Evaluating a string of simple mathematical expressions [closed]

Answer 1

JavaScript (Not IE compatible)

Number of characters: 268/260

Fully obfuscated function:

function e(x){x=x.replace(/ /g,'')+')'
function P(n){return x[0]=='('?(x=x.substr(1),E()):(n=/^[-+]?[\d.]+/(x)[0],x=x.substr(n.length),+n)}function E(a,o,b){a=P()
for(;;){o=x[0]
x=x.substr(1)
if(o==')')return a
b=P()
a=o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:a/b}}return E()}

or, in JavaScript 1.8 (Firefox 3+), you can save a few characters by using expression closures:

e=function(x,P,E)(x=x.replace(/ /g,'')+')',P=function(n)(x[0]=='('?(x=x.substr(1),E()):(n=/^[-+]?[\d.]+/(x)[0],x=x.substr(n.length),+n)),E=function(a,o,b){a=P()
for(;;){o=x[0]
x=x.substr(1)
if(o==')')return a
b=P()
a=o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:a/b}},E())

Clear/semi-obfuscated function:

function evaluate(x) {
    x = x.replace(/ /g, "") + ")";
    function primary() {
        if (x[0] == '(') {
            x = x.substr(1);
            return expression();
        }

        var n = /^[-+]?\d*\.?\d*/.exec(x)[0];
        x = x.substr(n.length);
        return +n;
    }

    function expression() {
        var a = primary();
        for (;;) {
            var operator = x[0];
            x = x.substr(1);

            if (operator == ')') {
                return a;
            }

            var b = primary();
            a = (operator == '+') ? a + b :
                (operator == '-') ? a - b :
                (operator == '*') ? a * b :
                                    a / b;
        }
    }

    return expression();
}

Neither version will work in IE, because they use array-style subscripting on the string. If you replace both occurrences of x[0] with x.charAt(0), the first one should work everywhere.

I cut out some more characters since the first version by turning variables into function parameters and replacing another if statement with the conditional operator.

Answer 2

PHP

Number of characters: 284

obfuscated:

function f($m){return c($m[1]);}function g($n,$m){$o=$m[0];$m[0]=' ';return$o=='+'?$n+$m:($o=='-'?$n-$m:($o=='*'?$n*$m:$n/$m));}function c($s){while($s!=($t=preg_replace_callback('/\(([^()]*)\)/',f,$s)))$s=$t;preg_match_all('![-+/*].*?[\d.]+!',"+$s",$m);return array_reduce($m[0],g);}

readable:

function callback1($m) {return c($m[1]);}
function callback2($n,$m) {
    $o=$m[0];
    $m[0]=' ';
    return $o=='+' ? $n+$m : ($o=='-' ? $n-$m : ($o=='*' ? $n*$m : $n/$m));
}
function c($s){ 
    while ($s != ($t = preg_replace_callback('/\(([^()]*)\)/','callback1',$s))) $s=$t;
    preg_match_all('![-+/*].*?[\d.]+!', "+$s", $m);
    return array_reduce($m[0], 'callback2');
}


$str = '  2.45/8.5  *  -9.27   +    (   5   *  0.0023  ) ';
var_dump(c($str));
# float(-2.66044117647)

Should work with any valid input (including negative numbers and arbitrary whitespace)

Answer 3

Python (without importing anything)

Number of characters: 222

I stole many tricks from Dave's answer, but I managed to shave off some more characters.

def e(s,l=0,n=0,f='+'):
 if s:l=[c for c in s+')'if' '!=c]
 while f!=')':
  p=l.pop;m=p(0)
  if m=='(':m=e(0,l)
  while l[0]not in'+-*/)':m+=p(0)
  m=float(m);n={'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[f];f=p(0)
 return n

Commented version:

def evaluate(stringexpr, listexpr=0, n=0, f_operation='+'):
    # start out as taking 0 + the expression... (or could use 1 * ;)

    # We'll prefer to keep the expression as a list of characters,
    # so we can use .pop(0) to eat up the expression as we go.
    if stringexpr:
        listexpr = [c for c in stringexpr+')' if c!=' ']

    # use ')' as sentinel to return the answer
    while f_operation != ')':
        m_next = listexpr.pop(0)
        if m_next == '(':
            # lists are passed by reference, so this call will eat the (parexp)
            m_next = evaluate(None, listexpr)

        else:
            # rebuild any upcoming numeric chars into a string
            while listexpr[0] not in '+-*/)':
                m_next += listexpr.pop(0)

        # Update n as the current answer.  But never divide by 0.
        m = float(m_next)
        n = {'+':n+m, '-':n-m, '*':n*m, '/':n/(m or 1)}[f_operation]

        # prepare the next operation (known to be one of '+-*/)')
        f_operation = listexpr.pop(0)

    return n

Answer 4

Ruby

Number of characters: 217 179

This is the shortest ruby solution up to now (one heavily based on RegExp yields incorrect answers when string contains few groups of parenthesis) -- no longer true. Solutions based on regex and substitution are shorter. This one is based on stack of accumulators and parses whole expression from left to right. It is re-entrant, and does not modify input string. It could be accused of breaking the rules of not using eval, as it calls Float's methods with identical names as their mathematical mnemonics (+,-,/,*).

