我们需要创建一个可用于“秘密圣诞老人”游戏的程序：
from random import *
people=[]
while True:
person=input("Enter a person participating.(end to exit):\n")
if person=="end": break
people.append(person)
shuffle(people)
for i in range(len(people)//2):
print(people[0],"buys for",people[1])
这是我开发的程序。截至目前，如果我输入 3 个人（例如 Bob、Ben、Bill），它将返回“Ben 为 Bill 购买”，而没有人为 Ben 或 Bob 购买。我目前正试图让它输出“Bob 为 Ben 购买，Ben 为 Bill 购买，Bill 为 Bob 购买”，但到目前为止还没有成功。如果有人可以给我一个提示/基础来设置它，将不胜感激。另外，如果我的代码中有任何错误不允许我完成此操作，您能告诉我吗？谢谢。
i第一个提示，在for循环中使用像 0 和 1 这样的常量是没有意义的。
for i in range(len(people)):
print(people[i],"buys for",people[(i+1)%(len(people))])
但是，此实现不会为您提供秘密圣诞老人提供的所有可能性。
假设您输入“Alice”、“Bob”、“Claire”、“David”，您永远不会遇到以下情况：
你只会得到循环排列，即：
等等。
你需要一些额外的工作来制作一个完美的秘密圣诞老人:)
您正在索引 0 和 1，因此它始终是第一个和第二个被打印的人。你真正想要的是：
shuffle(people)
offset = [people[-1]] + people[:-1]
for santa, receiver in zip(people, offset):
print(santa, "buys for", receiver)
from random import *
prompt = "Enter a person participating.(end to exit):\n"
people = list(iter(lambda:input(prompt), "end"))
shuffle(people)
people.append(people[0])
for i in range(len(people) - 1):
print(people[i],"buys for", people[i + 1])
样品运行
Enter a person participating.(end to exit):
A
Enter a person participating.(end to exit):
B
Enter a person participating.(end to exit):
C
Enter a person participating.(end to exit):
end
A buys for B
B buys for C
C buys for A
你可以更换
people=[]
while True:
person=input("Enter a person participating.(end to exit):\n")
if person=="end": break
people.append(person)
和
prompt = "Enter a person participating.(end to exit):\n"
people = list(iter(lambda:input(prompt), "end"))
他们都在做同样的事情。iter函数将继续执行我们作为第一个参数传递的函数，直到第二个值匹配。
然后，你这样做
people.append(people[0])
它是一个循环的东西，最后一个人必须为第一个人购买。append将在最后插入。
for i in range(len(people) - 1):
我们这样做len(people) - 1，因为如果n有人，就会有n购买。在这种情况下，我们在最后添加了第一个人，所以我们减去了一个。最后
print(people[i],"buys for", people[i + 1])
每个人都必须为列表中的下一个人购买。所以，people[i]为people[i + 1].
