Briefest code to find vowels and print consonants

Question

Challenge: Read a string from user, print number of each vowel, then place all remaining chars in a line. Shortest answer wins!

Output must look like this:

Enter String: Input

Vowel Count: a: #, e: #, i: #, o: #, u: #

Remaining characters: Vowel-less Input

For example:

Enter String: LoremIpsumDolorSitAmetConsecuteurAdipscingElit!
Vowel Count: a: 0, e: 4, i: 4, o: 4, u: 3
Remaining characters: LrmIpsmDlrStAmtCnsctrAdpscngElt!

Answer 1

05AB1E, 62 bytes

Code:

”‡Á•Ë: ÿ”,”Vowelš‹: ”?"aeiou"©vy": "¹y¢««}", "ý,“«¶šÞ: “ª¹®-J,

Explanation:

There is a lot of string compression going on here. Here is a list of all strings that are compressed:

• ”‡Á•Ë” is equivalent to Enter String.
• ”Vowelš‹” is equivalent to Vowel Count.
• “«¶šÞ“ª is equivalent to Remaining characters.

To see how the actual string compression works, see the Hello, World! answer, or an even more elaborate explanation at the bottom of the Code Review question. Without the string compression, the code looks like this:

"Enter String: ÿ","Vowel Count: "?"aeiou"©vy": "¹y¢««}", "ý,"Remaining characters: "¹®-J,

Or a more readable 3-part version:

"Enter String: ÿ",
"Vowel Count: "?"aeiou"©vy": "¹y¢««}", "ý,
"Remaining characters: "¹®-J,

Part 1:

We first push the string "Enter String: ÿ". The ÿ here is used for string interpolation. But since the stack is empty, implicit input is requested. The ÿ will be substituted with the user input. We print this with a newline using ,.

Part 2:

We first print the "Vowel Count: " string without a newline using ?. After that, we push the aeiou string, ©opy that to the register and map over the string using v and do the following:

vy": "¹y¢««}    # Mapping loop

v          }    # The loop itself
 y              # Push the current letter (which is a, e, i, o or u)
  ": "          # Push this string
      ¹         # Push the first input again
       y        # Push the letter again
        ¢       # Count the number of occurences of that letter in the input
         ««     # Concatenate to one single string, (e.g. "a: 4")

This leaves 5 different string onto the stack, the stack looks like this now:

stack: ['a: 0', 'e: 1', 'i: 0', 'o: 2', 'u: 0']

We join each element in the stack with ", " using ý. After this, we print it with a newline using ,.

Part 3:

We first push the string "Remaining characters: ". We push the first input again using ¹. We then push "aeiou" which we retrieve from the register using ®. After this, we remove all letters in aeiou using - and Join this with the Remaining characters-string. Finally, we output this with a newline using ,.

Uses CP-1252 encoding. Try it online!.

Answer 2

Python, 168 bytes

s=input('Enter String:')
v='aeiou'
print('Vowel Count: '+', '.join([a+':'+str(s.count(a)) for a in v])+'\nRemaining characters: '+''.join([a for a in s if a not in v]))

https://repl.it/CMzO/1

Answer 3

JavaScript (ES6), 165

Fixed error in output, thx @Neil
3 bytes saved thx @Dom Hastings

a=prompt`Enter String:`.replace(/[aeiou]/g,v=>(V[v]=-~V[v],''),V=[]);alert("Vowel Count: "+[...'aeiou'].map(v=>v+': '+(V[v]|0)).join`, `+`
Remaining characters: `+a)

Answer 4

Java, 272 bytes

class L{public static void main(String[]v){String p=v[0],o="Enter String: "+p+"\nVowel Count:",d=" ",e="aeiou";for(char c:e.toCharArray()){o+=d+c+": "+p.replaceAll("[^"+c+"]","").length();d=", ";}System.out.println("\nRemaining characters: "+p.replaceAll("["+e+"]",""));}}

Ungolfed

class L {
    public static void main(String[] v) {
        String input = v[0], result = "Enter String: " + input + "\nVowel Count:", delimiter = " ", vowels = "aeiou";
        for (char c : vowels.toCharArray()) {
            result += delimiter + c + ": " + input.replaceAll("[^" + c + "]", "").length();
            delimiter = ", ";
        }
        System.out.println("\nRemaining characters: " + input.replaceAll("[" + vowels + "]", ""));
    }
}

Notes

• Input via program arguments: Save as L.java, compile with javac L.java, run with java L LoremIpsumDolorSitAmetConsecuteurAdipscingElit!

Output

Enter String: LoremIpsumDolorSitAmetConsecuteurAdipscingElit!
Vowel Count: a: 0, e: 4, i: 4, o: 4, u: 3
Remaining characters: LrmIpsmDlrStAmtCnsctrAdpscngElt!

