评估字符串中的数学表达式

stringExp = "2^4"
intVal = int(stringExp)      # Expected value: 16

This returns the following error:

Traceback (most recent call last):  
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int()
with base 10: '2^4'

I know that eval can work around this, but isn’t there a better and – more importantly – safer method to evaluate a mathematical expression that is being stored 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

eval is evil

eval("__import__('os').remove('important file')") # arbitrary commands
eval("9**9**9**9**9**9**9**9", {'__builtins__': None}) # CPU, memory

Note: even if you use set __builtins__ to None it still might be possible to break out using introspection:

eval('(1).__class__.__bases__[0].__subclasses__()', {'__builtins__': None})

Evaluate arithmetic expression using ast

import ast
import operator as op

# supported operators
operators = {ast.Add: op.add, ast.Sub: op.sub, ast.Mult: op.mul,
             ast.Div: op.truediv, ast.Pow: op.pow, ast.BitXor: op.xor,
             ast.USub: op.neg}

def eval_expr(expr):
    """
    >>> eval_expr('2^6')
    4
    >>> eval_expr('2**6')
    64
    >>> eval_expr('1 + 2*3**(4^5) / (6 + -7)')
    -5.0
    """
    return eval_(ast.parse(expr, mode='eval').body)

def eval_(node):
    if isinstance(node, ast.Num): # <number>
        return node.n
    elif isinstance(node, ast.BinOp): # <left> <operator> <right>
        return operators[type(node.op)](eval_(node.left), eval_(node.right))
    elif isinstance(node, ast.UnaryOp): # <operator> <operand> e.g., -1
        return operators[type(node.op)](eval_(node.operand))
    else:
        raise TypeError(node)

You can easily limit allowed range for each operation or any intermediate result, e.g., to limit input arguments for a**b:

def power(a, b):
    if any(abs(n) > 100 for n in [a, b]):
        raise ValueError((a,b))
    return op.pow(a, b)
operators[ast.Pow] = power

Or to limit magnitude of intermediate results:

import functools

def limit(max_=None):
    """Return decorator that limits allowed returned values."""
    def decorator(func):
        @functools.wraps(func)
        def wrapper(*args, **kwargs):
            ret = func(*args, **kwargs)
            try:
                mag = abs(ret)
            except TypeError:
                pass # not applicable
            else:
                if mag > max_:
                    raise ValueError(ret)
            return ret
        return wrapper
    return decorator

eval_ = limit(max_=10**100)(eval_)

Example

>>> evil = "__import__('os').remove('important file')"
>>> eval_expr(evil) #doctest:+IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
TypeError:
>>> eval_expr("9**9")
387420489
>>> eval_expr("9**9**9**9**9**9**9**9") #doctest:+IGNORE_EXCEPTION_DETAIL
Traceback (most recent call last):
...
ValueError:

Some safer alternatives to eval() and sympy.sympify().evalf()^*:

• asteval
• numexpr

^*SymPy sympify is also unsafe according to the following warning from the documentation.

Warning: Note that this function uses eval, and thus shouldn’t be used on unsanitized input.

Okay, so the problem with eval is that it can escape its sandbox too easily, even if you get rid of __builtins__. All the methods for escaping the sandbox come down to using getattr or object.__getattribute__ (via the . operator) to obtain a reference to some dangerous object via some allowed object (''.__class__.__bases__[0].__subclasses__ or similar). getattr is eliminated by setting __builtins__ to None. object.__getattribute__ is the difficult one, since it cannot simply be removed, both because object is immutable and because removing it would break everything. However, __getattribute__ is only accessible via the . operator, so purging that from your input is sufficient to ensure eval cannot escape its sandbox.
In processing formulas, the only valid use of a decimal is when it is preceded or followed by [0-9], so we just remove all other instances of ..

import re
inp = re.sub(r"\.(?![0-9])","", inp)
val = eval(inp, {'__builtins__':None})

Note that while python normally treats 1 + 1. as 1 + 1.0, this will remove the trailing . and leave you with 1 + 1. You could add ),, and EOF to the list of things allowed to follow ., but why bother?

You can use the ast module and write a NodeVisitor that verifies that the type of each node is part of a whitelist.

import ast, math

locals =  {key: value for (key,value) in vars(math).items() if key[0] != '_'}
locals.update({"abs": abs, "complex": complex, "min": min, "max": max, "pow": pow, "round": round})

class Visitor(ast.NodeVisitor):
    def visit(self, node):
       if not isinstance(node, self.whitelist):
           raise ValueError(node)
       return super().visit(node)

    whitelist = (ast.Module, ast.Expr, ast.Load, ast.Expression, ast.Add, ast.Sub, ast.UnaryOp, ast.Num, ast.BinOp,
            ast.Mult, ast.Div, ast.Pow, ast.BitOr, ast.BitAnd, ast.BitXor, ast.USub, ast.UAdd, ast.FloorDiv, ast.Mod,
            ast.LShift, ast.RShift, ast.Invert, ast.Call, ast.Name)

def evaluate(expr, locals = {}):
    if any(elem in expr for elem in '\n#') : raise ValueError(expr)
    try:
        node = ast.parse(expr.strip(), mode='eval')
        Visitor().visit(node)
        return eval(compile(node, "<string>", "eval"), {'__builtins__': None}, locals)
    except Exception: raise ValueError(expr)

Because it works via a whitelist rather than a blacklist, it is safe. The only functions and variables it can access are those you explicitly give it access to. I populated a dict with math-related functions so you can easily provide access to those if you want, but you have to explicitly use it.

