# Calculate average characters of string

Your task is to produce string that contains average characters of string. First character of result would be average character of first character (which is first character) and second character average of two first characters and so on.

# What is average character?

Strings are arrays of bytes. Average character of string can be found by calculating the average of the ASCII values of characters in string and taking corresponding ASCII character.

For example string "Hello!" can be written as byte sequence 72 101 108 108 111 33. Average of ascii values is 533/6 = 88.833... and when it's rounded to nearest integer we get 89 which is ascii code for captial letter Y.

# Rules

• You can assume that input contains only printable ASCII characters
• Input can be read from stdin or as command line arguments or as function arguments
• Output must be stdout. If your program is function, you can also return the string you would otherwise print.
• It must be whole program or function, not snippet
• Standard loopholes apply
• Integers are rounded by function floor(x+0.5) or similar function.

# How do I win?

This is , so shortest answer (in bytes) in wins.

# Examples

• Hello!HW^adY
• testtmop
• 4243
• StackExchangeSdccd_ccccddd
• Edited question. Now it should be clear: you have to round halves upwards. Jul 24 '15 at 9:46
• "Input can be read from stdin or as command line arguments": or as function arguments (since you allow functions), right? Jul 24 '15 at 12:17
• Of course, edited again. Jul 24 '15 at 13:36
• Sorry to bother you once again, but do functions actually have to print the output to STDOUT or can they return the desired string? Jul 24 '15 at 14:20
• Sorry, forgot to edit that before. Now it should be ok. Jul 24 '15 at 20:41

## Brainfuck 106 bytes

,[>,<.[->+<]>>++<[>[->+>+<<]>[-<<-[>]>>>[<[>>>-<<<[-]]>>]<<]>>>+<<[-<<+>>]<<<]>[-]>>>>[-<<<<<+>>>>>]<<<<<]


This is my first participation in a code-golf, please be gentle! It does work but brainfuck can't handle floats (not that i know of) so the rounded value is always the bottom one (might fix my algorithm later).

Also, the algorithm averages the values 2 by 2, meaning it could be innacurate in some spots. And I need to fix a bug that is printing a number at the end of the output too.

# Pyth, 16 bytes

smCs+.5csaYCdlYz


Pretty straightforward. Using s+.5 instead of rounding, because for some reason round(0.5, 0) is 0 in Python.

• Python 3 rounds half towards even, which introduces less bias. The question doesn't explicitly specify how halves should be rounded, so I've requested clarification from the OP. Jul 24 '15 at 4:35
• Edited question. 0.5 rounded should be 1. Jul 24 '15 at 7:54

# Q, 15 12 bytes

12 bytes as an expression

"c"$avgs"i"$

q)"c"$avgs"i"$"Hello!"
q)"c"$avgs"i"$"test"
"tmop"
q)"c"$avgs"i"$"42"
"43"
q)"c"$avgs"i"$"StackExchange"
"Sdccd_ccccddd"


or 15 bytes as a function

{"c"$avgs"i"$x}

q){"c"$avgs"i"$x} "Hello!"
q){"c"$avgs"i"$x} "test"
"tmop"
q){"c"$avgs"i"$x} "42"
"43"
q){"c"$avgs"i"$x} "StackExchange"
"Sdccd_ccccddd"


1. the "i"$cast to convert a string (list of characters) to a list of integers 2. the avgs function, which computes the running average of a list as a list of floats 3. the "c"$ cast to convert a list of floats to a list of characters, and which automatically rounds each float to the nearest integer before doing so [i.e. ("c"$99.5) = ("c"$100) and ("c"$99.4) = ("c"$99) ]
• Does Q require the function wrapper here or can you get away with just the tacit expression "c"$avgs"i"$ ? I don't think a solution could get much more straightforward than that. :) Jul 24 '15 at 15:57
• you're correct -- no need for the function wrapper, as "c"$avgs"i"$ "Hello!" works fine Jul 24 '15 at 16:15
• I think you can save 2 bytes by changing "c" to c and "i" to i. Jul 24 '15 at 20:10
• unfortunately, i don't think that works. To use the symbol type representation for casting I'd have to use char and int as per code.kx.com/wiki/JB:QforMortals2/… I considered using 10h and 6h instead of "c" and "i" but that wouldn't save any bytes -- 10h is the same length as "c" and substituting 6h for "i" requires a trailing space, making them the same length also. Jul 24 '15 at 20:25

# CJam, 19 bytes

Uq{i+_U):Ud/moco}/;


Try it online in the CJam interpreter.

