6

我正在尝试创建一个程序，该程序使用 sympy 获取一组变量并在这些变量的域上评估符号逻辑表达式。问题是在它吐出真值表后，我无法让 python 来评估表达式。

这是代码：

    from sympy import *
from sympy.abc import p, q, r

def get_vars():
    vars = []
        print "Please enter the number of variables to use in the equation"
        numVars = int(raw_input())
    print "please enter each of the variables on a newline"
        for i in xrange(numVars):
        vars.append(raw_input())
    return vars

def get_expr():
    print "Please enter the expression to use"
    return str(raw_input())

def convert_to_expr(inputStr):
    return eval(inputStr)

def main():
    vars = get_vars()
    expr = get_expr()

    print("recieved input: " + str(vars) + " expr " + str(expr))

    print "Truth table for " + str(len(vars)) + "variable(s)"
    for i in enumerate(truth_table(vars, expr)):
        print i

def fixed_table(numvars):
    """
    Generate true/false permutations for the given number of variables.
    So if numvars=2
    Returns (not necessarily in this order):
        True, True
        True, False
        False, False
        False, True
    """
    if numvars is 1:
        yield [True]
        yield [False]
    else:
        for i in fixed_table(numvars-1):
            yield i + [True]
            yield i + [False]


def truth_table(vars, expr):
    """
    Takes an array of variables, vars, and displays a truth table
    for each possible value combination of vars.
    """
    for cond in fixed_table(len(vars)):
        values=dict(zip(vars,cond))
        yield cond + [eval(expr)]   

if __name__ == "__main__":
    main()

如果我执行以下操作，输出如下：

    Please enter the number of variables to use in the equation
3
please enter each of the variables on a newline
p
q
r
Please enter the expression to use
p&q&r
recieved input: ['p', 'q', 'r'] expr p&q&r
Truth table for 3variable(s)
(0, [True, True, True, And(p, q, r)])
(1, [True, True, False, And(p, q, r)])
(2, [True, False, True, And(p, q, r)])
(3, [True, False, False, And(p, q, r)])
(4, [False, True, True, And(p, q, r)])
(5, [False, True, False, And(p, q, r)])
(6, [False, False, True, And(p, q, r)])
(7, [False, False, False, And(p, q, r)])

如果存在执行此任务的某些软件，我真的很想知道它:-)

提前致谢。

4

1 回答 1

9

你真的很亲近！一旦你有了And(p, q, r)你的真值表，你就可以使用该subs方法将你的valuesdict 推送到表达式中：即

    yield cond + [eval(expr).subs(values)]

给

p&q&r
recieved input: ['p', 'q', 'r'] expr p&q&r
Truth table for 3variable(s)
(0, [True, True, True, True])
(1, [True, True, False, False])
(2, [True, False, True, False])
(3, [True, False, False, False])
(4, [False, True, True, False])
(5, [False, True, False, False])
(6, [False, False, True, False])
(7, [False, False, False, False])

但我认为有一种更简单的方法可以做到这一点。该sympify函数已经可以从字符串生成表达式：

In [7]: expr = sympify("x & y | z")

In [8]: expr
Out[8]: Or(z, And(x, y))

我们也可以得到变量：

In [9]: expr.free_symbols
Out[9]: set([x, z, y])

plusitertools.product可以生成值（并且cartes是它的别名sympy）：

In [12]: cartes([False, True], repeat=3)
Out[12]: <itertools.product at 0xa24889c>

In [13]: list(cartes([False, True], repeat=3))
Out[13]: 
[(False, False, False),
 (False, False, True),
 (False, True, False),
 (False, True, True),
 (True, False, False),
 (True, False, True),
 (True, True, False),
 (True, True, True)]

结合这些，基本上只是sympify用来获取表达式和避免eval，使用内置的笛卡尔积，并添加.subs()到使用你的values字典，我们得到：

def explore():
    expr_string = raw_input("Enter an expression: ")
    expr = sympify(expr_string)
    variables = sorted(expr.free_symbols)
    for truth_values in cartes([False, True], repeat=len(variables)):
        values = dict(zip(variables, truth_values))
        print sorted(values.items()), expr.subs(values)

这使

In [22]: explore()
Enter an expression: a & (b | c)
[(a, False), (b, False), (c, False)] False
[(a, False), (b, False), (c, True)] False
[(a, False), (b, True), (c, False)] False
[(a, False), (b, True), (c, True)] False
[(a, True), (b, False), (c, False)] False
[(a, True), (b, False), (c, True)] True
[(a, True), (b, True), (c, False)] True
[(a, True), (b, True), (c, True)] True

这比你的短，但它完全使用你的方法。

于 2012-09-17T16:10:38.363 回答