Obfuscated code (old version, tweaked below):

def f(p);a,o=[0],['+']
p.sub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n|
q,w=n;case w;when'(';a<<0;o<<'+';when')';q=a.pop;else;o<<w
end if q.nil?;a[-1]=a[-1].method(o.pop).call(q.to_f) if !q.nil?};a[0];end

More obfuscated code:

def f(p);a,o=[0],[:+]
p.scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n|q,w=n;case w
when'(';a<<0;o<<:+;when')';q=a.pop;else;o<<w;end if !q
a<<a.pop.send(o.pop,q.to_f)if q};a[0];end

Clean code:

def f(p)
  accumulators, operands = [0], ['+']
  p.gsub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each do |n|
    number, operand = n
    case operand
      when '('
        accumulators << 0
        operands << '+'
      when ')'
        number = accumulators.pop
        operands.pop
      else 
        operands[-1] = operand
    end if number.nil?
    accumulators[-1] = accumulators.last.method(operands[-1]).call(number.to_f) unless number.nil?
  end
  accumulators.first
end

Answer 5

Haskell

Number of characters: 182

No attempt at cleverness, just some compression: 4 lines, 312 bytes.

import Data.Char;import Text.ParserCombinators.Parsec
q=either(error.show)id.runParser t id"".filter(' '/=);t=do
s<-getState;a<-fmap read(many1$oneOf".-"<|>digit)<|>between(char '('>>setState id)(char ')'>>setState s)t
option(s a)$choice(zipWith(\c o->char c>>return(o$s a))"+-*/"[(+),(-),(*),(/)])>>=setState>>t

And now, really getting into the golf spirit, 3 lines and 182 bytes:

q=snd.(`e`id).filter(' '/=)
e s c|[(f,h)]<-readsPrec 0 s=g h(c f);e('(':s)c=g h(c f)where(')':h,f)=e s id
g('+':h)=e h.(+);g('-':h)=e h.(-);g('*':h)=e h.(*);g('/':h)=e h.(/);g h=(,)h

Exploded:

-- Strip spaces from the input, evaluate with empty accumulator,
-- and output the second field of the result.
q :: String -> Double
q = snd . flip eval id . filter (not . isSpace)

-- eval takes a string and an accumulator, and returns
-- the final value and what’s left unused from the string.
eval :: (Fractional a, Read a) => String -> (a -> a) -> (String, a)

-- If the beginning of the string parses as a number, add it to the accumulator,
-- then try to read an operator and further.
eval str accum | [(num, rest)] <- readsPrec 0 str = oper rest (accum num)

-- If the string starts parentheses, evaluate the inside with a fresh
-- accumulator, and continue after the closing paren.
eval ('(':str) accum = oper rest (accum num) where (')':rest, num) = eval str id

-- oper takes a string and current value, and tries to read an operator
-- to apply to the value.  If there is none, it’s okay.
oper :: (Fractional a, Read a) => String -> a -> (String, a)

-- Handle operations by giving eval a pre-seeded accumulator.
oper ('+':str) num = eval str (num +)
oper ('-':str) num = eval str (num -)
oper ('*':str) num = eval str (num *)
oper ('/':str) num = eval str (num /)

-- If there’s no operation parsable, just return.
oper str num = (str, num)

Answer 6

Python

Number of characters: 382

Yet another Python solution, heavily using regular expression replacement. Each run through the loop the simplest expressions are computed and the results are put back into the string.

This is the unobfuscated code, unless you consider regular expressions to be obfuscated.

import re
from operator import *    
operators = dict(zip("+-/*", (add, sub, truediv, mul)))    
def compute(s):
    def repl(m):
        v1, op, v2 = m.groups()
        return str(operators[op](float(v1), float(v2)))
    while not re.match("^\d+\.\d+$", s):
        s = re.sub("([.\d]+)\s*([+-/*])\s*([.\d]+)", repl, s)
        s = re.sub("\(([.\d]+)\)", r"\1", s)
    return s

Had this idea just as I was turning in and couldn't let it go until I wrote it down and made it work.