看了老师给的博客，学习了一下，我学的确实很菜啊，仿照写了一个可以实现任意数字个数的加括号运算，目前只支持加一个括号。。。。

收获真的挺大

def to_rpn(f4):    def compare(x, y):        v = {
   '+': 0, '-': 0, '*': 1, '/': 1, '(': 2}        return 1 if y == '(' else v[x] - v[y]    rpn, operators = [], []    for i in f4:        if i not in ['+', '-', '*', '/', '(', ')']:            rpn.append(i)        else:            if i == ')':                while operators[-1] != '(':                    rpn.append(operators.pop())                operators.pop()            elif not operators or 0 < compare(i, operators[-1]):                operators.append(i)            else:                while operators and compare(i, operators[-1]) < 1:                    rpn.append(operators.pop())                operators.append(i)    while operators:        rpn.append(operators.pop())    return rpndef rpn_to_answer(f4):    answer_func = [lambda x, y: x + y, lambda x, y: x - y, lambda x, y: x * y, lambda x, y: x / y]    answer_func_re = [lambda x, y: y + x, lambda x, y: (y - x) * -1, lambda x, y: y * x]    number = []    for i in f4:        if i not in ['+', '-', '*', '/']:            number.append(i)        else:            x, y, z = number.pop(), number.pop(), None            try:                z = answer_func[['+', '-', '*', '/'].index(i)](y, x)            except TypeError:                z = answer_func_re[['+', '-', '*', '/'].index(i)](y, x)            number.append(z)    return number[0]def random_f4(_op):#操作数    from random import randint as ran    r = []    j = _op*2 - 1    for i in range(j):        if i % 2 == 1:            r.append(['+', '-', '*', '/'][ran(0, 3)])        if i % 2 == 0:            r.append(ran(1, 9))    x = ran(0,_op-2)    r.insert(x*2,'(')    y = ran(x+2,_op)    r.insert(y*2,')')    return (r)if __name__ == "__main__":    from fractions import Fraction as F    ops = input("数字个数：")    ops = int(ops)    r = random_f4(_op=ops)    print(r)    r = to_rpn(r)    print(r)    r = rpn_to_answer(r)    r = F(r).limit_denominator(1000)    print(r)

 

截图：

 

 

现在python学的还是很菜啊。。。。。。。。。