Python甚至组分配

Question

我是python的初学者，正在研究随机团队生成器。我遇到的问题是我不确定如何使它产生甚至团队。下面是代码以及示例输出。

import random


def main():
    run = True
    while run:

        try:
            print("Welcome to this group picker, follow the instructions and your groups will be picked.")
            groupnum = int(input("How many groups do you want?"))
            peoplenum = int(input("How many people are there?"))
            print("Okay, assign everyone a number from 0 to", peoplenum - 1, ".")
            nums = []
            for i in range(0, peoplenum):
                nums.append(i)

            for i in nums:
                print("Number", i, "is in group", random.randint(1, groupnum))

            break

        except:
            print("Error, please follow instructions and enter only numbers.")
            break


main()

示例输出：

 Welcome to this group picker, follow the instructions and your groups
 will be picked.
 How many groups do you want?2 
 How many people are there?8
 Okay, assign everyone a number from 0 to 7 . 
 Number 0 is in group 1 
 Number 1 is in group 2 
 Number 2 is in group 1 
 Number 3 is in group 2 
 Number 4 is in group 1 
 Number 5 is in group 1 
 Number 6 is in group 2 
 Number 7 is in group 1

Answer 1

仅使用标准库，我将这样解决：

import random
from itertools import accumulate


def print_groups(n, g):
    # Prepare group separators
    size = n // g
    rem = n % g
    separators = list(accumulate([0] + [size+1] * rem + [size] * (g - rem)))

    # Make raw data
    items = list(range(n))
    random.shuffle(items)

    # Iterate and print
    for i, s in enumerate(zip(separators, separators[1:])):
        group = items[slice(*s)]
        print(f'Group {i+1}: {group} (size {len(group)})')

如果您的人数可被组的数量整除，则所有组的大小将相同，否则前n % g个组将获得一个额外的成员。

示例1：

print_groups(12, 4)

Group 1: [6, 11, 10] (size 3)
Group 2: [7, 2, 5] (size 3)
Group 3: [3, 1, 9] (size 3)
Group 4: [4, 8, 0] (size 3)

示例2：

print_groups(14, 4)

Group 1: [8, 3, 4, 6] (size 4)
Group 2: [1, 11, 0, 12] (size 4)
Group 3: [7, 5, 9] (size 3)
Group 4: [13, 10, 2] (size 3)

Answer 2

问题是为每个球员随机挑选一支球队。由于random.randint产生均等的价值分配，因此每个球员都有被分配给任何给定团队的相同机会，因此您可以最终成为同一团队中的每个人。

相反，您应该考虑遍历各个团队并为其分配一个随机的玩家。

该想法的实施不佳会

>>> import random
>>> 
>>> groupnum = 2
>>> peoplenum = 8
>>> 
>>> people = [i for i in range(peoplenum)]
>>> 
>>> for i in range(peoplenum):
...     group = i % groupnum
...     person = random.choice(people)
...     people.remove(person)
...     print('Number {} assigned to {}'.format(person, group))
... 
Number 6 assigned to 0
Number 7 assigned to 1
Number 4 assigned to 0
Number 3 assigned to 1
Number 2 assigned to 0
Number 5 assigned to 1
Number 1 assigned to 0
Number 0 assigned to 1

这将解决问题，但它依靠致电remove来避免重复团队成员。为避免这种情况，您可以随机排列玩家列表（使其随机排列），并在团队列表中zip将结果{}。

>>> import random
>>>
>>> players = [i for i in range(peoplenum)]
>>> teams = [i for i in range(groupnum)]
>>> 
>>> random.shuffle(players)
>>> 
>>> [x for x in zip(players, teams)]
[(7, 0), (5, 1)]

这显然行不通。这是因为zip将在最短的迭代器停止时停止，在本例中为groups。我们想要的是在有玩家的情况下重复播放。我们可以使用itertools.cycle实现这种功能：

>>> import random
>>> import itertools
>>> 
>>> players = [i for i in range(peoplenum)]
>>> teams = itertools.cycle(range(groupnum))
>>> 
>>> random.shuffle(players)
>>> 
>>> [x for x in zip(players, teams)]
[(7, 0), (2, 1), (0, 0), (1, 1), (5, 0), (4, 1), (3, 0), (6, 1)]

Answer 3

根据组数调整每个组的概率。使用while循环可连续分配组，并且仅在组的总和相等时才返回。

Answer 4

我将使用numpy.random.choice进行此操作，并将replace设置为false。 See relevant numpy documentation here

尝试使用此代码：

import numpy as np

def assign_people_to_teams(people_count = 8, team_count = 2):
    """
        selects people_count random numbers in the range 0 to people_count-1 without replacement
        then make a team assignment decision based on desired number of teams.
    """
    return [
        element % team_count for element in list(
            np.random.choice(
                people_count,
                people_count,
                replace=False
            )
        )
    ]

team_assignment = assign_people_to_teams()

print(team_assignment)

结果：

[0, 0, 1, 0, 0, 1, 1, 1]

Answer 5

您将必须为每个组使用一组，然后为该人选择随机组。一旦每个组都用尽，然后重复直到所有组都填满。

import random
import sys


def main():
    while True:

        try:
            print("Welcome to this group picker, follow the instructions and your groups will be picked.")
            groupnum = int(input("How many groups do you want?"))
            peoplenum = int(input("How many people are there?"))
            if groupnum == 'Q':
                sys.exit(0)
            print("Okay, assign everyone a number from 1 to", peoplenum, ".")

            nums = []
            for i in range(1, peoplenum+1):
                nums.append(i)