假设你有三个人：['Bob', 'Ben', 'Bill']。
In [1]: people = ['Bob', 'Ben', 'Bill']
现在，当您获得此列表的长度时，您正在执行除以 2。这将导致：
In [2]: len(people) // 2
Out[2]: 1
这就是为什么你只能得到一行输出。
你怎么能得到你想要的秘密圣诞老人结果？以下是实现此功能的简单方法的一些提示：
我意识到我的回答有点晚了，这是我过去几年一直在使用的秘密圣诞老人项目，我的回答可能会有所帮助。我从我的剧本中剪掉了重要的部分。
def pick_recipient(group,recipients,single_flag):
for person in group:
gift = random.choice(recipients)
if single_flag == 0:
while gift in group:
gift = random.choice(recipients)
else:
while gift in person:
gift = random.choice(recipients)
mail_list.append( '%s=%s' %(person,gift))
recipients.remove(gift)
return recipients
if __name__ == "__main__":
global mail_list
mail_list = []
#create lists of people, group couples at beginning or end of list and the singles opposite
all_recipients = ['name_1-CoupleA: name_1-CoupleA@gmail.com','name_2-CoupleA: name_2-CoupleA@gmail.com',
'name_3-CoupleB: name_3-CoupleB@hotmail.com','name_4: name_4CoupleB@hotmail.com',
'name_5-Single: name_5-Single@gmail.com','name_6-Single: name_6-Single@gmail.com']
#create couples and lists of singles to make sure couples don't get their other half
#modify the groups to match the list of people from above
coupleA = all_recipients [0:2]
coupleB = all_recipients [2:4]
single = all_recipients [4:]
#keep initial list in tact
possible_recipients = all_recipients
#modify the groups to match what the input list is
possible_recipients = pick_recipient(coupleA,possible_recipients,single_flag=0)
possible_recipients = pick_recipient(coupleB,possible_recipients,single_flag=0)
possible_recipients = pick_recipient(single,possible_recipients,single_flag=1)
print mail_list
https://github.com/savitojs/secret-santa-finder-script
要使用此脚本，您需要在文件中添加您正在玩的候选人的名称secret_santa.py。_from 和 _to 列表应该相同以获得更好的结果。
更改_from和_to列表
$ ./secret_santa.py
Hi.. What is your nick name? [ENTER] savsingh
Hey savsingh!, Hold on!! Searching...
Any guess... ?
Your secret santa is: bar
Clear the screen [ENTER]:
很快...
脚本在这里：
#!/usr/bin/python
#===============================================================================
#
# FILE: secret_santa.py
#
# USAGE: ./secret_santa.py
#
# DESCRIPTION: It can help you to find your secret santa, add participants name in _from and _to list
#
# REQUIREMENTS: python 2.6+, _from list and _to list should be same to work perfectly.
# AUTHOR: Savitoj Singh (savitojs@gmail.com)
# VERSION: 1.1
# CREATED: 12/21/2016 05:13:38 AM IST
# REVISION: * Initial release
# * Fixed minor bugs
#=============================================================================== import random import os import time import sys '''Add participants to list'''
_from = ['bob','foo','bar','savsingh','tom','jack','mac','hex'] '''Add participants to list'''
_to = ['bob','foo','bar','savsingh','tom','jack','mac','hex']
class bcolors:
HEADER = '\033[95m'
OKBLUE = '\033[94m'
OKGREEN = '\033[92m'
WARNING = '\033[93m'
FAIL = '\033[91m'
ENDC = '\033[0m'
BOLD = '\033[1m'
UNDERLINE = '\033[4m'
for i in range(len(_from)):
try:
user_name = raw_input('Hi.. What is your nick name? [ENTER] ')
if user_name not in _from and user_name not in _to:
try:
print("")
print(bcolors.FAIL + bcolors.BOLD + "Nick name doesn't seem to be in `_from` or `_to` list" + bcolors.ENDC)
print(bcolors.HEADER + "\nDo you expect me to look in galaxy ? It may take many x'mas to retrive results :P" + bcolors.ENDC)
break
except:
pass
else:
print('Hey ' + bcolors.OKBLUE + bcolors.BOLD + str(user_name) + bcolors.ENDC + '!, Hold on!! Searching...')
random.shuffle(_from)
_from.remove(user_name)
print(bcolors.WARNING + 'Any guess... ?' + bcolors.ENDC)
print('')
time.sleep(1)
a = random.choice(_to)
while str(a) == str(user_name):
a = random.choice(_to)
_to.remove(a)
print(bcolors.BOLD + 'Your secret santa is: ' + bcolors.ENDC + bcolors.OKGREEN + bcolors.BOLD + str(a) + bcolors.ENDC)
raw_input('Clear the screen [ENTER]: ')
os.system('reset')
except:
print(bcolors.FAIL + bcolors.BOLD + "\n\nOops!!, Something went wrong..." + bcolors.ENDC)
sys.exit(1)