Evaluate String with Math Operators

Evaluate string with math operators

Use Ncalc:

Expression e = new Expression("(4+8)*2");
Debug.Assert(24 == e.Evaluate());   

http://ncalc.codeplex.com/

Also, this question had been previously asked and has some interesting answers including Ncalc : Evaluating string "3*(4+2)" yield int 18

Math operations from string

Warning: this way is not a safe way, but is very easy to use. Use it wisely.

Use the eval function.

print eval('2 + 4')

Output:

6

You can even use variables or regular python code.

a = 5
print eval('a + 4')

Output:

9

You also can get return values:

d = eval('4 + 5')
print d

Output:

9

Or call functions:

def add(a, b):
    return a + b

def subtract(a, b):
    return a - b

a = 20
b = 10    
print eval('add(a, b)')
print eval('subtract(a, b)')

Output:

30
10

In case you want to write a parser, maybe instead you can built a python code generator if that is easier and use eval to run the code. With eval you can execute any Python evalution.

Why eval is unsafe?

Since you can put literally anything in the eval, e.g. if the input argument is:

os.system(‘rm -rf /’)

It will remove all files on your system (at least on Linux/Unix).
So only use eval when you trust the input.

How to evaluate a math expression given in string form?

With JDK1.6, you can use the built-in Javascript engine.

import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;
import javax.script.ScriptException;

public class Test {
  public static void main(String[] args) throws ScriptException {
    ScriptEngineManager mgr = new ScriptEngineManager();
    ScriptEngine engine = mgr.getEngineByName("JavaScript");
    String foo = "40+2";
    System.out.println(engine.eval(foo));
    } 
}

Evaluating a string as a mathematical expression in JavaScript

I've eventually gone for this solution, which works for summing positive and negative integers (and with a little modification to the regex will work for decimals too):

function sum(string) {
  return (string.match(/^(-?\d+)(\+-?\d+)*$/)) ? string.split('+').stringSum() : NaN;
}   

Array.prototype.stringSum = function() {
    var sum = 0;
    for(var k=0, kl=this.length;k<kl;k++)
    {
        sum += +this[k];
    }
    return sum;
}

I'm not sure if it's faster than eval(), but as I have to carry out the operation lots of times I'm far more comfortable runing this script than creating loads of instances of the javascript compiler

Evaluating a mathematical expression in a string

Pyparsing can be used to parse mathematical expressions. In particular, fourFn.py
shows how to parse basic arithmetic expressions. Below, I've rewrapped fourFn into a numeric parser class for easier reuse.

from __future__ import division
from pyparsing import (Literal, CaselessLiteral, Word, Combine, Group, Optional,
                       ZeroOrMore, Forward, nums, alphas, oneOf)
import math
import operator

__author__ = 'Paul McGuire'
__version__ = '$Revision: 0.0 $'
__date__ = '$Date: 2009-03-20 $'
__source__ = '''http://pyparsing.wikispaces.com/file/view/fourFn.py
http://pyparsing.wikispaces.com/message/view/home/15549426
'''
__note__ = '''
All I've done is rewrap Paul McGuire's fourFn.py as a class, so I can use it
more easily in other places.
'''

class NumericStringParser(object):
    '''
    Most of this code comes from the fourFn.py pyparsing example

    '''

    def pushFirst(self, strg, loc, toks):
        self.exprStack.append(toks[0])

    def pushUMinus(self, strg, loc, toks):
        if toks and toks[0] == '-':
            self.exprStack.append('unary -')

