Shorten the Java Package

Question

Briefing

Given a Fully Qualified Java Class/ Package name, you must shorten it as such:

Each part of the dot separated package will be shortened to its first letter, apart from the last section and the class (if it exists).

package names will be all lower case, and the class (if it exists) will start with a capital letter and be UpperCamelCase. packages come in the form of:

foo.bar.foo

and

foo.bar.foo.Class

Examples

(No Class)
Input com.stackoverflow.main
Output c.s.main

(Class)
Input com.google.parser.Gson
Output c.g.parser.Gson

(Class)
Input com.google.longer.package.TestClass
Output c.g.l.package.TestClass

Rules

• Shortest code in bytes wins
• Standard loopholes apply

Answer 1

Retina, 17 bytes

\B\w+(\.[a-z])
$1

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Explanation

\B         # Start from a position that isn't a word boundary. This ensures that
           # the first letter of the package name is skipped.
\w+        # Match one or more word characters. This is the remainder of the
           # package name which we want to remove.
(          # Capture the next part in group 1, because we want to keep it...
  \.[a-z]  #   Match a period and a lower-case letter. This ensures that we
           #   don't match the package that precedes the class, or the package or
           #   class at the end of the input.
)

This is replaced with $1, which is the period and lower case letter which shouldn't be removed.

Answer 2

JavaScript (ES6), 68 53 bytes

s=>s.split`.`.map((x,y,z)=>z[y+1]>"["?x[0]:x).join`.`

• 15 bytes saved thanks to Arnauld.

See my other solution here.

---

Try it

f=
s=>s.split`.`.map((x,y,z)=>z[y+1]>"["?x[0]:x).join`.`
i.addEventListener("input",_=>o.innerText=f(i.value))
console.log(f("com.stackoverflow.main"))
console.log(f("c.g.parser.Gson"))
console.log(f("com.google.longer.package.TestClass"))

<input id=i><pre id=o>

Answer 3

Python 2, 88 81 bytes

f=lambda s,d=".":s.count(d)>(s.split(d)[-1]<"[")and s[0]+d+f(s[s.find(d)+1:])or s

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Answer 4

Java 8, 66 41 bytes

s->s.replaceAll("\\B\\w+(\\.[a-z])","$1")

Port from @MartinEnder's amazing Retina answer.

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Answer 5

Mathematica, 75 bytes

#[[;;-3]]~StringTake~1~Join~#[[-2;;]]~StringRiffle~"."&[#~StringSplit~"."]&

Anonymous function. Takes a string as input and returns a string as output.

Answer 6

Japt, 30 27 25 bytes

¡Y>Zl -('[>ZgJ)-2?X:Xg}'.

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Answer 7

Python 2, 76 73 bytes

q=input().split('.')
i=~(q[-1]<'_')
q[:i]=zip(*q[:i])[0]
print'.'.join(q)

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Answer 8

V, 9 bytes

Í쓃…®õÀ!

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Hexdump:

00000000: cdec 9383 85ae f5c0 21                   ........!

This is a wonderful example of V's signature regex compression.

Explanation:

Í          " Remove every match on every line:
 ì         " A lower case letter
  “ …      "   *ONLY MATCH THIS PART:*
   ƒ       "   As few characters as possible
      ®    " Followed by a dot
       õÀ! " Not followed by an uppercase letter

Answer 9

Haskell, 58 bytes

f s|[(a:t,p:x:r)]<-lex s=a:concat[t|x<'a']++p:f(x:r)|1<3=s

Try it online! Usage: f "some.string".

lex parses a string as Haskell tokens, so lex "some.string" returns [("some",".string")]. f recurses over the tokens in the string and always appends the first char a of the current token, but the rest t of the token only if the remaining string after the colon p starts with an uppercase char, that is x<'a'. If the pattern match failed, we have reached the last token and simply return s.

Answer 10

JavaScript (ES6), 36 bytes

Another port of Martin's Retina answer. See my other solution here.

s=>s.replace(/\B\w+(\.[a-z])/g,"$1")

f=
s=>s.replace(/\B\w+(\.[a-z])/g,"$1")
i.addEventListener("input",_=>o.innerText=f(i.value))
console.log(f("com.stackoverflow.main"))
console.log(f("c.g.parser.Gson"))
console.log(f("com.google.longer.package.TestClass"))

<input id=i><pre id=o>

Answer 11

sed, 57 22 bytes

I expected sed solution to be a little shorter than this...

Edit:

The shorter solution uses regex from Martin Ender's answer.

21 bytes sourcecode + 1 byte for -r flag (or -E flag for BSD sed).

s|\B\w+(\.[a-z])|\1|g

Answer 12

Python 2, 108 97 89 bytes

i=input().split(".")
for x in range(len(i)+~(i[-1][0]<"[")):i[x]=i[x][0]
print".".join(i)

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-8 with many thanks to @ovs for the tip

Answer 13

Go, 101 bytes

import."regexp"
func f(B string)string{return MustCompile(`\B\w+(\.[a-z])`).ReplaceAllString(B,`$1`)}

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Port of Martin's Retina answer.

Answer 14

Pip, 22 19 bytes

aR`\B\w+(\.[a-z])`B

-3 thanks to DLosc!

Yet another port of Martin's Regex.

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---

First Attempt(Non-Competing)

Pip, 45 bytes

d:"."c:dNae:XU Na?c-1ca:a^dFi,e{a@i:a@i@0}aJd

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Answer 15

05AB1E, 20 15 bytes

ΔA3ãvy'.«ÀDõ1ǝ:

Try it online or verify all test cases. (Extremely slow, so the A has been replaced with Dálê (duplicate; keep letters; convert to lowercase; sorted-uniquify) in the TIO-links.)

Original 20 bytes approach:

'.©¡Rć¬.uisć}s€н`r®ý

Try it online or verify all test cases.

Explanation:

I basically keep replacing every ab.c to a.c, where abc are any combination of lowercase letter including duplicates:

Δ                 # Loop until the result no longer changes,
                  # using the (implicit) input-string in the first iteration
 A                #  Push the lowercase alphabet
  3ã              #  Create all possible triplets of lowercase letters with the
                  #  cartesian power of 3
    v             #  Loop over each triplet `y`:
     y'.«        '#   Append a "." to the triplet
         À        #   Rotate it once towards the right: "abc." to "ab.c"
          D       #   Duplicate it
           õ1ǝ    #   Replace the character at index 1 with "": "ab.c" to "a.c"
              :   #   Replace all "ab.c" in the string to "a.c"
                  # (after which the resulting string is output implicitly)

'.               '# Push "."
  ©               # Store it in variable `®` (without popping)
   ¡              # Pop and split the (implicit) input by this "."
    R             # Reverse this list
     ć            # Extract head; pop and push remainder-list and first item separated
                  # to the stack
      ¬           # Get the first character (without popping the string)
       .ui        # Pop, and if it's uppercase:
          s       #  Swap so the remainder-list is at the top of the stack
           ć      #  Extract its head again
         }        # Close the if-statement
          s       # Swap so the remainder-list is at the top of the stack
           €      # Map over each string in this list:
            н     #  Pop and only leave its first character
             `    # Pop and dump these characters to the stack
              r   # Reverse the stack
               ®ý # Join the stack by the "."-delimiter from variable `®`
                  # (after which the result is output implicitly)