Answer 5

Lua, 242 Bytes

For a moment, I thought that it will be longer than SQL... still a monster for such a simple task.

t={}x="aeiou"i=io.write 
x:gsub(".",function(c)t[c]=0 end)i("Enter string: ")z=io.read()s="Remaining characters: "..z:gsub(".",function(c)if x:find(c)then
t[c]=t[c]+1return""end end)i("Vowel Count: ")x:gsub(".",function(c)i(c..": "..t[c]..(c=="u"and"\n"or", "))end)i(s)

The job is done by the part

z:gsub(".",function(c)if x:find(c)then t[c]=t[c]+1return""end end)

Everything else is mainly for the purpose of outputing in the right format.

Ungolfed

t={}                             -- create an empty array
x="aeiou"                        -- list of vowels
i=io.write                       -- shorthand for outputing
x:gsub(".",function(c)t[c]=0 end)-- create a celle at 0 for each vowel
i("Enter string: ")              -- output some boilerplate
z=io.read()                      -- take the input
s="Remaining characters: "..     -- save in s the concatenation of this string
  z:gsub(".",function(c)         -- and the result of gsubing on each character of the input 
  if x:find(c)                   -- if the current char is a vowel
  then
    t[c]=t[c]+1                  -- increment its counter
    return""                     -- suppress it from the input
  end
end)

i("Vowel Count: ")               -- output some more boilerplate
x:gsub(".",function(c)           -- iterate over the vowels
  i(c..": "..t[c]..              -- output "l: #" appended with
    (c=="u"
        and"\n"                  -- "\n" if the current char is "u"
      or", "))                   -- ", " otherwise
end)
i(s)                             -- output the content of s

Answer 6

Perl 5, 127 bytes

$_=<>;print"Enter String: ${_}Vowel Count:";for$s(a,e,i,o,u){print" $s: ".~~s/$s//g,$s!~u&&","}print"
Remaining characters: $_"

Many thanks to Dom Hastings (in a comment hereon) for 14 bytes.

Answer 7

PostgreSQL, 295, 263 bytes

SELECT v "Enter string:", string_agg(r,', 'ORDER BY n) "Vowel Count:",TRANSLATE(v,'aeyiou','') "Remaining characters:"
FROM (SELECT 'LoremIpsumDolorSitAmetConsecuteurAdipscingElit!'::text v)s
,LATERAL(SELECT LENGTH(v)l)a
,LATERAL(SELECT c||': '||l-LENGTH(REPLACE(v,c,''))r,n FROM (VALUES('a',1),('e',2),('i',3),('o',4),('u',5))t(c,n))b
GROUP BY v

SqlFiddleDemo

Output:

╔══════════════════════════════════════════════════╦═══════════════════════════════╦══════════════════════════════════╗
║                  Enter string:                   ║         Vowel Count:          ║      Remaining characters:       ║
╠══════════════════════════════════════════════════╬═══════════════════════════════╬══════════════════════════════════╣
║ LoremIpsumDolorSitAmetConsecuteurAdipscingElit!  ║ a: 0, e: 4, i: 4, o: 4, u: 3  ║ LrmIpsmDlrStAmtCnsctrAdpscngElt! ║
╚══════════════════════════════════════════════════╩═══════════════════════════════╩══════════════════════════════════╝    

Input: (SELECT '...'::text v)s

How it works:

• (SELECT 'LoremIpsumDolorSitAmetConsecuteurAdipscingElit!'::text v)s input
• ,LATERAL(SELECT LENGTH(v)l)a - length of input
• (VALUES('a',1),('e',2),('i',3),('o',4),('u',5)) - vowels with oridinal
• SELECT c||': '||l-LENGTH(REPLACE(v,c,''))r,n - calculate number of occurences
• string_agg(r,', 'ORDER BY n) ... GROUP BY v combine result
• TRANSLATE(v,'aeyiou','') string without vowels

EDIT 3:

Removed additional LATERAL, input moved to table.