    def __init__(self):
        """
        expop   :: '^'
        multop  :: '*' | '/'
        addop   :: '+' | '-'
        integer :: ['+' | '-'] '0'..'9'+
        atom    :: PI | E | real | fn '(' expr ')' | '(' expr ')'
        factor  :: atom [ expop factor ]*
        term    :: factor [ multop factor ]*
        expr    :: term [ addop term ]*
        """
        point = Literal(".")
        e = CaselessLiteral("E")
        fnumber = Combine(Word("+-" + nums, nums) +
                          Optional(point + Optional(Word(nums))) +
                          Optional(e + Word("+-" + nums, nums)))
        ident = Word(alphas, alphas + nums + "_$")
        plus = Literal("+")
        minus = Literal("-")
        mult = Literal("*")
        div = Literal("/")
        lpar = Literal("(").suppress()
        rpar = Literal(")").suppress()
        addop = plus | minus
        multop = mult | div
        expop = Literal("^")
        pi = CaselessLiteral("PI")
        expr = Forward()
        atom = ((Optional(oneOf("- +")) +
                 (ident + lpar + expr + rpar | pi | e | fnumber).setParseAction(self.pushFirst))
                | Optional(oneOf("- +")) + Group(lpar + expr + rpar)
                ).setParseAction(self.pushUMinus)
        # by defining exponentiation as "atom [ ^ factor ]..." instead of
        # "atom [ ^ atom ]...", we get right-to-left exponents, instead of left-to-right
        # that is, 2^3^2 = 2^(3^2), not (2^3)^2.
        factor = Forward()
        factor << atom + \
            ZeroOrMore((expop + factor).setParseAction(self.pushFirst))
        term = factor + \
            ZeroOrMore((multop + factor).setParseAction(self.pushFirst))
        expr << term + \
            ZeroOrMore((addop + term).setParseAction(self.pushFirst))
        # addop_term = ( addop + term ).setParseAction( self.pushFirst )
        # general_term = term + ZeroOrMore( addop_term ) | OneOrMore( addop_term)
        # expr <<  general_term
        self.bnf = expr
        # map operator symbols to corresponding arithmetic operations
        epsilon = 1e-12
        self.opn = {"+": operator.add,
                    "-": operator.sub,
                    "*": operator.mul,
                    "/": operator.truediv,
                    "^": operator.pow}
        self.fn = {"sin": math.sin,
                   "cos": math.cos,
                   "tan": math.tan,
                   "exp": math.exp,
                   "abs": abs,
                   "trunc": lambda a: int(a),
                   "round": round,
                   "sgn": lambda a: abs(a) > epsilon and cmp(a, 0) or 0}

    def evaluateStack(self, s):
        op = s.pop()
        if op == 'unary -':
            return -self.evaluateStack(s)
        if op in "+-*/^":
            op2 = self.evaluateStack(s)
            op1 = self.evaluateStack(s)
            return self.opn[op](op1, op2)
        elif op == "PI":
            return math.pi  # 3.1415926535
        elif op == "E":
            return math.e  # 2.718281828
        elif op in self.fn:
            return self.fn[op](self.evaluateStack(s))
        elif op[0].isalpha():
            return 0
        else:
            return float(op)

    def eval(self, num_string, parseAll=True):
        self.exprStack = []
        results = self.bnf.parseString(num_string, parseAll)
        val = self.evaluateStack(self.exprStack[:])
        return val

You can use it like this

nsp = NumericStringParser()
result = nsp.eval('2^4')
print(result)
# 16.0

result = nsp.eval('exp(2^4)')
print(result)
# 8886110.520507872

How to mathematically evaluate a string like 2-1 to produce 1?

I know this question is old, but I came across it last night while searching for something that wasn't quite related, and every single answer here is bad. Not just bad, very bad. The examples I give here will be from a class that I created back in 2005 and spent the past few hours updating for PHP5 because of this question. Other systems do exist, and were around before this question was posted, so it baffles me why every answer here tells you to use eval, when the caution from PHP is:

The eval() language construct is very dangerous because it allows execution of arbitrary PHP code. Its use thus is discouraged. If you have carefully verified that there is no other option than to use this construct, pay special attention not to pass any user provided data into it without properly validating it beforehand.

Before I jump in to the example, the places to get the class I will be using is on either PHPClasses or GitHub. Both the eos.class.php and stack.class.php are required, but can be combined in to the same file.

The reason for using a class like this is that it includes and infix to postfix(RPN) parser, and then an RPN Solver. With these, you never have to use the eval function and open your system up to vulnerabilities. Once you have the classes, the following code is all that is needed to solve a simple (to more complex) equation such as your 2-1 example.

require_once "eos.class.php";
$equation = "2-1";
$eq = new eqEOS();
$result = $eq->solveIF($equation);

That's it! You can use a parser like this for most equations, however complicated and nested without ever having to resort to the 'evil eval'.

Because I really don't want this only only to have my class in it, here are some other options. I am just familiar with my own since I've been using it for 8 years. ^^

Wolfram|Alpha API

Sage

A fairly bad parser

phpdicecalc

Not quite sure what happened to others that I had found previously - came across another one on GitHub before as well, unfortunately I didn't bookmark it, but it was related to large float operations that included a parser as well.

Anyways, I wanted to make sure an answer to solving equations in PHP on here wasn't pointing all future searchers to eval as this was at the top of a google search. ^^

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