Take first letter of each word, leave spaces and punctuation

Question

Shrink each word in a string of group of strings to single letters delineated by spaces or punctuation.

Example

I'm a little teapot,  short and stout. Here is my handle, here is my spout. When I get all steamed up - hear me shout!   Tip me over and pour me out. 

becomes

I' a l t, s a s. H i m h, h i m s. W I g a s u - h m s! T m o a p m o. 

Edit - if there are multiple spaces, preserve only one space. All punctuation should be preserved, I missed the apostrophe. Yes this is code golf :-).

Answer 1

CJam, 13 bytes

r{(\'A,f&Sr}h

Works if I can consider only the common punctuation characters, and the output can have trailing spaces. (Thanks to Dennis.)

This question needs much more clarification...

CJam, 17 16 bytes

r{(\eu_elf&Sr}h&

Try it online.

Explanation

r          e# Read one word from input.
{          e# While it is not EOF:
    (\     e# Extract the first character.
    eu     e# Convert the rest to uppercase.
    _el    e# And lowercase.
    f&     e# Delete characters in the first string if not in the second string.
    S      e# Append a space.
    r      e# Read the next word.
}h
&          e# Discard the last space by intersecting with empty string.

Answer 2

Pyth, 14 bytes

jdm+hd-rtd0Gcz

Try it online: Demonstration

Explanation:

                 implicit: z = input string
            cz   split z by spaces
  m              map each word d to:
    hd              first letter of d
   +                +
       rtd0         (lowercase of d[1:]
      -    G         but remove all chars of "abc...xyz")
jd               join resulting list by spaces and print

Answer 3

Python 3.4, 94 92 82 77 bytes

print(*[w[0]+''.join(c[c.isalpha():]for c in w[1:])for w in input().split()])

I'm new to code golfing but I thought I'd give it a try! This one's not a winner, but it was fun.

This just splits the string, taking the first character of each word along with any punctuation in the rest of the word.

*edited with changes by FryAmTheEggman, DLosc

Answer 4

Jelly, 14 bytes

Ḣ;ḟØB$
Ḳ¬ÐfÇ€K

Try it online!

Solution done with Mr. Xcoder and caird coinheringaahing in Jelly Hypertraining chat room.

Alternative 14 bytes: Try it online!

Answer 5

sed (39 chars)

Just a couple of regular expressions:

sed 's+\<\(.\)[A-Za-z]*+\1+g;s+  *+ +g'

Answer 6

Lua - 126 characters

Lua isn't much of a code golfing language, but I gave it a shot:

a=''for b in string.gmatch( c, '%S+%s?' )do d=(b:match('%w')or''):sub(1,1)e=b:match('[^%s%w]')or''a=a..d..e..' 'end print( a )

This assumes that c is the string.

Here it is cleaned up for readability:

local string = [[I'm a little teapot,  short and stout. Here is my handle, here is my 
spout. When I get all steamed up - hear me shout!   Tip me over and pour me out.]]

local final = ''
for word in string.gmatch( string, '%S+%s?' ) do 
    local first = ( word:match( '%w' ) or '' ):sub( 1, 1 )
    local second = word:match( '[^%s%w]' ) or ''
    final = final .. first .. second .. ' '
end
print( final )

You can test it here (copy and paste it. For the first on you also have to do c = "I'm a little ....) For some reason the online demo of Lua won't let you input variables using io.read...

Answer 7

PowerShell, 56 bytes

%{($_-split' '|%{$_[0]+($_-replace'[a-z]','')})-join' '}

Answer 8

Javascript (ES6) 72 68 bytes

f=x=>x.split(/\s+/).map(x=>x.replace(/^(.)|[a-z]/gi,'$1')).join(' ')

<input id="input" value="I'm a little teapot,  short and stout. Here is my handle, here is my spout. When I get all steamed up - hear me shout!   Tip me over and pour me out. " />
<button onclick="output.innerHTML=f(input.value)">Run</button>
<br /><pre id="output"></pre>

Commented:

f=x=>
    x.split(/\s+/). // split input string by 1 or more spaces
    map(x=> // map function to resulting array
        x.replace(/^(.)|[a-z]/gi, '$1') // capture group to get the first character
                                        // replace all other letters with empty string
    ).
    join(' ') // join array with single spaces

Answer 9

C99 - 170 169 bytes

main(_,a)char**a;{for(char*b=a[1],*c=b,*e,*d;*c++=*b;){for(e=b;*++b&&*b-32;);for(*b=0,d=strpbrk(e,"!',-."),d&&d-e?*c++=*d:0;b[1]==32;++b);++b;*c++=32;*c=0;}puts(a[1]);}