# Perl: 31 30 characters

(29 characters code + 1 character command line option.)

s!.!chr.5+($s+=ord$&)/++$c!ge  Sample run: bash-4.3$ perl -pe 's!.!chr.5+($s+=ord$&)/++$c!ge' <<< 'StackExchange' Sdccd_ccccddd  # C# 189 135 134 106 Bytes var x=s.Select((t,i)=>Math.Round(s.Select(a=>(int)a).Take(i+1).Average())).Aggregate("",(m,c)=>m+(char)c);  Can be seen here First time golfer # K, 36 bytes 0:_ci_.5+{(+/x)%#x}'.0+1_|(-1_)\_ic  Usage:  0:_ci_.5+{(+/x)%#x}'.0+1_|(-1_)\_ic"Hello!" HW^adY 0:_ci_.5+{(+/x)%#x}'.0+1_|(-1_)\_ic"test" tmop 0:_ci_.5+{(+/x)%#x}'.0+1_|(-1_)\_ic"42" 43 0:_ci_.5+{(+/x)%#x}'.0+1_|(-1_)\_ic"StackExchange" Sdccd_ccccddd  _ci and _ic convert ascii to chars and vice versa, respectively. {(+/x)%#x} is a classic K idiom for calculating a mean. Pretty straightforward overall. Edit: oh, misread the spec. 0: is needed to print the result to stdout. Waiting for clarification on input re. Dennis' question. • If you use K5, this can be shortened to 35 bytes: {[s]0:c${_.5+(+/u)%#u:x#s}'1+!#s}. Jul 24 '15 at 14:22
• Or 30: 0:c$_.5+{(+/x)%#x}'1_|(-1_)\  Jul 24 '15 at 14:51 • c$_.5+{(+\x)%+\~^x}i$ for 24. It could be shorter (c$i${(+\x)%+\~^x}i$) but your REPL (johnearnest.github.io/ok/index.html) isn't rounding correctly when casting from float to int. I'd hesitate to call this a k solution since _ci and _ic are nowhere in the K5 spec as far as I can tell, while 0: doesn't print to stdout but instead reads a txt file from disk. Jul 24 '15 at 15:10
• @tmartin: correct- _ci and _ic are entirely replaced in K5 with the forms like c$. The original solution I posted is compatible with Kona, which is based on K2/K3. I generally try not to post solutions with oK specifically because the semantics are still shifting around and partially inaccurate. Jul 24 '15 at 15:18 • Ah i see, makes sense to me. I figured this was another K5 solution. Here's a 28 char kona solution 0:_ci_0.5+{(+\x)%1.+!#x}_ic Jul 24 '15 at 15:30 # Mathematica, 75 bytes FromCharacterCode@Floor[.5+Accumulate@#/Range@Length@#]&@ToCharacterCode@#&  # Julia, 85 81 bytes s->(i=[int(c)for c=s];print(join([char(iround(mean(i[1:j])))for j=1:length(i)])))  This creates an unnamed function that accepts a string and creates a vector of its ASCII code points. Means are taken for each sequential group, rounded to integers, converted to characters, joined into a string, and printed to STDOUT. # Ruby, 46 s=0.0$<.bytes{|b|s+=b;$><<'%c'%(0.5+s/$.+=1)}


With apologies to w0lf, my answer ended up different enough that it seemed worth posting.

$<.bytes iterates over each byte in stdin, so we print the rolling average in each loop. '%c' converts a float to a character by rounding down and taking the ASCII, so all we have to do is add 0.5 to make it round properly. $. is a magic variable that starts off initialized to 0--it's supposed to store the line count, but since here we want byte count we just increment it manually.

Mathcad is mathematical application based on 2D worksheets comprised of "regions" each of which can be text, a mathematical expression, program, plot or scripted component.

A mathematical or programming instruction is picked from a palette toolbar or entered using a keyboard shortcut. For golfing purposes, an operation ("byte") is taken to be the number of keyboard operations necessary to create a name or expression (for example, to set the variable a to 3, we would write a:=3. The definition operator := is a single keypress ":", as are a and 3 giving a total of 3 "bytes". The programming for operator requires typing ctl-shft-# (or a single click on the programming toolbar) so again is equivalent to 1 byte.

In Mathcad the user enters programming language commands using keyboard shortcuts (or picking them from the Programming Toolbar) rather than writing them in text. For example, typing ctl-] creates a while-loop operator that has two "placeholders" for entering the condition and a single line of the body, respectively. Typing = at the end of a Mathcad expressions causes Mathcad to evaluate the expression.

