Split into three!

Question

Given a string, you've to split it into groups of three characters (by padding _ at the end).

Related, but still it's different.

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Sample I/O:

• 'abcde' -> 'abc', 'de_'

• '1234' -> '123', '4__'

• '' -> [] or falsey value

• 'Three spree!' -> 'Thr', 'ee ', 'spr', 'ee!'

• 6 MB string

This is a code-golf, so fewest bytes will win!

Edit: Finally, output is unrestricted.

Answer 1

Haskell, 40 bytes

f(a:b:c:t)=[a,b,c]:f t
f _=[]
f.(++"__")

Answer 2

JavaScript (Babel Node), 45 bytes

_=>_.match(/.{1,3}/g).map(a=>a.padEnd(3,`_`))

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Answer 3

Jelly, 10 bytes

s3;€⁾__ḣ€3

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It feels like an 8 or 9 should be possible, but haven’t found one yet.

Answer 4

Attache, 34 23 bytes

PadRight&"_"&3=>@Chop&3

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Explanation (outdated)

{On[#_-1,PadRight&"_"&3,_]}@Chop&3
                           @Chop&3    chop the input string into groups of 3s
{On[#_-1               ,_]}           on the last group
        ,PadRight                         pad it
                     &3                   to length three
                 &"_"                     with underscores

Answer 5

Charcoal, 10 bytes

Ｅ⪪Ｓ³…⁺ι__³

Try it online! Link is to verbose version of code. Explanation:

  Ｓ         Input string
 ⪪ ³        Split into groups of up to 3 characters
Ｅ           Map over each group
      ι     Current group
     ⁺      Concatenated with
       __   Literal `__`
    …    ³  Moulded to length 3
            Each group implicitly printed on its own line

Answer 6

Scala, 31 bytes

(s+"___").grouped(3).toSeq.init

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Explanation

(s+"___")       // add "___"
  .grouped(3)   // make groups of 3
  .toSeq        // change to Seq
  .init         // take all but last

Answer 7

Befunge-93, 30 29 bytes

~:1+%,~:1+!"`"*+,~:1+!"`"*+,,

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Outputs nothing for an empty input, otherwise outputs strings of length 3 separated by NUL characters.

Explanation:

~                               Push character of input to stack
 :1+                            Duplicate input and add 1
    %                           Modulo top 2 values of stack and push result
                                If input was -1 (end of stream), calculate -1%0 -> halt
                                Otherwise calculate input%(input+1) -> input
     ,                          Pop and output character
      ~                         Push character of input to stack
       :1+!                     Duplicate input, add 1 and perform logical NOT
           "`"*                 Multiply by 96 (ASCII code for '`')
                                This returns 96 or 0 depending on the result of the NOT
               +                Add the top 2 values from the stack and push the result
                                If the input was -1 (end of stream), pushes -1+96=95, or the ASCII code for '_'
                                Otherwise pushes input+0
                ,               Pop and output character
                 ~:1+!"`"*+,    Same block again to handle the third character
                            ,   Output a NUL character (stack is empty, so popping from stack just returns 0)

The instruction pointer wraps around at the end, looping the entire line.

Answer 8

Pyth, 10 9 bytes

Pc+z*3\_3

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-1 from u_ndefined.

Is it ok if it throws an error on empty input? (does that count as falsey)

Answer 9

Rust, 93 bytes

Works for ASCII input strings. Empties the input string and returns a Vec<String>.

|mut s:String|{s+="__";let mut v=vec![];while s.len()>2{let t=s.split_off(3);v.push(s);s=t}v}

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Answer 10

GoLang, 84 bytes

func(x string)[]string{return regexp.MustCompile("...").FindAllString(x+"__",-1)}(s)

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Answer 11

Haskell, 62 61 59 bytes

z(a:b:c:d)=[a,b,c]:z d
f s=takeWhile(/="___")$z$s++cycle"_"

Infinite lists and strings!

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Answer 12

SNOBOL4 (CSNOBOL4), 93 bytes

	S =INPUT
	S =S DUPL('_',REMDR(3 - REMDR(SIZE(S),3),3))
R	S LEN(3) . OUTPUT REM . S	:S(R)
END

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SNOBOL has a size limit on the length of its input (1023 bytes).

Answer 13

Julia 1.0, 40 bytes

s->[(s*"__")[i:i+3] for i=1:3:length(s)]

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Answer 14

GolfScript, 10 bytes

"___"+3/);

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Answer 15

Ruby, 27 21 bytes

p"#{$_}__".scan /.../

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-6 bytes thanks to vrintle

Answer 16

Uiua ^SBCS, 8 bytes

⬚@_↯¯1_3

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Answer 17

Nibbles, 8 bytes (16 nibbles)

In literal form:

`/3;_<3:$"__"

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`/             chunks of
  3              3 characters
   ;_            for input string
               then, apply map(\chunk->...):
     <3          take first 3 characters of
       :$"__"    concatenation of chunk and "__"

Answer 18

Rattle, 21 bytes

\__|!II^/3P[[gb>]3B]`

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Outputs groups of 3 separated by newlines.

Explanation:

\__|                  take input and concatenate "__"
    !                 disable implicit output
     I                save characters of the input in consecutive memory slots
      I^              get length of the input
        /3            divide by 3 to get n
          [...]`      loop n times (ignoring any values after the decimal point)

[...]3                loop 3 times
 g                    get the character at the pointer
  b                   add this character to the print buffer
   >                  move pointer right
      B               print contents of buffer and reset