是否有一种方法可以自动生成有效的算术表达式？

Question

我目前正在尝试创建一个Python脚本，它将自动生成有效的空格分隔的算术表达式。但是，我得到的示例输出如下：( 32 - 42 / 95 + 24 ( ) ( 53 ) + ) 21

虽然空括号完全可以，但我不能在计算中使用这个自动生成的表达式，因为24和53之间没有运算符，并且结尾21之前的+没有第二个参数。在

我想知道的是，有没有一种方法可以使用python解决方案来解释/修复这些错误？（在有人指出这一点之前，我将首先承认，我在下面发布的代码可能是我推出的最差的代码，并且符合……好吧，几乎没有Python的核心原则。）

import random
parentheses = ['(',')']
ops = ['+','-','*','/'] + parentheses

lines = 0

while lines < 1000:
    fname = open('test.txt','a')
    expr = []
    numExpr = lines
    if (numExpr % 2 == 0):
        numExpr += 1
    isDiv = False # Boolean var, makes sure there's no Div by 0

    # isNumber, isParentheses, isOp determine whether next element is a number, parentheses, or operator, respectively
    isNumber = random.randint(0,1) == 0 # determines whether to start sequence with number or parentheses
    isParentheses = not isNumber
    isOp = False
    # Counts parentheses to ensure parentheses are matching
    numParentheses = 0
    while (numExpr > 0 or numParentheses > 0):
        if (numExpr < 0 and numParentheses > 0):
            isDiv = False
            expr.append(')')
            numParentheses -= 1
        elif (isOp and numParentheses > 0):
            rand = random.randint(0,5)
            expr.append(ops[rand])
            isDiv = (rand == 3) # True if div op was just appended
            # Checks to see if ')' was appended
            if (rand == 5):
                isNumber = False
                isOp = True
                numParentheses -= 1
            # Checks to see if '(' was appended
            elif (rand == 4):
                isNumber = True
                isOp = False
                numParentheses += 1
            # All other operations go here
            else:
                isNumber = True
                isOp = False
        # Didn't add parentheses possibility here in case expression in parentheses somehow reaches 0
        elif (isNumber and isDiv):
            expr.append(str(random.randint(1,100)))
            isDiv = False
            isNumber = False
            isOp = True
        # If a number's up, decides whether to append parentheses or a number
        elif (isNumber):
            rand = random.randint(0,1)
            if (rand == 0):
                expr.append(str(random.randint(0,100)))
                isNumber = False
                isOp = True
            elif (rand == 1):
                if (numParentheses == 0):
                    expr.append('(')
                    numParentheses += 1
                else:
                    rand = random.randint(0,1)
                    expr.append(parentheses[rand])
                    if rand == 0:
                        numParentheses += 1
                    else:
                        numParentheses -= 1
            isDiv = False
        numExpr -= 1

    fname.write(' '.join(expr) + '\n')
    fname.close()
    lines += 1

Answer 1

我在类似的任务中发现了这个线程，即为符号计算的单元测试生成随机表达式。在我的版本中，我包含了一元函数并允许符号是任意字符串，也就是说，可以使用数字或变量名。在

from random import random, choice

UNARIES = ["sqrt(%s)", "exp(%s)", "log(%s)", "sin(%s)", "cos(%s)", "tan(%s)",
           "sinh(%s)", "cosh(%s)", "tanh(%s)", "asin(%s)", "acos(%s)",
           "atan(%s)", "-%s"]
BINARIES = ["%s + %s", "%s - %s", "%s * %s", "%s / %s", "%s ** %s"]

PROP_PARANTHESIS = 0.3
PROP_BINARY = 0.7

def generate_expressions(scope, num_exp, num_ops):
    scope = list(scope) # make a copy first, append as we go
    for _ in xrange(num_ops):
        if random() < PROP_BINARY: # decide unary or binary operator
            ex = choice(BINARIES) % (choice(scope), choice(scope))
            if random() < PROP_PARANTHESIS:
                ex = "(%s)" % ex
            scope.append(ex)
        else:
            scope.append(choice(UNARIES) % choice(scope))
    return scope[-num_exp:] # return most recent expressions

从前面的答案中复制，我只是提出了一些关于概率为PROP_PARANTHESIS的二元运算符（这有点作弊）。二元运算符比一元运算符更常见，因此我也将其留作配置（PROP_BINARY）。例如：

^{pr2}$

这将产生类似于：

e / acos(tan(a)) / a * acos(tan(a)) ** (acos(tan(a)) / a + a) + (d ** b + a)
(a + (a ** sqrt(e)))
acos((b / acos(tan(a)) / a + d) / (a ** sqrt(e)) * (a ** sinh(b) / b))
sin(atan(acos(tan(a)) ** (acos(tan(a)) / a + a) + (d ** b + a)))
sin((b / acos(tan(a)) / a + d)) / (a ** sinh(b) / b)
exp(acos(tan(a)) / a + acos(e))
tan((b / acos(tan(a)) / a + d))
acos(tan(a)) / a * acos(tan(a)) ** (acos(tan(a)) / a + a) + (d ** b + a) + cos(sqrt(e))
(acos(tan(a)) / a + acos(e) * a + e)
((b / acos(tan(a)) / a + d) - cos(sqrt(e))) + sinh(b)

放入PROP_BINARY = 1.0并使用

scope = range(100)

把我们带回到输出

43 * (50 * 83)
34 / (29 / 24)
66 / 47 - 52
((88 ** 38) ** 40)
34 / (29 / 24) - 27
(16 + 36 ** 29)
55 ** 95
70 + 28
6 * 32
(52 * 2 ** 37)