无法接受用户输入并在没有错误的情况下使用它

Question

好吧，我已经做了几天了，但我自己无法修复它。

Python3和下面的代码片段。我的问题是变量类型和用户输入。这个代码是一个更大代码的一部分，我希望用户(大多数是我(能够使用有意义的变量名称(地球质量的"m_earth"和类似的量(作为计算"m_"的输入；F1"；，对于其他人，我想使用。。。。如果…我可以先把这个部分做好。

我尝试过使用"；浮动"；以及"；int"；它真的只是造成了其他我无法修复的错误。

下面包含的代码是使用此代码时会发生的情况。

任何人都有一些建议。。。请

感谢

G = float(6.674E-11)
# All masses are in Kg
m_earth = 5.972E24
m_moon = 7.345E22
m1 = input('Enter the first mass  ')
print(m1, 'Kg')
m2 = input('Enter the second mass  ')
print(m2, 'Kg')
r = input('Enter the center-to-center distance from mass m1 to mass m2  ')
F1 = G * m1 * m2 / r ** 2
print('The gravitational force is  ', F1, "N")

---

Enter the first mass  m_earth
m_earth Kg
Enter the second mass  m_moon
m_moon Kg
Enter the center-to-center distance from mass m1 to mass m2  3E12
Traceback (most recent call last):
File "C:/Users/xyz/PycharmProjects/My Python Programs/try it.py", line 15, in <module>
F1 = G * m_earth * m_moon / r ** 2
TypeError: unsupported operand type(s) for ** or pow(): 'str' and 'int'

进程结束，退出代码为1

Answer 1

由于input()总是返回一个字符串，因此会出现此错误。不能向ints或floats添加任意字符串的原因是，在大多数情况下，这是没有意义的。

考虑3 + 'Hello'。你预计结果会是什么？

那么我们该如何处理呢？我们将其转换为python所理解的允许执行算术函数的东西。在python中，这些类型是int(整数(和float小数。为了论证起见，还有很多其他的，但你不需要它们作为你的榜样。

由于我们不能保证用户已经输入了数字，我们需要进行某种类型的检查以确保他们输入了数字。

这就是我们可以使用try/except块的地方。

G = float(6.674E-11)
# All masses are in Kg
m_earth = 5.972E24
m_moon = 7.345E22
m1 = input('Enter the first mass  ')
print(m1, 'Kg')
m2 = input('Enter the second mass  ')
print(m2, 'Kg')
r = input('Enter the center-to-center distance from mass m1 to mass m2  ')
# any Error that occurs in the try block
# will hand control to the except block
# if the type of Exception in except is 
# the type of error that occurred
try:
# here we try to convert the variable m1 of type str
# to a variable of type float
# since float() "returns" a float and does not convert in place
# we need to catch the output it gives into the same variable
m1 = float(m1)
m2 = float(m2)
r = float(r)
F1 = G * m1 * m2 / r ** 2
print('The gravitational force is  ', F1, "N")
# any conversion error here will result in a ValueError
# We can catch that specific error and alert the user they 
# mis-entered some data
except ValueError as v:
print('Some of your input is not valid.')

但是您希望能够指定变量的名称。为此，您可以使用pythondict。dict将键映射到某个值。键可以是多种类型，但最重要的是，它们可以是str，即input()返回的类型。在这种情况下，它可以是m_earth。

# we need sys to do an early exit in the program
# if the data entered is not a key in the dict
import sys
G = float(6.674E-11)
# All masses are in Kg
my_dict = {'m_earth': 5.972E24,
'm_moon': 7.345E22}
m1 = input('Enter the first mass  ')
# this will raise a KeyError
# if the key is not in your dict
try:
m1 = my_dict[m1]
except KeyError as k:
sys.exit('That name is invalid.')
print(m1, 'Kg')
m2 = input('Enter the second mass  ')
try:
m2 = my_dict[m2]
except KeyError as k:
sys.exit('That name is invalid.')
r = input('Enter the center-to-center distance from mass m1 to mass m2  ')
try:
r = float(r)
F1 = G * m1 * m2 / r ** 2
print('The gravitational force is  ', F1, "N")
except ValueError as v:
print('Some of your input is not valid.')