calculate math expression from a string using eval
While I don't suggest using eval
for this (it is not the solution), the problem is that eval
expects complete lines of code, not just fragments.
$ma ="2+10";
$p = eval('return '.$ma.';');
print $p;
Should do what you want.
A better solution would be to write a tokenizer/parser for your math expression. Here's a very simple regex-based one to give you an example:
$ma = "2+10";
if(preg_match('/(\d+)(?:\s*)([\+\-\*\/])(?:\s*)(\d+)/', $ma, $matches) !== FALSE){
$operator = $matches[2];
switch($operator){
case '+':
$p = $matches[1] + $matches[3];
break;
case '-':
$p = $matches[1] - $matches[3];
break;
case '*':
$p = $matches[1] * $matches[3];
break;
case '/':
$p = $matches[1] / $matches[3];
break;
}
echo $p;
}
How to evaluate a math expression given in string form?
With JDK1.6, you can use the built-in Javascript engine.
import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;
import javax.script.ScriptException;
public class Test {
public static void main(String[] args) throws ScriptException {
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine engine = mgr.getEngineByName("JavaScript");
String foo = "40+2";
System.out.println(engine.eval(foo));
}
}
Evaluating a mathematical expression in a string
Pyparsing can be used to parse mathematical expressions. In particular, fourFn.py
shows how to parse basic arithmetic expressions. Below, I've rewrapped fourFn into a numeric parser class for easier reuse.
from __future__ import division
from pyparsing import (Literal, CaselessLiteral, Word, Combine, Group, Optional,
ZeroOrMore, Forward, nums, alphas, oneOf)
import math
import operator
__author__ = 'Paul McGuire'
__version__ = '$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 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
Calculate string value in javascript, not using eval
This exactly the place where you should be using eval, or you will have to loop through the string and generate the numbers. You will have to use isNaN method to do it.
Evaluating a mathematical expression without eval() on Python3
You should use ast.parse
:
import ast
try:
tree = ast.parse(expression, mode='eval')
except SyntaxError:
return # not a Python expression
if not all(isinstance(node, (ast.Expression,
ast.UnaryOp, ast.unaryop,
ast.BinOp, ast.operator,
ast.Num)) for node in ast.walk(tree)):
return # not a mathematical expression (numbers and operators)
result = eval(compile(tree, filename='', mode='eval'))
Note that for simplicity this allows all the unary operators (+
, -
, ~
, not
) as well as the arithmetic and bitwise binary operators (+
, -
, *
, /
, %
, //
**
, <<
, >>
, &
, |
, ^
) but not the logical or comparison operators. If should be straightforward to refine or expand the allowed operators.
Where to enter the string in evaluate simple math expression
You need to pass your String
to the eval
function like this:
String myEquation = "8+4/2";
double answer = eval(myEquation);
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