Python TypeError:不支持的操作数类型/:'NoneType'和'float' [英] Python TypeError: unsupported operand type(s) for /: 'NoneType' and 'float'

问题描述

这是我们老师给我们的作业.我们应该使用辛普森规则对函数进行数值积分 f(x) = x*cos(third_root(x))

This is a assignment we got from our teacher. We are supposed to use Simpson's Rule to do a numerical integration of a functions f(x) = x*cos(third_root(x))

但是我们不允许使用cos的内置函数或使用x**(1.0/3.0)来求第三根.

But we are not allowed to use the built in function of cos or use x**(1.0/3.0) to find the third root.

我收到错误:

Traceback (most recent call last):
  File "path", line 104, in <module>
    print simpson(f, 1.0, 50.0, 10)
  File "path", line 91, in simpson
    I += 2 * f(x) + (4.0 * f(x + h))
  File "path", line 101, in f
     return x*final_cos(final_3root(x))
  File "path", line 72, in final_cos
    x = float_mod(x, 2 * pi)
  File "path", line 42, in float_mod
    k = int(x / a)
TypeError: unsupported operand type(s) for /: 'NoneType' and 'float'

Process finished with exit code 1

这是我的代码:

import math


def final_3root(a):
    q, m = math.frexp(a)

    if 0.5 > q or q > 1.0:
        raise ValueError('Math domain error')

    x = 0.8968521468804229452995486
    factor_1 = 0.6299605249474365823836053
    factor_2 = 0.7937005259840997373758528

    q_croot = (q / (x * x) + 2.0 * x) / 3.0
    q_croot = (q / (q_croot * q_croot) + 2.0 * q_croot) / 3.0
    q_croot = (q / (q_croot * q_croot) + 2.0 * q_croot) / 3.0
    q_croot = (q / (q_croot * q_croot) + 2.0 * q_croot) / 3.0

    if m % 3.0 == 0.0:
        m /= 3
        answer = math.ldexp(q_croot, m)

    elif m % 3 == 1:
        m += 2
        m /= 3
        answer = factor_1 * math.ldexp(q_croot, m)

    elif m % 3 == 2:
        m += 1
        m /= 3
         answer = factor_2 * math.ldexp(q_croot, m)

    fasit = a ** (1.0 / 3.0)

#----------------------------------------------

def float_mod(x, a):
    k = int(x / a)
    if (x * a) < 0:
        k -= 1
    return x - float(k) * a


def ratio_based_cosinus(x):
    epsilon = 1.0e-16

    previous_Value = 1
    return_Value = 1
    n = -1
    while True:
        n += 1
        ratio = (-x * x) / (((2 * n) + 1) * ((2 * n) + 2))

        previous_Value *= ratio
        return_Value += previous_Value

        if abs(previous_Value) < epsilon:
            break
    return return_Value


def final_cos(x):
    if isinstance(x, int):
        x += 0.0

    pi = 3.1415926

    x = float_mod(x, 2 * pi)

    if x > pi:
        return ratio_based_cosinus(-x)
   else:
        return ratio_based_cosinus(x)

 #----------------------------------------------


def simpson(f, a, b, N):
    if N & 1:
        raise ValueError('Ugyldig tall')

    I = 0
    h = float((b - a) / N)
    x = float(a)

    for i in range(0, N / 2):
        I += 2 * f(x) + (4.0 * f(x + h))
        x += 2 * h

    I += float(f(b) - f(a))
    I *= h / 3

    print "The sum is: ", I


def f(x):


     return x*final_cos(final_3root(x))


print simpson(f, 1.0, 50.0, 10)

推荐答案

final_3root 缺少 return 语句.

final_3root is missing a return statement.

仔细查看错误.x 是 None.如果回溯它，您会看到使用了该函数的返回值，但它从不返回任何内容.

Look closely at the error. x is None. If you trace it back, you'll see that the return value of that function is used, but it never returns anything.

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