python - 将参数传递给python中字典调用的函数

Question

我是 python 和实现堆栈的新手。我正在使用字典调用堆栈函数。但是，push() 要求我传递一个参数。我该怎么做？

class stack():

    def __init__(self):
        self.items = []

    def push(self,item):
        self.items.append(item)

    def pop(self):
        return self.items.pop()

    def isEmpty(self):
        return self.items == []

    def getStack(self):
        return print(self.items)

s = stack()
switcher = {
           '1' : s.push,
           '2' : s.pop,
           '3' : s.isEmpty,
           '4' : s.getStack,
        }

def dictionaryCall(key):
    switcher[key]()

while(1):
    key = input('enter choice 1.push 2. pop 3.isEmpty 4 getStack 5.exit: ')
    if key == '5':
        break
    dictionaryCall(key)

Answer 1

这是一个有效的实现。如您所见，您必须向 dictionaryCall 添加一个可选的第二个参数(此处称为 element)，然后将其传递给选定的方法。

class stack():

    def __init__(self):
        self.items = []

    def push(self,item):
        self.items.append(item)

    def pop(self):
        print(self.items.pop())

    def isEmpty(self):
        # I dropped the `return` here due to syntax error
        print (self.items == [])

    def getStack(self):
        # I dropped the `return` here due to syntax error
        print (self.items)

s = stack()
switcher = {
            '1' : s.push,
            '2' : s.pop,
            '3' : s.isEmpty,
            '4' : s.getStack,
        }

# I added an optional parameter for the `push` method
def dictionaryCall(key, element = None):
    method = switcher[key]

    # call the method with element if it exists
    if element == None: method()
    else: method(element)

# call from command line
if __name__ == '__main__':
    while(1):
        key = input('enter choice 1.push 2. pop 3.isEmpty 4 getStack 5.exit: ')
        if key == '5':
            break

        print ('> {0}'.format(key))

        if key == '1':
            element = input('enter an element to push onto the stack: ')
            dictionaryCall(key, element)
        else: dictionaryCall(key)

示例:

enter choice 1.push 2. pop 3.isEmpty 4 getStack 5.exit: '1'
> 1
enter an element to push onto the stack: 'foo'
enter choice 1.push 2. pop 3.isEmpty 4 getStack 5.exit: '3'
> 3
False
enter choice 1.push 2. pop 3.isEmpty 4 getStack 5.exit: '4'
> 4
['foo']
enter choice 1.push 2. pop 3.isEmpty 4 getStack 5.exit:

---

需要注意的几点:

• return print (...) 是无效语法(只写 print (...))
• 通过 input 从命令行读取的参数被评估(这意味着你必须输入例如 '1' 来推送一个元素，因为你的字典有字符串键<
• 要从命令行运行 python 程序，您需要将其包装在 if __name__ == '__main__': ... 否则文件将被解析但不会执行
• 正如@Pm2Ring 在评论中所说，pop 应该打印弹出的元素