Python 3.x 尝试更正函数

我对用python（或任何东西）编写代码非常陌生。我必须创建一个函数，该函数将根据制动时车辆打滑的距离返回车辆的速度。当我尝试运行doctests时，我得到一个“float”对象不可调用错误。我试过用几种方法更改变量的名称，但似乎都不正确。可能遗漏了一些简单的东西

import math

def car_speed(distance_of_skid):
    '''
Calculate the speed in MPH of a car that skidded
d feet on dry concrete when the brakes were applied

args:
    distance_of_skid (float): the distance of the skid in feet

returns:
    an estimate of the speed the car was going when the brakes were applied (float)

formula:
    speed in MPH equals the square root of (24 * d)

examples/doctest:

the car didn't skid at all
>>> round(car_speed(0), 2)
0.0

the car skid 1 foot
>>> round(car_speed(1), 2)
4.9

the car skid 10 feet
>>> round(car_speed(10), 2)
15.49

the car skid 33.33 feet
>>> round(car_speed(33.33), 2)
28.28

the car skid 12345 feet
>>> round(car_speed(12345), 2)
544.32

'''
    d = distance_of_skid
    car_speed = math.sqrt (24 * d)

    return (round(car_speed(d), 2))

---

感觉我把事情弄得更复杂了（教授之前提到过这一点）。该函数的输出应该是速度（以英里/小时为单位），四舍五入到小数点后2位。

多亏了藏红花指甲，我才能够计算出来。我使用函数名作为常量，这导致了问题。将常量名称更改为其他名称修复了它。将车辆速度改为返回线中的仅速度

> '''
>     to run tests on mac: python3 -m doctest car_speed.py -v
>     to run tests on Win: python -m doctest car_speed.py -v ''' import math
> 
> def car_speed(distance_of_skid):
>     '''
>     Calculate the speed in MPH of a car that skidded
>     d feet on dry concrete when the brakes were applied
> 
>     args:
>         distance_of_skid (float): the distance of the skid in feet
> 
>     returns:
>         an estimate of the speed the car was going when the brakes were applied (float)
> 
>     formula:
>         speed in MPH equals the square root of (24 * d)
> 
>     examples/doctest:
> 
>     the car didn't skid at all
>     >>> round(car_speed(0), 2)
>     0.0
> 
>     the car skid 1 foot
>     >>> round(car_speed(1), 2)
>     4.9
> 
>     the car skid 10 feet
>     >>> round(car_speed(10), 2)
>     15.49
> 
>     the car skid 33.33 feet
>     >>> round(car_speed(33.33), 2)
>     28.28
> 
>     the car skid 12345 feet
>     >>> round(car_speed(12345), 2)
>     544.32
> 
>     '''
>     # TO DO: Add your code here
>     d = distance_of_skid
>     speed = math.sqrt (24 * d)
> 
>     return (round(speed, 2))

---

你期望车速（d）返回多少？它需要返回车速（以英里/小时为单位），四舍五入到2位。啊，我浏览了文件的开头。我的错。DX那么看起来你是想从函数内部调用函数？在function.yep的最后一行中，这就是错误。只要我更改了常量名称并重新运行，它就工作了。我知道这很简单。谢谢很高兴听到你把它修好了！您是否介意创建一个解释您所做更改的答案，包括显示修改后的代码，以方便将来的读者？