(Count bytes) By looking at it from a user input perspective and equating one Mathcad input operation (keyboard usually, mouse-click on toolbar if no kbd shortcut) to a character and interpreting this as a byte. csort = 5 bytes as it's typed char-by-char as are other variable/function names. The for operator is a special construct that occupies 11 characters (including 3 blank "placeholders" and 3 spaces) but is entered by ctl-shft-#, hence = 1 byte (similar to tokens in some languages). Typing ' (quote) creates balanced parentheses (usually) so counts as 1 byte. Indexing v = 3 bytes (type v[k).

# JavaScript (ES6), 75 bytes

let f =
s=>s.replace(/./g,x=>String.fromCharCode((t+=x.charCodeAt())/++i+.5),i=t=0)
<input oninput="O.value=f(this.value)" value="Hello!"><br>
<input id=O value="HW^adY" disabled>

I can't believe there's no JS answer with this technique yet...

# Python 3, 65 bytes

n=t=0
for c in input():n+=1;t+=ord(c);print(end=chr(int(.5+t/n)))


Try it online!

If I use round() instead of int(.5+ etc., it saves one character but is technically not in compliance with the challenge: Python's round() rounds halves to the nearest even integer, not upwards. However, it works correctly on all sample inputs.

• If you increment n before printing, you can avoid adjusting it by 1.
– xnor
Jul 24 '15 at 16:35
• @xnor: Face, palm. Palm, face. Thanks for pointing that out. Jul 24 '15 at 17:26
• do print(end=chr(int(...)) to save some bytes Dec 11 '16 at 20:12
• @Flp.Tkc: Thanks! Answer updated. Dec 20 '16 at 17:34

## Python 2, 71

i=s=0
r=''
for c in input():s+=ord(c);i+=1.;r+=chr(int(s/i+.5))
print r


With each new character, updates the character sum s and the number of characters i to compute and append the average character.

• Almost exactly the same approach as mine, only Python 2 instead of 3, and posted hours earlier: +1. (Also, I found I saved a few bytes printing each character as it came rather than storing them for one final print. Will that work with Python 2? I forget just now how to suppress newlines in the print statement... comma makes it a space instead, right?) Jul 25 '15 at 6:06
• Python 2 can do print _, to leave a space rather than newline, but no good way to omit the space. Good call with Python 3's end argument, I'd forgotten about that.
– xnor
Jul 25 '15 at 6:17
• @TimPederick Maybe the backspace control could be justified for a terminal that uses it, as a hack to delete the space: print'\b'+_,
– xnor
Jul 25 '15 at 6:25

### Ruby 59 61

->w{s=c=0.0;w.chars.map{|l|s+=l.ord;(s/c+=1).round.chr}*''}

• c+=1;(s/c)(s/c+=1) ideone.com/H2tB9W Jul 24 '15 at 9:38
• @manatwork Well spotted! Thanks! I applied the change. Jul 24 '15 at 10:13

# Java, 100

Much like many other answers here, I'm summing and averaging in a loop. Just here to represent Java :)

void f(char[]z){float s=0;for(int i=0;i<z.length;System.out.print((char)Math.round(s/++i)))s+=z[i];}


My original code is a 97, but it only returns the modified char[] rather than printing it:

char[]g(char[]z){float s=0;for(int i=0;i<z.length;z[i]=(char)Math.round(s/++i))s+=z[i];return z;}


Now, it's just long enough for scrollbars to appear for me, so here's a version with some line breaks, just because:

void f(char[]z){
float s=0;
for(int i=0;
i<z.length;
System.out.print((char)Math.round(s/++i)))
s+=z[i];
}

• Interesting. Can you show us a call sample too? My Java is very rusty. Jul 24 '15 at 13:49
• As in how to call it? Assuming test is a char array, just use f(test);. If it's a String object, then you'd use f(test.toCharArray());. String literals are fine like that, too: f("Hello!".toCharArray()); Jul 24 '15 at 13:53
• Oh. Sure. toCharArray() Me stupid, I tried to violate it with some casting. Thank you. Jul 24 '15 at 13:58
• It'd be way too easy to just cast it. The Java gods would be furious :P Jul 24 '15 at 13:59

# C, 62 bytes

c;t;main(n){for(;(c=getchar())>0;n++)putchar(((t+=c)+n/2)/n);}


The results are slightly different from the OP's examples, but only because this code rounds 0.5 down instead of up. Not any more!

