#1] Skript Challenge - Number Format

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Do you like Skript Challenges like this?

  • Yes

    Votes: 41 78.8%

  • Kinda, they need work

    Votes: 7 13.5%

  • No

    Votes: 4 7.7%
  • Total voters
    52
  • Poll closed .
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code_language.skript: 
command /skriptchallenge [<text>]:
    trigger:
        if arg-1 is set:
            set {_number::*} to arg-1 split at ""
            loop size of {_number::*} times:
                add last element of {_number::*} to {_number2::*}
                set {_last} to size of {_number::*}
                delete {_number::%{_last}%}


            set {_time} to 1
            set {_n} to -1
            set {_integer} to ""
            loop {_number2::*}:
                add 1 to {_n}
                set {_integer} to "%loop-value%%{_integer}%"
                if ({_n} / 3) = {_time}:
                    add 1 to {_time}
                    if {_n} + 1 isn't size of {_number2::*}:
                        set {_integer} to ",%{_integer}%"

            send "%{_integer}%"
            stop
upload_2017-1-30_9-18-0.png
 
Last edited by a moderator:
iRegalia said:
code_language.skript: 
command /skriptchallenge [<text>]:
    trigger:
        if arg-1 is set:
            set {_number::*} to arg-1 split at ""
            loop size of {_number::*} times:
                add last element of {_number::*} to {_number2::*}
                remove last element of {_number::*} from {_number::*}
         
         
            set {_time} to 1
            set {_n} to -1
            set {_integer} to ""
            loop {_number2::*}:
                add 1 to {_n}
                set {_integer} to "%loop-value%%{_integer}%"
                if ({_n} / 3) = {_time}:
                    add 1 to {_time}
                    if {_n} + 1 isn't size of {_number2::*}:
                        set {_integer} to ",%{_integer}%"
         
            send "%{_integer}%"
View attachment 163
Click to expand...
i count 5 1's on the end of that number, why's it returning 6 :O
 
Tlatoani said:
Code: 
function a(b: text) :: text:
  if length of {_b} > 3:
    return "%a(first length of {_b} - 3 characters of {_b})%,%last 3 characters of {_b}%"
  return {_b}

command /c <integer>:
  trigger:
    send a("%arg%")

Not sure whether the command was required, but it's there.

EDIT: Fixed the code formatting issue.
Click to expand...
I'm not that much good at math specially when i try to make it in english.. because i'm not english so, i might use your code in some of my skripts (if you don't mind) and i will try to understand it more :emoji_slight_smile: Good work
 
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Reactions: Tlatoani
Code: 
regex replace "(?<=\d)(?=(\d{3})+(?!\d))" with "," in {_integer}
1 less character :emoji_wink:
 
it_twit said:
Code: 
regex replace "(?<=\d)(?=(\d{3})+(?!\d))" with "," in {_integer}
1 less character :emoji_wink:
Click to expand...
That's just using regex. Not really using the ability of Skript in any way. I don't really think that the challenge was about that, but it works.
 
LimeGlass said:
That's just using regex. Not really using the ability of Skript in any way. I don't really think that the challenge was about that, but it works.
Click to expand...
This one was about the shortest amount of code. I will however be making some focused on code design instead of character count :emoji_slight_smile:. Those ones I expect will show a completely different type of answer.
 
Got in here with high hopes, hoping I've learned something and that I'd be able to join this challenge... saw numbers, saw math, headache hit, too dumb for this. GG, must admit, every code I saw (especially Snow-Pyon's(cuz shortest)) was pretty damn good. Anyone wants to teach me how to math?
 
CrimZ0N said:
Got in here with high hopes, hoping I've learned something and that I'd be able to join this challenge... saw numbers, saw math, headache hit, too dumb for this. GG, must admit, every code I saw (especially Snow-Pyon's(cuz shortest)) was pretty damn good. Anyone wants to teach me how to math?
Click to expand...
First you start doing math. Realise it's stupid, and stop.

That's how.
 
Gatt said:
First you start doing math. Realise it's stupid, and stop.

That's how.
Click to expand...

You don't really need that much math, just the modulus operator. All you need to do is do a reverse iteration through every char in the string, and when the char you're on is evenly divisible by 3 (I.E. char mod 3 = 0), add a comma.

An example in Java would be:
Code: 
    public static String addCommas(String s){
        String ret = "";
        for(int i = s.length() - 1, j = 1; i > -1; i--, j++){//reverse iteration
            ret = s.charAt(i) + ret;
            if (j % 3 == 0){ //if evenly divisble by 3
                ret = "," + ret;
            }
        }
        return ret;
    }

I would hardly call it difficult, Should be able to do this within your first year of taking a programming course.

Doing this in skript would be annoying because it doesn't have for loops, but it can be done:

Code: 
//assume {s} is the string
    {_ret} = ""
    {_i} = {s}'s length
    {_j} = 1
    while {_i} > 0:
        ret = text of {s} from i to i + {_ret}
        if j mod 3 = 0:
            ret = "'" + {_ret}
        {_i} = i - 1
        {_j} = j + 1
    {s} = {_ret}
    #Keep going

I haven't actually tested the above skript code, but if I didn't make any mistakes and the documentation is correct, it should work.
 
Last edited by a moderator:
Gamebuster said:
If you want to call basic arithmetic a lot of math, then sure, I guess...
Click to expand...
He's not calling it a lot of math. He was only responding to you stating that you don't need a lot of math, just the modulus operator, while the modulus operator is math.
 
mathhulk said:
He's not calling it a lot of math. He was only responding to you stating that you don't need a lot of math, just the modulus operator, while the modulus operator is math.
Click to expand...
That's why I said "You don't really need that much math, just the modulus operator." One operation is not a lot of math, like I said.
 
Here's my code, with no addons :
code_language.skript: 
set {_r} to "%{_number}%"
set {_t} to ""
loop round up length of {_r} / 3 times:
    set {_l} to the last 3 characters of {_r}
    set {_r} to the first length of {_r} - 3 characters of {_r}
    if loop-number = 1:
        set {_t} to "%{_l}%%{_t}%"
    else:
        set {_t} to "%{_l}%,%{_t}%"

Formatted number will be in {_t}
 
Entries close on Monday, at whatever time I get up. I'll try and find the shortest answer but you lot argued about maths so much, I have no idea.
 
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