Best and Shortest Way to Evaluate Mathematical Expressions

Best and shortest way to evaluate mathematical expressions

Further to Thomas's answer, it's actually possible to access the (deprecated) JScript libraries directly from C#, which means you can use the equivalent of JScript's eval function.

using Microsoft.JScript;        // needs a reference to Microsoft.JScript.dll
using Microsoft.JScript.Vsa;    // needs a reference to Microsoft.Vsa.dll

// ...

string expr = "7 + (5 * 4)";
Console.WriteLine(JScriptEval(expr));    // displays 27

// ...

public static double JScriptEval(string expr)
{
    // error checking etc removed for brevity
    return double.Parse(Eval.JScriptEvaluate(expr, _engine).ToString());
}

private static readonly VsaEngine _engine = VsaEngine.CreateEngine();

What is the best way to evaluate mathematical expressions in C++?

Boost.Spirit is a C++ parser library.

Examples:

• in its distribution: classic version and current version (look for "calc");
• on Rosetta wiki;
• some applications using it.

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 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

Evaluating mathematical expressions

An alternative to implementing your own parser and expression evaluator would be to link against a library that provides one for you to use. An interesting choice would be an easily embedded scripting language such as Lua.

It is straightforward to set up a Lua interpreter instance, and pass it expressions to be evaluated, getting back a function to call that evaluates the expression. You can even let the user have variables...

Update: LE, A simple expression evaluator using Lua

Here is a sketchy implementation of a simple expression evaluator based on a Lua interpreter. I compiled this and tried it for a few cases, but it certainly should not be trusted in production code without some attention to error handling and so forth. All the usual caveats apply here.

I compiled and tested this on Windows using Lua 5.1.4 from Lua for Windows. On other platforms, you'll have to find Lua from your usual source, or from www.lua.org.

Public interface to LE

Here is the file le.h:

/* Public API for the LE library.
 */
int le_init();
int le_loadexpr(char *expr, char **pmsg);
double le_eval(int cookie, char **pmsg);
void le_unref(int cookie);
void le_setvar(char *name, double value);
double le_getvar(char *name);

Sample code using LE

Here is the file t-le.c, demonstrating a simple use of this library. It takes its single command-line argument, loads it as an expression, and evaluates it with the global variable x changing from 0.0 to 1.0 in 11 steps:

#include <stdio.h>
#include "le.h"

int main(int argc, char **argv)
{
    int cookie;
    int i;
    char *msg = NULL;

    if (!le_init()) {
    printf("can't init LE\n");
    return 1;
    }
    if (argc<2) {
    printf("Usage: t-le \"expression\"\n");
    return 1;
    }
    cookie = le_loadexpr(argv[1], &msg);
    if (msg) {
    printf("can't load: %s\n", msg);
    free(msg);
    return 1;
    }
    printf("  x    %s\n"
       "------ --------\n", argv[1]);
    for (i=0; i<11; ++i) {
    double x = i/10.;
    double y;

    le_setvar("x",x);
    y = le_eval(cookie, &msg);
    if (msg) {
        printf("can't eval: %s\n", msg);
        free(msg);
        return 1;
    }
    printf("%6.2f %.3f\n", x,y);
    }
}

Here is some output from t-le:

E:...>t-le "math.sin(math.pi * x)"
  x    math.sin(math.pi * x)
------ --------
  0.00 0.000
  0.10 0.309
  0.20 0.588
  0.30 0.809
  0.40 0.951
  0.50 1.000
  0.60 0.951
  0.70 0.809
  0.80 0.588
  0.90 0.309
  1.00 0.000

E:...>

Implementation of LE

Here is le.c, implementing the Lua Expression evaluator:

#include <lua.h>
#include <lauxlib.h>

#include <stdlib.h>
#include <string.h>

static lua_State *L = NULL;

/* Initialize the LE library by creating a Lua state.
 *
 * The new Lua interpreter state has the "usual" standard libraries
 * open.
 */
int le_init()
{
    L = luaL_newstate();
    if (L) 
    luaL_openlibs(L);
    return !!L;
}

/* Load an expression, returning a cookie that can be used later to
 * select this expression for evaluation by le_eval(). Note that
 * le_unref() must eventually be called to free the expression.
 *
 * The cookie is a lua_ref() reference to a function that evaluates the
 * expression when called. Any variables in the expression are assumed
 * to refer to the global environment, which is _G in the interpreter.
 * A refinement might be to isolate the function envioronment from the
 * globals.
 *
 * The implementation rewrites the expr as "return "..expr so that the
 * anonymous function actually produced by lua_load() looks like:
 *
 *     function() return expr end
 *
 *
 * If there is an error and the pmsg parameter is non-NULL, the char *
 * it points to is filled with an error message. The message is
 * allocated by strdup() so the caller is responsible for freeing the
 * storage.
 * 
 * Returns a valid cookie or the constant LUA_NOREF (-2).
 */
int le_loadexpr(char *expr, char **pmsg)
{
    int err;
    char *buf;

    if (!L) {
    if (pmsg)
        *pmsg = strdup("LE library not initialized");
    return LUA_NOREF;
    }
    buf = malloc(strlen(expr)+8);
    if (!buf) {
    if (pmsg)
        *pmsg = strdup("Insufficient memory");
    return LUA_NOREF;
    }
    strcpy(buf, "return ");
    strcat(buf, expr);
    err = luaL_loadstring(L,buf);
    free(buf);
    if (err) {
    if (pmsg)
        *pmsg = strdup(lua_tostring(L,-1));
    lua_pop(L,1);
    return LUA_NOREF;
    }
    if (pmsg)
    *pmsg = NULL;
    return luaL_ref(L, LUA_REGISTRYINDEX);
}