SELECT v"Enter string:",string_agg(r,', 'ORDER BY n)"Vowel Count:",TRANSLATE(v,'aeyiou','')"Remaining characters:"
FROM s,LATERAL(SELECT c||': '||LENGTH(v)-LENGTH(REPLACE(v,c,''))r,n FROM(VALUES('a',1),('e',2),('i',3),('o',4),('u',5))t(c,n))b
GROUP BY v

SqlFiddleDemo

Answer 8

Python, 220 201 bytes

This is my first submission on codegolf so go easy. :P

def f(y):
  global x;c=x.count(y);x=x.replace(y,'');return c;
x=input('Enter string: ');print('Vowel Count: a: {}, e: {}, i: {}, o: {}, u: {}'.format(*map(f,"aeiou")));print('Remaining characters: '+x)

Ungolfed:

def f(y):
  global x
  c=x.count(y)
  x=x.replace(y,'')
  return c;
x=input('Enter string: ')
print('Vowel Count: a: {}, e: {}, i: {}, o: {}, u: {}'.format(*map(f,"aeiou")))
print('Remaining characters: '+x)

Output:

Enter string: LoremIpsumDolorSitAmetConsecuteurAdipscingElit!
Vowel Count: a: 0, e: 4, i: 4, o: 4, u: 3
Remaining characters: LrmIpsmDlrStAmtCnsctrAdpscngElt!

Answer 9

JavaScript, 190 186 bytes

(s,a=[..."aeiou"].map(v=>(s.match(v,"g")||[]).length))=>`Enter String: ${s}
Vowel Count: ${[..."aeiou"].map((v,i)=>v+": "+a[i]).join`, `}
Remaining characters: `+s.replace(/[aeiou]/g,"")

I just remembered that literal newlines are only 1 byte! Also remembered to golf something I missed before.

Answer 10

PowerShell v2+, 196 149 bytes

$z=read-host "Enter string"
"Vowel Count: "+(([char[]]"aeiou"|%{$_+": "+($z.length-($z=$z-creplace$_).length)})-join', ')
"Remaining characters: "+$z

Takes input via read-host with a prompt, save into $z.

We then output a string literal "Vowel Count: " concatenated with the result of a loop/join through the vowels to construct the a: #, e: #,... string. This is done by taking the current string's .length minus the new string's .length after we perform a case-sensitive replacement with -creplace (i.e., how many did we remove). In addition, we re-save the result of the replace back into $z for the next loop.

We then also output the second string literal concatenated with whatever is still left in $z.

This keeps the current specification intact, in that capital vowels are distinguished from lowercase vowels, since -creplace is explicitly case sensitive.

Examples

PS C:\Tools\Scripts\golfing> .\briefest-code-find-vowels.ps1
Enter string: LoremIpsumDolorSitAmetConsecuteurAdipscingElit!
Vowel Count: a: 0, e: 4, i: 4, o: 4, u: 3
Remaining characters: LrmIpsmDlrStAmtCnsctrAdpscngElt!

PS C:\Tools\Scripts\golfing> .\briefest-code-find-vowels.ps1
Enter string: ProgrammingPuzzlesAndCodeGolf
Vowel Count: a: 1, e: 2, i: 1, o: 3, u: 1
Remaining characters: PrgrmmngPzzlsAndCdGlf

Answer 11

Retina, 133 bytes

(a()|e()|i()|o()|u()|.)+
Enter string: $&¶Vowel count: a: $#2, e: $#3, i:$#4, o:$#5, u:$#6¶Remaining characters: $&
(?<=s:.+)[aeiou]
<empty-line>

Try it online!

Who is more fit for this challenge than Retina?

Answer 12

Groovy, 181 177 bytes

online

​{i->r=[:].withDefault{0};o=i.toLowerCase().replaceAll(/[aeiou]/,{r[it]++;''})
"Enter String: $i\nVowel Count: ${r.collect{k,v->"$k: $v"}.join(' ')}\nRemaining characters: $o"}

Answer 13

Tcl, 184 (counting majuscule vowels) / 180 (ignoring them)

puts -nonewline {Enter String: }
flush stdout
gets stdin s
set c {Vowel Count:}
foreach v {a e i o u} {append c " $v: [regsub -all (?i)$v $s {} s]"}
puts "$c\nRemaining Characters: $s"

Explanation

Most of the code is actually dealing with the strict I/O requirements. The remainder works by doing a regular expression substitution (deletion) on the input string $s for every vowel and appending the vowel $v and associated number of substitutions to the output string $c.

The above code works for mixed-case vowels. Remove the (?i) option from the regular expression on the penultimate line for strictly minuscule vowels.

Answer 14

Python 162 Bytes

v,s="aeiou",input()
print "String:"+s+"\n"+"Vowel Count"+" ".join([c+":"+str(s.count(c)) for c in v])+"\n"+"Remaining Characters:"+filter(lambda x:v.count(x)<1,s)

Answer 15

C# 317 bytes

string t=Console.ReadLine();Dictionary v = new Dictionary<char,int>{{'a',0},{'e',0},{'i',0},{'o',0},{'u',0}};string o="";foreach(char c in t){if(v.Keys.Contains(c)){v[c]++;}else{o+=c;}}Console.WriteLine("a: "+v['a']+", e: "+v['e']+", i: "+v['i']+", o: "+v['o']+", u: "+v['u']);Console.WriteLine("Remaining Characters: "+o);

Any suggestions, trying to learn to golf with C#. 1KB is obviously way bigger than the other answers though.