Ungolfed:

main(int argc, char**a) {
    char*b=a[1],*c=b,*e,*d;
    while(*c++=*b){
        for(e=b;*++b&&*b-32;); //scan for first space or end of word
        *b=0; //mark end of word
        for(;b[1]==32;++b); //skip following spaces
        d=strpbrk(e,"!',-."); //find punctuation
        if(d&&d-e) //append punctuation if any, and it's not the word itself
            *c++=*d;
        *c++=32; //append space
        b++;
    }
    *c=0; //mark end of line
    puts(a[1]);
}

Usage:

gcc -std=c99 test.c -o test
./test "I'm a little teapot,  short and stout. Here is my handle, here is my spout. When I get all steamed up - hear me shout!   Tip me over and pour me out."

Output:

I' a l t, s a s. H i m h, h i m s. W I g a s u - h m s! T m o a p m o.

Answer 10

Java 8, 87 bytes

s->{for(String x:s.split(" +"))System.out.print(x.replaceAll("^(.)|[a-z]+","$1")+" ");}

Explanation:

Try it here.

s->{                                  // Method with String parameter and no return-type
  for(String x:s.split(" +"))         //  Split the input on one or multiple spaces
                                      //  And loop over the substrings
    System.out.print(                 //   Print:
      x.replaceAll("^(.)|[a-z]+","$1")//    Regex to get the first letter + all non-letters
      +" ");                          //    + a space delimiter
                                      //  End of loop (implicit / single-line body)
}                                     // End of method

Regex explanation:

x.replaceAll("^(.)|[a-z]+","$1")
x.replaceAll("           ","  ") # Replace the match of String1 with String2, in String `x`
             "           "       # Regex to match:
              ^(.)               #  The first character of String `x`
                  |[a-z]+        #  Or any one or multiple lowercase letters
                           "  "  # Replace with:
                            $1   #  The match of the capture group (the first character)

So it basically removes all lowercase letters of a String, except the very first one.

Answer 11

Perl 5, 34 bytes

33 bytes code + 1 for -p.

s!\S\K\S+| \K +!$&=~s/\w| //gr!ge

Try it online!

Answer 12

R, 46 45 bytes

cat(gsub("^\\w\\W*\\K\\w*","",scan(,""),T,T))

This reads a line from STDIN and prints to STDOUT. It uses a regular expression to remove all characters after the first letter followed by any amount of punctuation.

Ungolfed + explanation:

# Read a string from STDIN and convert it to a character vector,
# splitting at spaces

input <- scan(what = "")

# Replace stuff with nothing using a regex.
# ^ start position anchor
# \w single "word" character
# \W* any amount of non-word (i.e. punctuation) characters
# \K move the match position forward
# \w* any number of word characters

replaced <- gsub("^\\w\\W*\\K\\w*", "", input, ignore.case = TRUE, perl = TRUE)

# Print the vector elements to STDOUT, separated by a space

cat(replaced, sep = " ")

Example:

> cat(gsub("^\\w\\W*\\K\\w*","",scan(,""),T,T))
1: I'm a little teapot,  short and stout. Here is my handle, here is my spout. When I get all steamed up - hear me shout!   Tip me over and pour me out. 

I' a l t, s a s. H i m h, h i m s. W I g a s u - h m s! T m o a p m o.

Answer 13

05AB1E, 13 bytes

#õKεćsDáмJ}ðý

Try it online!

Explanation:

#õKεćsDáмJ}ðý  Full program
#              Split on spaces
 õK            Remove empty strings from the list
   ε           For each...
    ć             push a[1:], a[0]
     s            Swap
      D           Duplicate
       á          Push only letters of a
        м         pop a,b => push a.remove(all elements of b)
         J        Join
          }    End for each
           ðý  Join with spaces

Answer 14

VBA (Excel), 141 133 Bytes

Using VBA Immediate Window, [A1] as Inputted Strings.

z=" "&[A1]:for a=2 to len(z):c=mid(z,a,1):[A2]=[A2]&IIF(mid(z,a-1,1)=" ",c,IIF(asc(lcase(c))>96 and asc(lcase(c))<123,"",c)):next:?[TRIM(A2)]

z=" "&[a1]:for a=2 to len(z):c=mid(z,a,1):e=asc(lcase(c)):[a2]=[a2]&iif(mid(z,a-1,1)=" ",c,IIF((e>96)*(e<123),"",c)):next:?[TRIM(A2)]

Answer 15

Perl 6, 15 bytes

s:g/(\w)\w+/$0/

Try it online!