# R, 135 127 Bytes

This got long real quick and I really got it wrong the first time:) Need to read the questions properly.

cat(sapply(substring(a<-scan(,''),1,1:nchar(a)),function(x)rawToChar(as.raw(round(mean(as.integer(charToRaw(x)))+.5)))),sep='')


Test Run

cat(sapply(substring(a<-scan(,''),1,1:nchar(a)),function(x)rawToChar(as.raw(round(mean(as.integer(charToRaw(x)))+.5)))),sep='')
1: Hello!
2:

• somebody posted a dupe challenge in the sandbox so I found this...This was a looong time ago, but utf8ToInt will help! I have a 68 byte golf of this if you want to update to it. Jan 18 '18 at 17:31
• @Giuseppe Go ahead and post it yourself if you would like. I suspect that it is significantly different to what I did here. Jan 18 '18 at 18:41

# Perl 5, 41 bytes

say map{$s+=ord;chr($s/++$c+.5)}pop=~/./g  run as $ perl -E 'say map{$s+=ord;chr($s/++$c+.5)}pop=~/./g' StackExchange Sdccd_ccccddd  ## TSQL, 118 bytes DECLARE @ varchar(400) = 'StackExchange' SELECT top(len(@))char(avg(ascii(stuff(@,1,number,''))+.5)over(order by number))FROM master..spt_values WHERE'P'=type  Returning characters vertical S d c c d _ c c c c d d d  # ><>, 30 bytes i:0(?v v &l~< \+l2(? \&,12,+o;  • The first line reads from stdin and puts the characters on the stack • The second will remove the EOL char, take the size of the stack and put it in the & register • The third line will add numbers on the stack while there are two or more of them • The fourth line will divide the resulting number by the register's value, then add 1/2, output the value as a character and stop. When faced with a float value when displaying a char, ><> will floor it, which is why we added 1/2 You can try it on the online interpreter but then you need to use the following version, because the online interpreter pads the code box to a rectangle and applies ? to spaces. i:0(?v v &l~< \+l2( ? \&,12,+o;  # 05AB1E, 15 bytes [Non-competing?] .pvyDSÇOsg/îç}J  Try it online! • -2 bytes by removing D and }, and replacing s with y. Aug 10 '18 at 12:39 • @KevinCruijssen for how old this answer is there's a lot more than that :P. .p is a 1byter now too Aug 10 '18 at 14:20 • Ah lol, forgot about η xD Aug 10 '18 at 14:22 # Japt, 13 bytes (non-competing) £T±Xc)/°Y r d  Test it online! ### How it works £ T± Xc)/° Y r d mXY{T+=Xc)/++Y r d} // Implicit: U = input string, T = 0 mXY{ } // Replace each char X and index Y in the string by this function: T+=Xc // Add X.charCodeAt() to T. )/++Y // Take T / (Y + 1). r d // Round, and convert to a character. // Implicit: output result of last expression  • Ah, nuts; I thought the "non-competing" filter had been removed from the leaderboard, so I didn't see this before posting this. Sep 19 '17 at 13:01 # Japt, 11 bytes c@V±X /°T r  Try it PHP, 176 byte <?=(implode('',array_reduce(str_split($argv[1]),function($c,$k){array_push($c[1],chr(floor(((ord($k)+$c[0])/(count($c[1])+1))+0.5)));return[ord($k)+$c[0],$c[1]];},[0,[]])[1]));  Example: >php cg.php Hello! HW^adY >php cg.php test tmop >php cg.php 42 43  The biggest solution so far, but based on php it can't get much shorter I think. 2 bytes could be saved by removing the newlines. • Hmm, good point. I thought I might leave them in the post for better readability. But yeah, this is code golf. I'll remove them ;) – cb0 Sep 24 '17 at 14:16 • You can always include an additional version with padding for readability along side your short one. I often do this when my code is too long to be entirely visible on most monitors. Sep 24 '17 at 14:50 # Ruby, 27 bytes putc gets.sum.quo(~/$/)+0.5


Solution produced through some collaborative golfing at work.

