String and ASCII [duplicate]

Question

The Challenge

Given a non-empty string containing only lowercase or uppercase letters and no spaces:

• Sum up the ASCII values for each character instance in the input string.

• The sum will be converted to binary.

• And, the result will be the number of ones in the binary value.

Input example:

abcDeFAaB

1. Sum up the ASCII values for each character instance, so for this example: 97+98+99+68+101+70+65+97+66 = 761.

2. Convert the sum (761) to binary. This will give 1011111001.

3. The final result will be the number of ones (7).

So the output for this example is 7.

Winning Criterion

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

Answer 1

Jelly, 4 bytes

OSBS

TryItOnline

How?

OSBS - Main link: s     e.g. "abcDeFAaB"
O    - ordinal (vectorises)  [97, 98, 99, 68, 101, 70, 65, 97, 66]
 S   - sum                   761
  B  - binary                [1, 0, 1, 1, 1, 1, 1, 0, 0, 1]
   S - sum                   7

Answer 2

Python3, 46 40 bytes

lambda s:bin(sum(map(ord,s))).count("1")

Answer 3

MATL, 3 bytes

sBs

Try it online!

Shortest language for this challenge, I think: sum, Binary, sum.

Answer 4

R, 65 53 43 42 bytes

Counting the number of 1s in a binary representation is calculating the Hamming weight of a number, a common procedure in cryptanalysis. There's an R package boolfun, described in this journal article, which handily has a function weight which computes the Hamming weight of any (base 10) integer. That makes our job a lot easier! The rest of the code just sums over the integer representation of STDIN, before sending that to weight.

boolfun::weight(sum(utf8ToInt(scan(,'')))

As an added bonus, this code works for UTF-8 as well as ASCII. There are 7 ones in the binary representation of the sum of 高尔夫 (that's golf in Mandarin).

A note for those playing along at home: unfortunately, boolfun is no longer hosted on CRAN, so if you don't already have it installed you'll have to do it manually with install.packages("https://cran.r-project.org/src/contrib/Archive/boolfun/boolfun_0.2.8.tar.gz", repos = NULL, type = "source"), rather than install.packages(boolfun).

If you really don't want to use boolfun, you can achieve the same result with only two more bytes:

sum(intToBits(sum(utf8ToInt(scan(,''))))==1)

Answer 5

05AB1E, 6 5 bytes

ÇObSO

Explanation

 Ç      # convert to list of ascii values
  O     # sum
   b    # convert to binary
    SO  # sum as a list

Try it online!

Answer 6

Bash + coreutils, 43 49 bytes

Edit: corrected version based on manatwork's comments.

od -vAn -tuC|tr '\n ' +|dc -e?2op|grep -o 1|wc -l

Explanation:

The result is printed on STDOUT, with some warnings given on STDERR which can be ignored.

od -vAn -tuC       # get the ASCII value for each string character
|tr '\n ' +        # replace each whitespace with a plus (in dc, a warning will be 
                   #raised for each consecutive plus)
|dc -e?2op         # execute input, calculating the sum, then convert to binary
|grep -o 1|wc -l   # count how many ones exist

Run:

echo -n "abcDeFAaB" | ./script.sh 2> /dev/null

Output:

7

A similar solution is presented below (51 bytes), that had the potential to be shorter, but with xxd I can't concatenate multiple flags and also dc only understands upper case hex letters.

xxd -c1 -u -p|tr \\n +|dc -e16i?2op|grep -o 1|wc -l

Answer 7

Java 7,80,76 bytes

Thanks to kevin for saving 4 bytes

int f(char[]a){int s=0,c=0;for(char i:a)s+=i;for(;s>0;s&=s-1,c++);return c;} 

Ungolfed

int f(char[]a){
    int s=0,c=0;
    for(char i:a)
        s+=i;
    for(;s>0;s&=s-1,c++);
        return c;
} 

Answer 8

PHP, 79 76 bytes

foreach(str_split($argv[1])as$c)$s+=ord($c);echo substr_count(decbin($s),1);

Test online

Ungolfed testing code:

$argv[1] = 'abcDeFAaB';

foreach(str_split($argv[1]) as $c){
    $s += ord($c);
}
echo substr_count(decbin($s),1);

Test online

Answer 9

R, 98 bytes

Was interested in how difficult this would be in R. Turns out you have to do a lot of conversions back and forth (in this solution at least) but I'm sure there should be a shortcut, especially if using packages (can't access any here at work).

sum(as.integer(rev(intToBits(sum(strtoi(sapply(strsplit(readline(),"")[[1]],charToRaw),16L)))))^2)

Could add an explanation if someone is interested.

