What's my middle name?

Question

Note: The winning answer will be selected on 4/12/17 the current winner is Jolf, 1 byte.

I'm surprised that we haven't had a what's my middle name challenge on this site yet. I did alot of searching but found nothing. If this is a dup, please flag it as such.

Your challenge

Parse a string that looks like Jo Jean Smith and return Jean.

Test cases

Input: Samantha Vee Hills
Output: Vee

Input: Bob Dillinger
Output: (empty string or newline)

Input: John Jacob Jingleheimer Schmidt
Output: Jacob Jingleheimer

Input: Jose Mario Carasco-Williams
Output: Mario

Input: James Alfred Van Allen
Output: Alfred Van 

(That last one is incorrect technically, but fixing that would be too hard.)

Notes:

• Names will always have at least 2 space-separated parts, with unlimited middle names between them or can be a list/array of strings.
• Names may contain the alphabet (case-insensitive) and - (0x2d)
• You may output a trailing newline.
• You may require input to have a trailing newline.
• Input from STDIN, a function parameter, or command-line argument is allowed, but hard-coding it in is not allowed.
• Standard loopholes forbidden.
• Output may be function return value, STDOUT, STDERR, etc.
• Trailing spaces/newlines/tabs in the output are allowed.
• Any questions? Comment below!

This is code-golf, so the shortest anwser in bytes wins!

Answer 1

Ruby, 17 bytes

$><<$*[1..-2]*' '

Full program, takes input as command line arguments, e.g.:

$ ruby middle.rb john jacob jingleheimer shmidt
jacob jingleheimer

If we don't mind having the result wrapped in quotation marks, can just use:

p$*[1..-2]*' '
$ ruby middle2.rb john jacob jingleheimer shmidt
"jacob jingleheimer"

Of course, we can also just take a list and return a list, which can be as short as a 13 byte lambda.

->s{s[1..-2]}
# usage:
y = ->s{s[1..-2]};
y.call ["john","jacob","jingleheimer","shmidt"] #=> ["jacob", "jingleheimer"]

Answer 2

Sed, 23 20 bytes

s/\w* (.*) \w*$/\1/

Unlike my last sed entry which involved a pipeline from one sed command to another, this one is just a simple, single command, so I'm only counting the bytes of the sed script.

Edited: Plus a 1-byte penalty for the -r flag. Thank you, Krixti Lithos, for pointing that out!

Answer 3

Pushy, 17 bytes

32K-$.;@$.;.@32+"

Try it online!

32 K-    \ Take away 32 from every character.
         \ Now we have the value 0 where there were spaces.
$        \ Until we reach a zero:
 .;      \   Pop a character from the end
         \ Everything after the last space is now removed
@        \ Reverse stack
$.;      \ Do the same
.        \ Pop the leading space
32+      \ Add 32 to everything to map the numbers back to their original characters
@"       \ Reverse and print

This is the first method I thought of, and looks like the shortest.

Answer 4

C#, 72 69 bytes

Takes a string, returns a string...

s=>string.Join(" ",s.Split(' ').Skip(1).TakeWhile(n=>!s.EndsWith(n)))

C# 43 bytes

Takes a string array, returns a string array...

s=>s.Skip(1).TakeWhile((n,i)=>i<s.Length-2)

Answer 5

c, 72 bytes

g(char*p){char*e,*s=0;for(;*p;p++)*p-32?0:s?(e=p):(s=p);*e=0;puts(s+1);}

Try it online

Answer 6

Clojure, 107 bytes

(fn[f](clojure.string/join" "(map #(apply str %)(drop 1(drop-last(take-nth 2(partition-by #(= % \ )f)))))))

Unfortunately drop/drop-last was shorter than subvec.

(defn middle-name [full-name]
  ; Partition the string into a list of names split on spaces, then take every second element
  ;  since partition keeps the spaces 
  (let [sub-names (take-nth 2
                    (partition-by #(= % \ ) full-name))]

    ; ... then join the middle names together on a space.
    (clojure.string/join " "
      ; Drop returns list of chars, so turn them back into strings...
      (map #(apply str %)
           ; Drop the first and last name
           (drop 1
             (drop-last sub-names))))))

Answer 7

JavaScript (ES6), 16 Bytes

x=>x.slice(1,-1)

Simple. Return the array passed from the second element to the second from last.

Answer 8

Lua, 39 bytes

print(io.read("*l"):match(" (.+) .+$"))

Answer 9

c, 53 bytes

main(a,b)char**b;{for(++b;--a>2;printf("%s ",*++b));}

Try it online

Answer 10

Noodel, 5 bytes

Input must be an array of the names.

Ẹ⁻2¤İ

Try it:)

---

How it works

      # Pushes the input on implicitly.
Ẹ⁻2¤İ
Ẹ     # Takes the last item of an array and pushes it to the back.
 ⁻2   # Pop the first two items off of the array.
   ¤  # Push the string "¤" onto the stack which represents a space.
    İ # Concatenates the array elements by the "¤" string.
      # Implicitly output the top of the stack to the screen.

<div id="noodel" code="Ẹ⁻2¤İ" input='["James", "Alfred", "Van", "Allen"]' cols="10" rows="2"/>

<script src="https://tkellehe.github.io/noodel/noodel-latest.js"></script>
<script src="https://tkellehe.github.io/noodel/ppcg.min.js"></script>

Answer 11

Javascript ES6 23Bytes

x=>x.pop()*x.shift()||x

use * instead of && to save 1 byte

Try it online

Answer 12

Tcl, 28 bytes

proc R n {lrange $n 1 end-1}

Try it online!

Answer 13

Japt, 7 3 bytes

s1J

Try it online!

This is essentially Input.slice(1,-1). The -1 wraps around to the last item in the input.

Answer 14

Gambit Scheme (gsi), 48 bytes

(lambda(x)(if(>(length x)2)(cadr(reverse x))""))

Try it online!

Answer 15

Arturo, 29 bytes

$->a[take drop a 1(size a)-2]

Try it