Question

我正在为我的介绍编程课程编写一个程序，我需要制作一个计算器来打印结果，就像历史一样。我这样做的方式是第一个数字，算术运算和第二个数字是列表中的单独对象，但我需要将它们组合成列表中的完整对象。我已经尝试了几种方法来做到这一点，但每次都会崩溃或者不能正常工作。

history = []
first_number = int(input("please input your first number"))
second_number = int(input("please input your second number"))
operator = input("ADD, SUBTRACT, DIVIDE, MULTIPLY, CHOOSE again or STOP? ").lower()

def add():
    return(first_number + second_number)

def problemAdd(): 
    return(first_number, "+", second_number)

while operator != "stop":
    if operator == "add":
        print("the problem was:", first_number, "+", second_number, "=", add())
        history.append (problem())
        print(history)
        first_number = int(input("please input your first number"))
        second_number = int(input("please input your second number"))
        operator = input("ADD, SUBTRACT, DIVIDE, MULTIPLY, CHOOSE again or STOP? ").lower()

现在这部分代码我希望它足以发现问题。这就是它的结果：

please input your first number2
please input your second number2
would you like to ADD, SUBTRACT, DIVIDE, MULTIPLY, CHOOSE again or STOP? add
the problem was: 2 + 2 = 4
[(2, '+', 2)]
please input your first number1
please input your second number2
would you like to ADD, SUBTRACT, DIVIDE, MULTIPLY, CHOOSE again or STOP? add
the problem was: 1 + 2 = 3
[(2, '+', 2), (1, '+', 2)]
please input your first number

我经历了两次，以显示每个问题的显示方式。

Answer 1

# lst = [(2, '+', 2)]
for tpl in lst:
   operation = " ".join(map(str,tpl))
   # map(str, tpl) returns an object where every element in tpl is mapped
   # using the str function. str(object) returns the value of object as a
   # string.

   result = eval(operation)
   # eval IS A BAD IDEA, BUT SIMPLE TO IMPLEMENT

   print("{} = {}".format(operation, result))

如果这是你正在尝试做的事情，我们有点不清楚。我不确定你问题中[ ..., "3 - 2 = 1"]部分的来源。也许这会有所帮助吗？

如果您只有两个操作数和一个运算符（例如，所有内容都是a ? b形式，其中?是运算符，那么这样更安全：

def do_operation(operation):
    import operator
    operand_1, operator, operand_2 = operation
    try:
        f = {"+": operator.add, "-": operator.sub,
             "/": operator.truediv, "*": operator.mul,
             "//": operator.floordiv, "%": operator.mod}[operator]
    except KeyError:
        raise ValueError("Invalid operator")
    return f(operand_1, operand_2)

for tpl in lst:
    operation = " ".join(map(str,tpl))
    result = do_operation(tpl)
    print("{} = {}".format(operation, result))

直接回答您的问题：

history = []

first_number = int(input("please input your first number"))    
second_number = int(input("please input your second number"))
# what do you do if the user doesn't enter a number? Your program crashes

operator = input("ADD, SUBTRACT, DIVIDE, MULTIPLY, CHOOSE again or STOP? ").lower()
op_mapping = {"add":"+", "subtract":"-", "divide":"/", "multiply":"*"}

if operator in op_mapping: # this will exclude 'choose' and 'stop'
    operator = op_mapping[operator]
elif operator == 'choose': # handle it
elif operator == 'stop': # handle it, these are up to you
else: # invalid input, handle it.

# NEW
operation = (first_number, operator, second_number)
# this makes it easier to refer to all three at once.

def calculate(operation):
    """Calculates the result based on which operator is used"""
    import operator
    operand_1, operator, operand_2 = operation
    try:
        f = {"+": operator.add, "-": operator.sub,
             "/": operator.truediv, "*": operator.mul,
             "//": operator.floordiv, "%": operator.mod}[operator]
    except KeyError:
        raise ValueError("Invalid operator")
    return f(operand_1, operand_2)

## while operator != "stop": # you're gonna handle this above

## if operator == "add": # we modified it so our mapping handles all if cases
##     print("the problem was:", first_number, "+", second_number, "=", add())  
##     history.append (problem())  
##     print(history)  
##     first_number = int(input("please input your first number"))    
##     second_number = int(input("please input your second number"))  
##     operator = input("ADD, SUBTRACT, DIVIDE, MULTIPLY, CHOOSE again or STOP? ").lower()

human_readable = " ".join(map(str,operation))
history.append(human_readable)
print("{} = {}".format(human_readable, calculate(operation))
print(history)

# LOOP AS NEEDED, IMPLEMENTED BY YOU.

请注意，如果我要进行完全重写，它可能会像：

def get_input():
    """Get user input for the calculation"""
    inputprompt = """Return the result of a calculation, based on user input.
Your input must be of the form X ? Y where X, Y are any number and ? is one of:
+, -, /, //, *, %
Please include spaces between operand and operator, or STOP to stop.

>> """
    return input(inputprompt)

def calculate(operation):
    """Calculates the result based on which operator is used"""
    import operator
    operand_1, operator, operand_2 = operation
    try:
        operand_1, operand_2 = float(operand_1), float(operand_2)
    except ValueError:
        raise ValueError("Invalid operand")
    try:
        f = {"+": operator.add, "-": operator.sub,
             "/": operator.truediv, "*": operator.mul,
             "//": operator.floordiv, "%": operator.mod}[operator]
    except KeyError:
        raise ValueError("Invalid operator")
    return f(operand_1, operand_2)

def main():
    from collections import deque
    history = deque()
    # a deque is a list that allows easy popping and appending from left OR right
    MAX_HISTORY_LENGTH = 10
    while True
        print("CALCULATOR:\n\n")
        readable = get_input()
        if "stop" in readable.lower():
            break
        operation = readable.split()
        history.append(readable)
        if len(history) > MAX_HISTORY_LENGTH:
            history.popleft()
        print("{} = {}".format(readable, calculate(operation))
        input(" ( press ENTER to continue ) ")

main()
# we love functional programming!

需要帮助将整数转换为列表中的字符串

1 个答案:

直接回答您的问题：