If the string attempts to call functions that haven’t been provided, or invoke any methods, an exception will be raised, and it will not be executed.

Because this uses Python’s built in parser and evaluator, it also inherits Python’s precedence and promotion rules as well.

>>> evaluate("7 + 9 * (2 << 2)")
79
>>> evaluate("6 // 2 + 0.0")
3.0

The above code has only been tested on Python 3.

If desired, you can add a timeout decorator on this function.

The reason eval and exec are so dangerous is that the default compile function will generate bytecode for any valid python expression, and the default eval or exec will execute any valid python bytecode. All the answers to date have focused on restricting the bytecode that can be generated (by sanitizing input) or building your own domain-specific-language using the AST.

Instead, you can easily create a simple eval function that is incapable of doing anything nefarious and can easily have runtime checks on memory or time used. Of course, if it is simple math, than there is a shortcut.

c = compile(stringExp, 'userinput', 'eval')
if c.co_code[0]==b'd' and c.co_code[3]==b'S':
    return c.co_consts[ord(c.co_code[1])+ord(c.co_code[2])*256]

The way this works is simple, any constant mathematic expression is safely evaluated during compilation and stored as a constant. The code object returned by compile consists of d, which is the bytecode for LOAD_CONST, followed by the number of the constant to load (usually the last one in the list), followed by S, which is the bytecode for RETURN_VALUE. If this shortcut doesn’t work, it means that the user input isn’t a constant expression (contains a variable or function call or similar).

This also opens the door to some more sophisticated input formats. For example:

stringExp = "1 + cos(2)"

This requires actually evaluating the bytecode, which is still quite simple. Python bytecode is a stack oriented language, so everything is a simple matter of TOS=stack.pop(); op(TOS); stack.put(TOS) or similar. The key is to only implement the opcodes that are safe (loading/storing values, math operations, returning values) and not unsafe ones (attribute lookup). If you want the user to be able to call functions (the whole reason not to use the shortcut above), simple make your implementation of CALL_FUNCTION only allow functions in a ‘safe’ list.

from dis import opmap
from Queue import LifoQueue
from math import sin,cos
import operator

globs = {'sin':sin, 'cos':cos}
safe = globs.values()

stack = LifoQueue()

class BINARY(object):
    def __init__(self, operator):
        self.op=operator
    def __call__(self, context):
        stack.put(self.op(stack.get(),stack.get()))

class UNARY(object):
    def __init__(self, operator):
        self.op=operator
    def __call__(self, context):
        stack.put(self.op(stack.get()))


def CALL_FUNCTION(context, arg):
    argc = arg[0]+arg[1]*256
    args = [stack.get() for i in range(argc)]
    func = stack.get()
    if func not in safe:
        raise TypeError("Function %r now allowed"%func)
    stack.put(func(*args))

def LOAD_CONST(context, arg):
    cons = arg[0]+arg[1]*256
    stack.put(context['code'].co_consts[cons])

def LOAD_NAME(context, arg):
    name_num = arg[0]+arg[1]*256
    name = context['code'].co_names[name_num]
    if name in context['locals']:
        stack.put(context['locals'][name])
    else:
        stack.put(context['globals'][name])

def RETURN_VALUE(context):
    return stack.get()

opfuncs = {
    opmap['BINARY_ADD']: BINARY(operator.add),
    opmap['UNARY_INVERT']: UNARY(operator.invert),
    opmap['CALL_FUNCTION']: CALL_FUNCTION,
    opmap['LOAD_CONST']: LOAD_CONST,
    opmap['LOAD_NAME']: LOAD_NAME
    opmap['RETURN_VALUE']: RETURN_VALUE,
}

def VMeval(c):
    context = dict(locals={}, globals=globs, code=c)
    bci = iter(c.co_code)
    for bytecode in bci:
        func = opfuncs[ord(bytecode)]
        if func.func_code.co_argcount==1:
            ret = func(context)
        else:
            args = ord(bci.next()), ord(bci.next())
            ret = func(context, args)
        if ret:
            return ret

def evaluate(expr):
    return VMeval(compile(expr, 'userinput', 'eval'))

Obviously, the real version of this would be a bit longer (there are 119 opcodes, 24 of which are math related). Adding STORE_FAST and a couple others would allow for input like 'x=5;return x+x or similar, trivially easily. It can even be used to execute user-created functions, so long as the user created functions are themselves executed via VMeval (don’t make them callable!!! or they could get used as a callback somewhere). Handling loops requires support for the goto bytecodes, which means changing from a for iterator to while and maintaining a pointer to the current instruction, but isn’t too hard. For resistance to DOS, the main loop should check how much time has passed since the start of the calculation, and certain operators should deny input over some reasonable limit (BINARY_POWER being the most obvious).

