python - 从队列数据结构列表中调用项目 (Python)

Question

我试图将 3 个人排入列表中，以显示每个人的结果，他们的所有姓名都位于顶部，但只能获得一个没有任何姓名的结果:

Contacting the following  
Phone answered: Yes  
Booked an appointment: No  
Reshedule an appointment again.

我想让输出显示他们的所有姓名和 3 个输出，每个人一个来自 'names' 中存储的信息，并且每个名称不会出现两次。

我想使用队列根据列表对它们进行优先级排序，因此我尝试将它们按顺序排列。 if 和 elif 是根据随机生成器属于任一类别的条件。现在，只是未定义在其中包含名称的方法。

代码

import random

class Queue:
    def __init__(self):
        self.container = []

    def isEmpty(self):
        return self.size() == 0  

    def enqueue(self, item):
        self.container.append(item)

    def dequeue(self):
        self.container.pop(0)

    def size(self):
        return len(self.container)

    def peek(self) :
        return self.container[0]

names = ["Alvin", "James", "Peter"]

# Enqueuing

q = Queue()
q.enqueue(random.choice(names))

# Dequeuing and Printing
print("Contacting the following:\n" + "\n".join(q.container))  # unsure about this




for i in range(q.size()):

    answered = random.randint(0,1)
    booked = random.randint(0, 1)

    if(answered == 1 and booked == 1):
        print("Now Calling -" + (q.names)) # unsure about this
        print("Phone answered: Yes")
        print("Booked an appointment: Yes")
        print("Booking successful.")

    elif(answered==1 and booked==0):
        print("Now Calling -" + (q.names)) # unsure about this
        print("Phone answered: Yes")
        print("Booked an appointment: No")
        print("Reshedule an appointment again.")

    elif(answered == 0):
        print("Now Calling -" + (q.names)) # unsure about this
        print("Phone answered: No")
        print("Reshedule a callback.")

    q.dequeue()

所需输出示例:

Contacting the following
Alvin
James
Peter

Now Calling - James
Phone answered: No
Reshedule a callback.

Answer 1

我对您的队列类进行了一些更改。首先，.dequeue 方法没有返回它弹出的项目，因此它返回默认值 None。

我还将 .size 方法更改为 __len__，以便您可以将 Queue 实例传递给内置 len函数。并为其提供了一个 iter 方法，您可以轻松地在 for 循环中使用它，或将其传递给 .join。我还将 .isEmpty 更改为 .is_empty 以符合 Python 的 PEP-0008 样式指南。

由于您希望将每个名称随机添加到队列中而不重复，因此我们不希望此处使用 random.choice。相反，我们可以使用random.shuffle；另一种选择是使用random.sample，尽管当您想从列表中进行部分选择时，这更合适。

from random import seed, shuffle, randrange

# Seed the randomizer so we can reproduce results while testing
seed(9)

class Queue:
    def __init__(self):
        self.container = []

    def __len__(self):
        return len(self.container)

    def is_empty(self):
        return len(self) == 0

    def enqueue(self, item):
        self.container.append(item)

    def dequeue(self):
        return self.container.pop(0)

    def peek(self) :
        return self.container[0]

    def __iter__(self):
        return iter(self.container)

names = ["Alvin", "James", "Peter"]

# Enqueuing
q = Queue()

# Make a temporary copy of the names that we can 
# shuffle without affecting the original list
temp = names.copy()
shuffle(temp)

# Put the shuffled names onto the queue
for name in temp:
    q.enqueue(name)

# Dequeuing and Printing
print("Contacting the following")
print('\n'.join(q))
#for name in q:
    #print(name)

while not q.is_empty():
    name = q.dequeue()
    print('\nNow Calling -', name)

    answered = randrange(2)
    booked = randrange(2)

    if answered:
        print("Phone answered: Yes")
        if booked:
            print("Booked an appointment: Yes")
            print("Booking successful.")
        else:
            print("Booked an appointment: No")
            print("Reshedule an appointment again.")
    else:
        print("Phone answered: No")
        print("Reshedule a callback.")

输出

 Contacting the following
Alvin
Peter
James

Now Calling - Alvin
Phone answered: Yes
Booked an appointment: No
Reshedule an appointment again.

Now Calling - Peter
Phone answered: No
Reshedule a callback.

Now Calling - James
Phone answered: Yes
Booked an appointment: Yes
Booking successful.

在上面的代码中我使用了

print('\n'.join(q))

打印所有名称，因为您在代码中使用.join来实现此目的。但我还展示了使用简单的 for 循环的替代方法，但我将其注释掉了:

for name in q:
    print(name)