Answer 10

><>, 34 bytes

01\
(?\i:@+$0
?!\:2%:}-2,r+$:
n;\~

Try it online!

Answer 11

Haskell, 47 bytes

g 0=0;g n=mod n 2+g(div n 2)
g.sum.map fromEnum

Answer 12

Perl, 36 bytes

Includes +1 for -p

Run with input string without newline on STDIN:

echo -n abcDeFAaB | strsum.pl; echo

strsum.pl:

#!/usr/bin/perl -p
$_=unpack"%b*",pack N,unpack"%32W*"

If the sum of the ASCII values fits in a 16-bit integer you can leave out the 32 and gain 2 more bytes.

Answer 13

Julia, 56 Bytes

f(s)=length(filter(i->i=='1',bin(sum(map(Int,[s...])))))

Convert to ascii value, sum up, filter for 1s, take length of result

Answer 14

Common Lisp, 53

(lambda(s)(logcount(reduce'+(map'list'char-code s))))

Answer 15

JavaScript ES6, 80 78 71 Bytes

g=
    s=>[...s].map(c=>r+=c.charCodeAt(),r=0)|r.toString(2).split`1`.length-1
;

console.log(g.toString().length);    // 71
console.log(g('abcDeFAaB'))          // 7

Saved 7 Bytes thanks to @BassdropCumberwubwubwub

Breakdown

 [...s]                              // Split string into array of chars
 .map(c=>r+=c.charCodeAt(),r=0)      // Sum up ASCII values
 |
 r.toString(2)                       // Convert to binary (represented as string)  
 .split`1`.length-1                  // Count occurences of 1

Answer 16

PHP, 73 bytes

<?=substr_count(decbin(array_sum(array_map(ord,str_split($_GET[s])))),1);

save to file, call in browser with <scriptpath>?s=<string>

or

<?=substr_count(decbin(array_sum(array_map(ord,str_split($argv[1])))),1);

save to file, run with php <scriptpath> <string>

---

To run without a file, replace <?= with echo<space> in the second version
and run with php -r '<code>' <string>.

Answer 17

C#, 51 50 bytes

s=>Convert.ToString(s.Sum(x=>x),2).Count(x=>x>48);

can be used like this:

Func<string, int> f=s=>Convert.ToString(s.Sum(x=>x),2).Count(x=>x>48);
f("abcDeFAaB") // == 7

would be so much shorter if Int32.ToString() had a to binary flag.

Edit: saved one byte by converting '0' to int

Answer 18

Javascript (ES6), 72 70 bytes

Saved 2 bytes thanks to ETHproductions:

f=
  (s,p=([c,...d])=>c?c.charCodeAt()+p(d):0,q=n=>n&&n%2+q(n>>1))=>q(p(s))
;
console.log(f('abcDeFAaB'));

Explanation:

(s,                                        //Input string
    p=([c,...d])=>c?c.charCodeAt()+p(d):0, //Recursive method to sum up the ASCII values
    q=n=>n&&n%2+q(n>>1)                    //Recursive method to count 1s in the binary form
)=>q(p(s))                                 

---

Previous solution (72 bytes):

(s,p=([c,...d])=>c?c.charCodeAt()+p(d):0,q=n=>n&&(n&1)+q(n>>1))=>q(p(s))

Answer 19

Pyth, 8 bytes

ssM.BsCM

Try it online!

Explanation:

      CM  (M)ap ordinals(C) over input
     s    (s)um
   .B     Binary string
 sM       map cast to integer
s         sum

Answer 20

Mathematica, 53 37 bytes

not the best solution but hey...built-in's ftw

Shortened. Thanks to Martin Ender

DigitCount[Tr@ToCharacterCode@#,2,1]&