Python TypeError：不支持的操作数类型/：'NoneType'和'float'
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Python TypeError: unsupported operand type(s) for /: 'NoneType' and 'float'
提问by Zahand
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))
这是我们老师给我们的作业。我们应该使用辛普森法则对函数进行数值积分f(x) = x*cos(third_root(x))
But we are not allowed to use the built in function of cosor use x**(1.0/3.0)to find the third root.
但是我们不允许使用cosor use的内置函数x**(1.0/3.0)来找到第三根。
I get the errors:
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
And here is my code:
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)
采纳答案by Jonathon Reinhart
final_3rootis missing a return statement.
final_3root缺少 return 语句。
Look closely at the error. xis None. If you trace it back, you'll see that the return value of that function is used, but it never returns anything.
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。