While this approach is somewhat longer than a simple grammar parser for simple expressions (see above about just grabbing the compiled constant), it extends easily to more complicated input, and doesn’t require dealing with grammar (compile take anything arbitrarily complicated and reduces it to a sequence of simple instructions).

I think I would use eval(), but would first check to make sure the string is a valid mathematical expression, as opposed to something malicious. You could use a regex for the validation.

eval() also takes additional arguments which you can use to restrict the namespace it operates in for greater security.

This is a massively late reply, but I think useful for future reference. Rather than write your own math parser (although the pyparsing example above is great) you could use SymPy. I don’t have a lot of experience with it, but it contains a much more powerful math engine than anyone is likely to write for a specific application and the basic expression evaluation is very easy:

>>> import sympy
>>> x, y, z = sympy.symbols('x y z')
>>> sympy.sympify("x**3 + sin(y)").evalf(subs={x:1, y:-3})
0.858879991940133

Very cool indeed! A from sympy import * brings in a lot more function support, such as trig functions, special functions, etc., but I’ve avoided that here to show what’s coming from where.

[I know this is an old question, but it is worth pointing out new useful solutions as they pop up]

Since python3.6, this capability is now built into the language, coined “f-strings”.

See: PEP 498 — Literal String Interpolation

For example (note the f prefix):

f'{2**4}'
=> '16'

Use eval in a clean namespace:

>>> ns = {'__builtins__': None}
>>> eval('2 ** 4', ns)
16

The clean namespace should prevent injection. For instance:

>>> eval('__builtins__.__import__("os").system("echo got through")', ns)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<string>", line 1, in <module>
AttributeError: 'NoneType' object has no attribute '__import__'

Otherwise you would get:

>>> eval('__builtins__.__import__("os").system("echo got through")')
got through
0

You might want to give access to the math module:

>>> import math
>>> ns = vars(math).copy()
>>> ns['__builtins__'] = None
>>> eval('cos(pi/3)', ns)
0.50000000000000011

Here’s my solution to the problem without using eval. Works with Python2 and Python3. It doesn’t work with negative numbers.

$ python -m pytest test.py

test.py

from solution import Solutions

class SolutionsTestCase(unittest.TestCase):
    def setUp(self):
        self.solutions = Solutions()

    def test_evaluate(self):
        expressions = [
            '2+3=5',
            '6+4/2*2=10',
            '3+2.45/8=3.30625',
            '3**3*3/3+3=30',
            '2^4=6'
        ]
        results = [x.split('=')[1] for x in expressions]
        for e in range(len(expressions)):
            if '.' in results[e]:
                results[e] = float(results[e])
            else:
                results[e] = int(results[e])
            self.assertEqual(
                results[e],
                self.solutions.evaluate(expressions[e])
            )

solution.py

class Solutions(object):
    def evaluate(self, exp):
        def format(res):
            if '.' in res:
                try:
                    res = float(res)
                except ValueError:
                    pass
            else:
                try:
                    res = int(res)
                except ValueError:
                    pass
            return res
        def splitter(item, op):
            mul = item.split(op)
            if len(mul) == 2:
                for x in ['^', '*', '/', '+', '-']:
                    if x in mul[0]:
                        mul = [mul[0].split(x)[1], mul[1]]
                    if x in mul[1]:
                        mul = [mul[0], mul[1].split(x)[0]]
            elif len(mul) > 2:
                pass
            else:
                pass
            for x in range(len(mul)):
                mul[x] = format(mul[x])
            return mul
        exp = exp.replace(' ', '')
        if '=' in exp:
            res = exp.split('=')[1]
            res = format(res)
            exp = exp.replace('=%s' % res, '')
        while '^' in exp:
            if '^' in exp:
                itm = splitter(exp, '^')
                res = itm[0] ^ itm[1]
                exp = exp.replace('%s^%s' % (str(itm[0]), str(itm[1])), str(res))
        while '**' in exp:
            if '**' in exp:
                itm = splitter(exp, '**')
                res = itm[0] ** itm[1]
                exp = exp.replace('%s**%s' % (str(itm[0]), str(itm[1])), str(res))
        while '/' in exp:
            if '/' in exp:
                itm = splitter(exp, '/')
                res = itm[0] / itm[1]
                exp = exp.replace('%s/%s' % (str(itm[0]), str(itm[1])), str(res))
        while '*' in exp:
            if '*' in exp:
                itm = splitter(exp, '*')
                res = itm[0] * itm[1]
                exp = exp.replace('%s*%s' % (str(itm[0]), str(itm[1])), str(res))
        while '+' in exp:
            if '+' in exp:
                itm = splitter(exp, '+')
                res = itm[0] + itm[1]
                exp = exp.replace('%s+%s' % (str(itm[0]), str(itm[1])), str(res))
        while '-' in exp:
            if '-' in exp:
                itm = splitter(exp, '-')
                res = itm[0] - itm[1]
                exp = exp.replace('%s-%s' % (str(itm[0]), str(itm[1])), str(res))

        return format(exp)