是否有一种方法可以将数组的内容配对，以便以后删除它自己？（秘密圣诞老人计划）

Question

我想做一个秘密圣诞老人节目。理想情况下，程序应该要求至少三个名字，然后程序会询问是否有更多的名字。如果是，打开另一个文本字段并将其添加到名称列表中。如果没有，打破循环，创建一个单独的列表，其中包含名称，打乱第二个名称列表，并将它们与原始列表中的名称配对。

例如，如果John、Elliot、Sarah和Jenny输入了他们的名字，程序将他们的名字打乱，它应该输出这样的内容：

[('John', 'Sarah'), ('Elliot', 'John'), ('Sarah', 'Jenny'), ('Jenny', 'Elliot')]

但我得到的是：

[('John', 'John'), ('Elliot', 'John'), ('Sarah', 'Elliot'), ('Jenny', 'Jenny')]

代码如下：

import sys
import random

names = []

class Santa:
    #The class that takes the values, shuffles them, pairs and then displays them
    def __init__(self, name):
        self.turns = name
        self.people = names
        self.final = []

    def sort_names(self):
        random.shuffle(self.people)
        for name in self.turns:
            pair = (name, random.choice(self.people))
            if pair[0] == [1]:
                pair = (name, random.choice(self.people))

            else:
                self.final.append(pair)

    def print_name(self):
        input("\nNames are ready! Press Enter to show the names.")
        print(self.final)

def main():
    # The function asking for user input
    name = input("\nType in your name. ")
    names.append(name)
    name = input("\nType in the next name. ")
    names.append(name)
    name = input("\nType in the next name. ")
    names.append(name)
    while True:
        next = input("\nIs this everyone? Y/N ")
        if next.upper() == "Y":
            break
        elif next.upper() == "N":
            name = input("\nType in the next name. ")
            names.insert(0, name)
        else:
            print("\nInvalid response, please try again.")
    print(names)
    start = Santa(names)
    start.sort_names()
    start.print_name()
    input("\nRecord these, and press enter to quit.")
    sys.exit()

main()

Answer 1

    for name in self.turns:
        pair = (name, random.choice(self.people)) 

        # you compare against [1] ? That is a list containing a 1 as only thing 
        # and it will never be True unless if you compare string vs. name
        if pair[0] == [1]:
            pair = (name, random.choice(self.people)) 
        else:
            self.final.append(pair)

您可以简化您的循环，并且在配对人员时必须处理奇数/偶数个名称。使用列表切片和压缩使得配对更容易。

例子：

import sys
import random
 
class Santa: 
    def __init__(self, names): 
        self.people = names 

    def shuffle(self):
        random.shuffle(self.people)

        self.final = []

        if len(self.people) < 2:
            print("Sorry, not enough people to gift stuff around")
            return

         # taken from the comment, because superior to what I had originally
         # see edit history for my solution ;o) - this one suggested by @Matthias
         self.final = [(n1, n2) for n1, n2 in zip(names, names[1:] + [names[0]])]
 
    def print_name(self): 
        print(self.final)


name = "dummy"
names = []
while name:
    name = input(f"Whats the {len(names)+1}. name? (Empty to end) ").strip()
    if not name:
        break
    names.append(name)
print(names)
santa = Santa(names)
santa.shuffle()
santa.print_name() 

输出（省略输入输出）：

# odd inputs
['Tim', 'John', 'Luise', 'Maria', 'Wasja']
[('Wasja', 'Maria'), ('Maria', 'John'), ('Luise', 'Tim')]

['Tim', 'John', 'Luise', 'Maria', 'Wasja']
[('Maria', 'Wasja'), ('Wasja', 'Tim'), ('John', 'Luise')]

# even inputs
['Tim', 'John', 'Luise', 'Maria']
[('John', 'Tim'), ('Maria', 'Luise')]

['Tim', 'John', 'Luise', 'Maria']
[('Luise', 'Tim'), ('John', 'Maria')]

您可以在此处阅读有关zip和列表切片的更多信息：

• Python中的Zip列表

Answer 2

您不需要整个类来设置这个，我已经尝试清理了一些阅读输入。

import random


names = []
for _ in range(3):
  name = input("\nType in your name. ")
  names.append(name)

while True:
  next = input("\nIs this everyone? Y/N ")
  if next.upper() == "Y":
    break
  elif next.upper() == "N":
    name = input("\nType in the next name. ")
    names.insert(0, name)
  else:
      print("\nInvalid response, please try again.")
print(names)

random.shuffle(names)
pairs = [(first_person, second_person) for first_person, second_person in zip(names, names[1:] + [names[0]])]
print(pairs)