Ungolfed:

string t=Console.ReadLine();
Dictionary v = new Dictionary<char,int>
{
    {'a',0},
    {'e',0},
    {'i',0},
    {'o',0},
    {'u',0}
};
string o="";
foreach(char c in t)
{
    if(v.Keys.Contains(c))
    {
        v[c]++;
    }
    else
    {
        o+=c;
    }
}
Console.WriteLine("a: "+v['a']+", e: "+v['e']+", i: "+v['i']+", o: "+v['o']+", u: "+v['u']);
Console.WriteLine("Remaining Characters: "+o);

Answer 16

Ruby, 202 bytes

print"Enter String: ";t=gets;c=->(x){t.chars.count(x)};a="aeiou".chars
s=a.zip(a.map(&c)).map{|e|"#{e[0]}: #{e[1]}"}.join(", ");r=t.gsub(/[aeiou]/,"")
puts"Vowel Count: #{s}\nRemaining characters: #{r}"

Answer 17

C++, 449 bytes

Golfed:

#include <iostream>
#include <string>
int main(){std::string x;std::cout<<"Enter String: ",std::cin>>x;int a=0,e=0,i=0,o=0,u=0;for(int c=0;c<x.length();c++){switch(x[c]){case 'a':a++;x.erase(c,1);break;case 'e':e++;x.erase(c,1);break;case 'i':i++;x.erase(c,1);break;case 'o':o++;x.erase(c,1);break;case 'u':u++;x.erase(c,1);break;}}std::cout<<"\n Vowel Count: a: "<<a<<", e: "<<e<<", i: "<<i<<", o: "<<o<<", u: "<<u<<"\n Remaining characters: "<<x;}

Ungolfed:

#include <iostream>
#include <string>

int main(){
    std::string x;
    std::cout<<"Enter String: ", std::cin>>x;
    int a=0,e=0,i=0,o=0,u=0;
    for(int c = 0; c < x.length(); c++){
        switch(x[c]){
            case 'a':
                a++;
                x.erase(c,1);
                break;
            case 'e':
                e++;
                x.erase(c,1);
                break;
            case 'i':
                i++;
                x.erase(c,1);
                break;
            case 'o':
                o++;
                x.erase(c,1);
                break;
            case 'u':
                u++;
                x.erase(c,1);
                break;
        }
    }
    std::cout<<"\n Vowel Count: a: "<<a<<", e: "<<e<<", i: "<<i<<", o: "<<o<<", u: "<<u<<"\n Remaining characters: "<<x;
}

Answer 18

C++17, 313 bytes

#include <cctype>
#include <iostream>
#include <string>
using namespace std;int main(){int n,h[6]={0};string s,v="aeiou",r;cout<<"Enter String: ";getline(cin,s);for(c:s){h[n=v.find(tolower(c))+1]++;if(!n)r+=c;}cout<<"Vowel Count:";n=1;for(c:v)cout<<" "<<c<<": "<<h[n++];cout<<"\nRemaining Characters: "<<r<<"\n";}

I suppose I could make this just a function to eliminate standard boilerplate from the byte count.

Ungolfed

#include <cctype>
#include <iostream>
#include <string>
using namespace std;

int main()
{
  int n;
  int histogram[6] = {0};   // { (consonants), A, E, I, O, U }
  string source;
  string vowels = "aeiou";  // vowel order matches the histogram
  string consonants;

  cout << "Enter String: ";
  getline( cin, source );

  for (c : source)
  {
    // n ← consonants=0, A=1, E=2, ...
    n = vowels.find( tolower(c) ) + 1;
    histogram[n]++;
    if (!n)
      consonants += c;
  }

  cout<<"Vowel Count:";
  n = 1;
  for (vowel : vowels)
    cout << " " << vowel << ": " << histogram[n++];

  cout << "\nRemaining Characters: " << consonants << "\n";
}

Explanation

Computes a histogram using the std::string lookup functions.

Again, you can eliminate 27 bytes from the count if only minuscule vowels matter. (Remove the first #include and the call to tolower().)

Answer 19

Ruby, 134 133 131 bytes

Because the other Ruby answer didn't respond to my suggestions, I golfed it down further myself.

-2 bytes from @manatwork

$><<"Enter String: "
r=gets.tr v='aeiou',''
puts"Vowel Count: #{v.chars.map{|e|e+": #{$_.count e}"}*', '}
Remaining characters: "+r