将操作员放入Python 3中的列表中

Question

我想将操作员作为列表，然后从列表中调用元素作为操作员。

如果我不在操作员周围放置报价，那么我会收到列表中逗号的语法错误：

  File "p22.py", line 24
    cat = [+,-,*]
            ^
SyntaxError: invalid syntax

如果我确实将报价放在周围，那么我似乎丢失了操作员的功能，如这种情况：

：

  File "p22.py", line 30
    a = (x which y)
               ^
SyntaxError: invalid syntax

这是完整的代码：

import random
def start():
    print('n________________________________________')
    print('|                                      |')
    print('|         Zach's Tutorifier!          |')
    print('|                                      |')
    print('|                                      |')
    print('|     Instructions:                    |')
    print('| Select the type of math you want     |')
    print('| to practice with:                    |')
    print('| 1-Add 2-Subtract 3-Multiply          |')
    print('|--------------------------------------|')
start()
math = int(input('> ')) - 1
cat = ['+','-','*']
def problems():
    which = cat[math]
    x = random.randint(0,9)
    y = random.randint(0,9)
    a = (x which y)
    print('What is %i %s %i?' % (x, which, y) )
    answer = input('> ')
    if answer == a:
        print('Congratulations! Try again? (Y/N)')
        again = input('> ')
        if again == 'Y' or again == 'y':
            problems()
        else:
            start()
    else: 
        print('Try again!')
problems()

Answer 1

为了正确地翻译 数学操作员的字符串表示，您实际上可以使用indentIn ocerator 模块来执行此操作你。简而言之，将字符串运算符映射到方法调用，然后相应地工作。这是一个示例，您应该能够弄清楚如何应用于您的代码：

from operator import add, sub, mul
operations = {'+': add, '-': sub, '*': mul}
op = input("enter +, - or *")
num1 = int(input("enter a number"))
num2 = int(input("enter another number"))
expected_result = int(input("what do you think the answer should be"))
result = operations[op](num1, num2)
if expected_result == result:
    print('you are right')
else:
    print('no, you are wrong')

提供有关此行的额外信息：

operations[op](num1, num2)

operations是一个字典，我们通过将输入的 op作为该字典的关键来使用字典上的 []访问该值。这样，您现在可以使用该方法，只需通过参数(num1，num2(来 call 。

Answer 2

可以使用Altough eval，他们说的是正确的：除非它是严格的必要条件，否则您没有其他选择，并且您没有最大的安全性。好吧，只是不要使用它，尽管它们需要更多代码，但有多种方法可以做到这一点。

我的主张正在映射操作员：

import random
# NEW CODE
def sum(a, b):
    return a + b;
def substract(a, b):
    return a - b;
def multiply(a, b):
    return a * b;
# END OF NEW CODE
def start():
    print('n________________________________________')
    print('|                                      |')
    print('|         Zach's Tutorifier!          |')
    print('|                                      |')
    print('|                                      |')
    print('|     Instructions:                    |')
    print('| Select the type of math you want     |')
    print('| to practice with:                    |')
    print('| 1-Add 2-Subtract 3-Multiply          |')
    print('|--------------------------------------|')
start()
math = int(input('> ')) - 1
cat = {
    "+": sum,
    "-": substract,
    "*": multiply
}
def problems():
    # NEW CODE
    operator_char= cat.keys()[math]
    operation = cat[math]
    # END OF NEW CODE
    x = random.randint(0,9)
    y = random.randint(0,9)
    print('What is %i %s %i?' % (x, operator_char, y) )
    answer = input('> ')
    # NEW CODE
    if answer == operation(x, y):
    # END OF NEW CODE
        print('Congratulations! Try again? (Y/N)')
        again = input('> ')
        if again == 'Y' or again == 'y':
            problems()
        else:
            start()
    else: 
        print('Try again!')
problems()