TypeError:'int'对象不可调用:Quadratic

Question

(在没有导入的二次方程中找到x的值.)每当我运行程序时,Python停在discriminant = (b ** 2) - 4(a * c)并显示TypeError:'int'对象不可调用.怎么了？

#------SquareRootDefinition---------#
def Square_Root(n, x):
if n > 0:
    y = (x + n/x) / 2
    while x != y:
        x = y
        return Square_Root(n, x)
    else:
        if abs(10 ** -7) > abs(n - x ** 2):
            return y
elif n == 0:
    return 0
else:
    return str(int(-n)) + "i"

#----------Quadratic Equation--------------#

a = input("Enter coefficient a: ")
while a == 0:
    print "a must not be equal to 0."
    a = input("Enter coefficient a: ")
b = input("Enter coefficient b: ")
c = input("Enter coefficient c: ")

def Quadratic(a, b, c):
    discriminant = (b ** 2) - 4(a * c)
    if discriminant < 0:
        print "imaginary"
    elif discriminant >= 0:
        Sqrt_Disc = Square_Root(discriminant)
        First_Root = (-b + Sqrt_Disc) / (2 * a)
        Second_Root = (-b - Sqrt_Disc) / (2 * a)

  return First_Root, Second_Root

X_1, X_2 = Quadratic(a, b, c)

Answer 1

4(a * c)是无效的Python.你的意思4 * a * c.您可以使用并置并省略数学符号中的乘法符号,但不能用Python(或大多数其他编程语言).

Answer 2

您正尝试将其4用作功能:

discriminant = (b ** 2) - 4(a * c)

你错过了*:

discriminant = (b ** 2) - 4 * (a * c)

此外,如果您的判别式低于0,您将获得一个未绑定的本地异常:

>>> X_1, X_2 = Quadratic(2, 1, 1)
imaginary
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 9, in Quadratic
UnboundLocalError: local variable 'First_Root' referenced before assignment

您需要添加一个return,或者更好的是,引发异常:

def Quadratic(a, b, c):
    discriminant = (b ** 2) - 4(a * c)
    if discriminant < 0:
        raise ValueError("imaginary")
    elif discriminant >= 0:
        Sqrt_Disc = Square_Root(discriminant)
        First_Root = (-b + Sqrt_Disc) / (2 * a)
        Second_Root = (-b - Sqrt_Disc) / (2 * a)

    return First_Root, Second_Root

您的Square_Root()函数缺少它的默认值x:

def Square_Root(n, x=1):

通过这些更改,您的功能实际上有效:

>>> Quadratic(1, 3, -4)
(1, -4)