/* Evaluate the loaded expression.
 * 
 * If there is an error and the pmsg parameter is non-NULL, the char *
 * it points to is filled with an error message. The message is
 * allocated by strdup() so the caller is responsible for freeing the
 * storage.
 * 
 * Returns the result or 0 on error.
 */
double le_eval(int cookie, char **pmsg)
{
    int err;
    double ret;

    if (!L) {
    if (pmsg)
        *pmsg = strdup("LE library not initialized");
    return 0;
    }
    lua_rawgeti(L, LUA_REGISTRYINDEX, cookie);
    err = lua_pcall(L,0,1,0);
    if (err) {
    if (pmsg)
        *pmsg = strdup(lua_tostring(L,-1));
    lua_pop(L,1);
    return 0;
    }
    if (pmsg)
    *pmsg = NULL;
    ret = (double)lua_tonumber(L,-1);
    lua_pop(L,1);
    return ret;
}

/* Free the loaded expression.
 */
void le_unref(int cookie)
{
    if (!L)
    return;
    luaL_unref(L, LUA_REGISTRYINDEX, cookie);    
}

/* Set a variable for use in an expression.
 */
void le_setvar(char *name, double value)
{
    if (!L)
    return;
    lua_pushnumber(L,value);
    lua_setglobal(L,name);
}

/* Retrieve the current value of a variable.
 */
double le_getvar(char *name)
{
    double ret;

    if (!L)
    return 0;
    lua_getglobal(L,name);
    ret = (double)lua_tonumber(L,-1);
    lua_pop(L,1);
    return ret;
}

Remarks

The above sample consists of 189 lines of code total, including a spattering of comments, blank lines, and the demonstration. Not bad for a quick function evaluator that knows how to evaluate reasonably arbitrary expressions of one variable, and has rich library of standard math functions at its beck and call.

You have a Turing-complete language underneath it all, and it would be an easy extension to allow the user to define complete functions as well as to evaluate simple expressions.

What is the best way to execute math expression?

You need a mathematical expression parser. The best way in my opinion is not reinventing the wheel. An existing open source solution NCalc is a good choice.

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

How to evaluate a mathematical expression using strings and arrays in C

Look into the shunting yard algorithm and convert the string into reverse polish notation.

It will help if you also separate out number detection from symbol detection by reading the string one character at a time instead of just jumping to the operators.

After that, it is rather easy to perform the computation, as the inputs are ordered in a manner that makes a stack based calculator easy to implement.

Yes, you could do a full fledged recursive descent parser; but, for the relatively simple matter of algebraic expressions, they are overkill.

--- Edited because I see that there's controversy over tools vs techniques ---

I see a lot of people mentioning lexx and yacc. They are great tools, they are also overkill for what you need. It's like saying to open a carpentry shop when you want to replace a board on your fence. You don't need to learn another language to handle your math language, and lexx and yacc will require you to learn their domain specific configuration language to build the parser for your domain specific language. That's a lot of languages for a simple, solved problem.

Lexx and Yacc are great if you want to build a mathematical tree of your data; however, RPN is a mathematical tree of your data stored in a list of two fundamental node types (data) and (operation).

(a)(b)(c) // three "data nodes"
(a)(b)(+) // two "data nodes" and one operation node.

This allows one to use a stack based machine (very easy to implement). The machine has an "instruction set" of

 if (x is data) { push X onto the top of the stack }
 if (x is operation) {
   pop the operation's required parameters.
   pass them to the operation.
   push the result on the stack
 }

so assuming you had a '+' operator, the expression 3 + 5 + 6 would convert to RPN 3 5 + 6 + and the stack would look like (during processing)

 <empty>
 (3)
 (3)(5)
 (8)
 (8)(6)
 (14)

The primary reason to convert to RPN is not because it's necessary, it's because it makes the rest of your program much easier to implement. Imagine 3 + 5 * 7 converted to RPN, you have 3 5 7 * + which means that you don't have to do anything special with evaluation "machine" language. Note that (3 + 5) * 7 converts to RPN 3 5 + 7 *. In short, you reduce the complexity of the evaluation engine by massaging your input data to be less ambiguous.

Lexx and Yacc provide a lot of "configurability" to allow you to accomplish the same thing; however, you are not going to be configuring lexx and yacc to do anything special here, so the configurable interface doesn't buy you much. It's like having a multiple selection knob when you only need an on-off switch.

Now, if you want to build any kind of parser, where you are not aware of any kind of future rules which might be added; then definately choose lexx and yacc. But a simple algebraic expression parser is far too "solved" a problem to bring in the big guns. Use a shunting algorthim and break your problem down into

simple scanner to identify the main types of "tokens", NUMBER, +, -, /, *, (, and )
shunting algorithim to convert the "list" of tokens into a RPN list
a stack to hold the "memory" of the evaluation "machine"
the "machine" pull / push / evaluate loop

This has the added benefits of being able to develop the pieces independently, so you can verify each step without the entire framework being in place. Lexx and Yacc promote adding your code to their framework, so unless you take steps to do otherwise, you might have to debug the entire thing as one big piece.

Better way to evaluate mathematical expression in python?

This approach is faster:

>>> import re
>>> split_numbers = re.compile(r'-?\d+').findall
>>> sum(int(x) for x in split_numbers('1+23-4+56-7'))
69

In my timings the sum expression takes 4.5 µs vs. 13 µs for eval('1+23-4+56-7')

Note, however, that it does not handle consecutive + and -, eg. 1-+2 or 1--2, or spaces: 1 - 2.

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