Taking first letter from each word

asked Nov 26, 2011 at 16:32

Easiest way without regex

var abbr = "Java Script Object Notation".split(' ').map(function(item){return item[0]}).join('');

answered Apr 3, 2015 at 6:23

To add to the great examples, you could do it like this in ES6

const x = "Java Script Object Notation".split(' ').map(x => x[0]).join('');
console.log(x);  // JSON

and this works too but please ignore it, I went a bit nuts here

const [j,s,o,n] = "Java Script Object Notation".split(' ').map(x => x[0]);
console.log(`${j}${s}${o}${n}`);

answered May 16, 2017 at 21:51

It’s important to trim the word before splitting it, otherwise, we’d lose some letters.

const getWordInitials = (word: string): string => {
  const bits = word.trim().split(' ');
  return bits
    .map((bit) => bit.charAt(0))
    .join('')
    .toUpperCase();
};

$ getWordInitials("Java Script Object Notation")

$ «JSON»

answered May 27, 2022 at 10:24

Try —

var text = '';
var arr = "Java Script Object Notation".split(' ');
for(i=0;i<arr.length;i++) {
    text += arr[i].substr(0,1)
}    
alert(text);

Demo — http://jsfiddle.net/r2maQ/

answered Nov 26, 2011 at 16:37

Using map (from functional programming)

'use strict';

function acronym(words)
{
    if (!words) { return ''; }

    var first_letter = function(x){ if (x) { return x[0]; } else { return ''; }};

    return words.split(' ').map(first_letter).join('');
}

answered Dec 8, 2015 at 11:26

Alternative 1:

you can also use this regex to return an array of the first letter of every word

/(?<=(s|^))[a-z]/gi

(?<=(s|^)) is called positive lookbehind which make sure the element in our search pattern is preceded by (s|^).

so, for your case:

// in case the input is lowercase & there's a word with apostrophe

const toAbbr = (str) => {
  return str.match(/(?<=(s|^))[a-z]/gi)
            .join('')
            .toUpperCase();
};

toAbbr("java script object notation"); //result JSON

(by the way, there are also negative lookbehind, positive lookahead, negative lookahead, if you want to learn more)

Alternative 2:

match all the words and use replace() method to replace them with the first letter of each word and ignore the space (the method will not mutate your original string)

// in case the input is lowercase & there's a word with apostrophe    

const toAbbr = (str) => {
  return str.replace(/(S+)(s*)/gi, (match, p1, p2) => p1[0].toUpperCase());
};

toAbbr("java script object notation"); //result JSON

// word = not space = S+ = p1 (p1 is the first pattern)
// space = s* = p2 (p2 is the second pattern)

answered Jul 21, 2018 at 14:42

If you came here looking for how to do this that supports non-BMP characters that use surrogate pairs:

initials = str.split(' ')
              .map(s => String.fromCodePoint(s.codePointAt(0) || '').toUpperCase())
              .join('');

Works in all modern browsers with no polyfills (not IE though)

answered Jun 4, 2020 at 10:30

Getting first letter of any Unicode word in JavaScript is now easy with the ECMAScript 2018 standard:

/(?<!p{L}p{M}*)p{L}/gu

This regex finds any Unicode letter (see the last p{L}) that is not preceded with any other letter that can optionally have diacritic symbols (see the (?<!p{L}p{M}*) negative lookbehind where p{M}* matches 0 or more diacritic chars). Note that u flag is compulsory here for the Unicode property classes (like p{L}) to work correctly.

To emulate a fully Unicode-aware b, you’d need to add a digit matching pattern and connector punctuation:

/(?<!p{L}p{M}*|[p{N}p{Pc}])p{L}/gu

It works in Chrome, Firefox (since June 30, 2020), Node.js, and the majority of other environments (see the compatibility matrix here), for any natural language including Arabic.

Quick test:

const regex = /(?<!p{L}p{M}*)p{L}/gu;
const string = "Żerard Łyżwiński";
// Extracting
console.log(string.match(regex));                        // => [ "Ż", "Ł" ]
// Extracting and concatenating into string
console.log(string.match(regex).join(""))                // => ŻŁ
// Removing
console.log(string.replace(regex, ""))                   // => erard yżwiński
// Enclosing (wrapping) with a tag
console.log(string.replace(regex, "<span>$&</span>"))    // => <span>Ż</span>erard <span>Ł</span>yżwiński

console.log("_Łukasz 1Żukowski".match(/(?<!p{L}p{M}*|[p{N}p{Pc}])p{L}/gu)); // => null

answered Jul 3, 2020 at 12:21

In ES6:

function getFirstCharacters(str) {
  let result = [];

  str.split(' ').map(word => word.charAt(0) != '' ? result.push(word.charAt(0)) : '');
  
  return result;
}

const str1 = "Hello4 World65 123 !!";
const str2 = "123and 456 and 78-1";
const str3 = " Hello World    !!";

console.log(getFirstCharacters(str1));
console.log(getFirstCharacters(str2));
console.log(getFirstCharacters(str3));

Output:

[ ‘H’, ‘W’, ‘1’, ‘!’ ]

[ ‘1’, ‘4’, ‘a’, ‘7’ ]

[ ‘H’, ‘W’, ‘!’ ]

This should do it.

var s = "Java Script Object Notation",
    a = s.split(' '),
    l = a.length,
    i = 0,
    n = "";

for (; i < l; ++i)
{
    n += a[i].charAt(0);
}

console.log(n);

The regular expression versions for JavaScript is not compatible with Unicode on older than ECMAScript 6, so for those who want to support characters such as «å» will need to rely on non-regex versions of scripts.

Event when on version 6, you need to indicate Unicode with u.

More details: https://mathiasbynens.be/notes/es6-unicode-regex

answered Oct 22, 2015 at 13:56

Yet another option using reduce function:

var value = "Java Script Object Notation";

var result = value.split(' ').reduce(function(previous, current){
    return {v : previous.v + current[0]};
},{v:""});


$("#output").text(result.v);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<pre id="output"/>

answered May 18, 2016 at 8:01

This is similar to others, but (IMHO) a tad easier to read:

const getAcronym = title =>
  title.split(' ')
    .map(word => word[0])
    .join('');

answered Sep 20, 2019 at 12:12

ES6 reduce way:

const initials = inputStr.split(' ').reduce((result, currentWord) => 
    result + currentWord.charAt(0).toUpperCase(), '');
alert(initials);

answered Jan 16, 2020 at 9:34

Try This Function

const createUserName = function (name) {
const username = name
    .toLowerCase()
    .split(' ')
    .map((elem) => elem[0])
    .join('');

    return username;
};

console.log(createUserName('Anisul Haque Bhuiyan'));

answered Mar 1, 2022 at 1:32

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 :-).

$begingroup$

r{('A,f&Sr}h

r{(eu_elf&Sr}h&

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.

$endgroup$

$begingroup$

jdm+hd-rtd0Gcz

                 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

$endgroup$

$begingroup$

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

$endgroup$

$begingroup$

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

$endgroup$

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…

answered May 25, 2015 at 2:21

PowerShell, 56 bytes

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

answered May 25, 2015 at 2:10

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

answered May 25, 2015 at 1:53

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.

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.

Perl 5, 34 bytes

33 bytes code + 1 for -p.

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

Try it online!

answered Oct 20, 2017 at 10:01

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.

answered May 25, 2015 at 18:04

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

answered Oct 19, 2017 at 19:02

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)]

answered Oct 20, 2017 at 8:21

$begingroup$

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

$endgroup$

Author John Walkenbach was Mr Spreadsheet until his retirement in June 2019.