## How this works

• Kernel#putc prints a single character (e.g. putc "Hello" outputs H). Passing an object to it prints the character whose code is the least-significant byte of the object. So putc 72 outputs H.
• Kernel#gets reads stdin and returns a string.
• String#sum returns a checksum. The result is the sum of the binary value of each byte in the string modulo 2**16 - 1. Since ASCII bytes comfortably fit in this space, this just returns the byte total.
• Because the byte total is an integer, we can’t do floating point calculations (in Ruby 10 / 3 is 3 whereas 10.to_f / 3 is 3.33…). Here’s where Numeric#quo comes in – 10.quo(3) returns the Rational (10/3).
• Regexp#~ matches a regular expression against the contents of $_ (the last string read by gets). It returns index of the match. • The Regexp /$/ matches the end of the string, so ~/$/ means “print the index of the end of the string” – in other words, the string length. This gives us the final is a short-hand for the length of the string read by gets. • This gives us the Rational (byte total/string length). Now we need to round it into an integer. +0.5 casts the Rational to a float. putc floors the float into an integer. # PHP 7, 79 bytes As I don't have enough reputation to comment on cb0's answer to provide him hints to improve his answer (mainly using the char index method in a string and while to loop through a string). But I still bothered to post it due to the large reduction in byte count. <?php$i=$b=0;while($a=ord($argv[1][$i++]??'')){$b+=$a;echo chr(round($b/$i));}


I didn't find a way though to use the <?= shortcut and avoid undefined errors without using command line flags (this is why i used $i=$b=0;). This answer does not work in PHP 5 due to the ?? syntax.

# JavaScript ES7, 122 bytes

s=>String.fromCharCode(...[for(i of s)i.charCodeAt()].map((l,i,a)=>Math.round(eval((t=a.slice(0,++i)).join+)/t.length)))


Mostly everything is happening in this bit

eval((t=a.slice(0,++i)).join+)/t.length)


The rest is looping / character code conversion

Split up:

s=>
String.fromCharCode(...                        ) // Converts average character code array to string, ... allows it to take an array
[for(i of s)i.charCodeAt()]                    // Converts string to char code array
.map((l,i,a)=>                             )   // Loops through each character
Math.round(                    /t.length)    // Rounds sum of previous char codes, divides by position + 1
eval(                       )              // evals string of char codes seperated with +
(                ).join+            // joins previous char codes with +
t=a.slice(0,++i)                     // creates an array with all the char codes


If functions aren't allowed:

alert(String.fromCharCode(...[for(i of prompt())i.charCodeAt()].map((l,i,a)=>Math.round(eval((t=a.slice(0,++i)).join+)/t.length))))


133 bytes

ES5 Snippet:

function _toConsumableArray(r){if(Array.isArray(r)){for(var e=0,t=Array(r.length);e<r.length;e++)t[e]=r[e];return t}return Array.from(r)}function _taggedTemplateLiteral(r,e){return Object.freeze(Object.defineProperties(r,{raw:{value:Object.freeze(e)}}))}var _templateObject=_taggedTemplateLiteral(["+"],["+"]),f,t=function t(s){return String.fromCharCode.apply(String,_toConsumableArray(function(){var r=[],e=!0,t=!1,a=void 0;try{for(var n,i=s[Symbol.iterator]();!(e=(n=i.next()).done);e=!0){var o=n.value;r.push(o.charCodeAt())}}catch(l){t=!0,a=l}finally{try{!e&&i["return"]&&i["return"]()}finally{if(t)throw a}}return r}().map(function(l,i,a){return Math.round(eval((f=a.slice(0,++i)).join(_templateObject))/f.length)})))};

// Demo
document.getElementById('go').onclick=function(){
document.getElementById('output').innerHTML = t(document.getElementById('input').value)
};
<div style="padding-left:5px;padding-right:5px;"><h2 style="font-family:sans-serif">Average of Words Snippet</h2><div><div  style="background-color:#EFEFEF;border-radius:4px;padding:10px;"><input placeholder="Text here..." style="resize:none;border:1px solid #DDD;" id="input"><button id='go'>Run!</button></div><br><div style="background-color:#EFEFEF;border-radius:4px;padding:10px;"><span style="font-family:sans-serif;">Output:</span><br><pre id="output" style="background-color:#DEDEDE;padding:1em;border-radius:2px;overflow-x:auto;"></pre></div></div></div>

# Python 2, 106 bytes

It's not short enough. Since it's python it's way too verbose, you can even read what it does by looking code. But it's working.

a=[.0]+[ord(i)for i in raw_input()]
print"".join([chr(int(.5+(sum(a[:i+1])/i)))for i in range(1,len(a))])

• "Since it's python it's way too verbose"... not compared to Java. And I disagree. Any less verbose and it wouldn't be as great as it is. Use Pyth if you want less verbose. Jul 24 '15 at 21:39

## Matlab, 43

Using an anonymous function:

f=@(s)char(round(cumsum(+s)./(1:numel(s))))


Examples:

>> f=@(s)char(round(cumsum(+s)./(1:numel(s))))
f =
@(s)char(round(cumsum(+s)./(1:numel(s))))

>> f('Hello!')
ans =
`