            ''' Validate number of people vs groups '''
            if peoplenum % groupnum:
                print("error: You have incorrect number of people to divide them equally")
                sys.exit(1)

            ''' People per team '''
            peoplePerTeam = int(peoplenum/groupnum)

            ''' Initialize the group'''
            group = {}
            for i in range(1,groupnum+1):
                group[i] = set()

            ''' Loop through the people'''
            for person in nums:
                while True:
                    ''' Find random team '''
                    randomGroup = random.randint(1,groupnum)

                    ''' If team is not filled up yet, then add otherwise find a new team'''
                    if len(group[randomGroup]) < peoplePerTeam:
                        group[randomGroup].add(person)
                        break

            ''' Display the team info'''    
            for i in range(1,groupnum+1):
                print("Team %s members are %s" %(i, group[i]))

        except:
            print("Error, please follow instructions and enter only numbers.")
            break


main()

该程序输出的示例运行如下：

Welcome to this group picker, follow the 
instructions and your groups will be picked.
How many groups do you want?2
How many people are there?8
Okay, assign everyone a number from 1 to 8 .
**Team 1 members are {1, 2, 5, 7}**
**Team 2 members are {8, 3, 4, 6}**
Welcome to this group picker, follow the 
instructions and your groups will be picked.
How many groups do you want?4
How many people are there?20
Okay, assign everyone a number from 1 to 20 .
**Team 1 members are {1, 18, 19, 20, 17}**
**Team 2 members are {16, 13, 11, 5, 7}**
**Team 3 members are {10, 9, 2, 3, 4}**
**Team 4 members are {8, 12, 15, 6, 14}**
Welcome to this group picker, follow the instructions and your groups will be picked.

Answer 6

解决方案

import random


def main():
    run = True
    while run:

            print("\nWelcome to this group picker, follow the" \
                "instructions and your groups will be picked.\n")

            group_amount = int(input("How many groups do you want?\n"))
            people_amount = int(input("How many people are there?\n"))

            print("\nOkay, assign everyone a number from 1 to " +
                str(people_amount) + " .\n")

            group = list(range(0, group_amount))
            person = list(range(0, people_amount))
            group_size = people_amount / group_amount


            if group_size % 2 != 0:
                reg_group_size = (people_amount - 1) / group_amount
                odd_group_size = ((people_amount - 1) / group_amount) + 1

            for i in group[0:-1]:
               group[i] = reg_group_size

            group[-1] = odd_group_size

            for p in person:
                r = random.randint(0, len(group)-1)
                while group[r] == 0:
                    r = random.randint(0, len(group)-1) 
                person[p] = r + 1
                group[r] -= 1
                print("Person " + str(p + 1) + " is in group " + 
                    str(person[p]) + ".")

main()

输出

(xenial)vash@localhost:~/pcc/12/alien_invasion_2$ python3 helping.py

Welcome to this group picker, follow theinstructions and your groups will be picked.

How many groups do you want?
3
How many people are there?
10

Okay, assign everyone a number from 1 to 10 .

Person 1 is in group 3.
Person 2 is in group 1.
Person 3 is in group 2.
Person 4 is in group 2.
Person 5 is in group 1.
Person 6 is in group 3.
Person 7 is in group 2.
Person 8 is in group 1.
Person 9 is in group 3.
Person 10 is in group 3.

评论

我同意所有上述解决方案，但是我认为大多数解决方案与您的原始代码有很大的出入。

我试图忠实于您的代码，同时实现了使代码按预期运行并保留原始输出所需的缺失部分。

主要问题是，random.randint值并不能说明将人员分配给一个组后每个组中可用空间的缩小情况。

步骤1：

        group = list(range(0, group_amount))
        person = list(range(0, people_amount))
        group_size = people_amount / group_amount

这将创建一个列表group和person，它们是各自金额的大小。另外，让我们创建group_size，它将为我们提供每个person中可以包含的group。

第2步：

        if group_size % 2 != 0:
            reg_group_size = (people_amount - 1) / group_amount
            odd_group_size = ((people_amount - 1) / group_amount) + 1

        for i in group[0:-1]:
           group[i] = reg_group_size

        group[-1] = odd_group_size

如果group的大小不相等，我们在此解决该问题。这将创建所有大小均匀的group，最后一个group可以保留剩余的person

第3步：

        for p in person:
            r = random.randint(0, len(group)-1)
            while group[r] == 0:
                r = random.randint(0, len(group)-1) 
            person[p] = r + 1
            group[r] -= 1
            print("Person " + str(p + 1) + " is in group " + 
                str(person[p]) + ".")

现在有趣的部分！此循环会将person[p]分配给随机的group，然后将该group的数量减少1。

让我们分解一下：

        for p in person:
            r = random.randint(0, len(group)-1)
            while group[r] == 0:
                r = random.randint(0, len(group)-1)

我们将遍历person来分配所有人口。 然后，我们选择一个随机的group来分配person，但我们要确保group[r]有可用空间，while group[r] == 0:进行检查以确保是否有空间，则会生成一个新的r，直到找到一个有可用空间的随机group为止。

            person[p] = r + 1
            group[r] -= 1
            print("Person " + str(p + 1) + " is in group " + 
                str(person[p]) + ".")

最后，我们给person[p] = r + 1赋予person[p]一个group的数字（使用r + 1是，因此我们可以消除“组0”，因为列表从0开始，对于演示文稿，我们想从1）开始。之后，group[r]的值将减少1，说明可用空间减少了。再次在print语句中，+ 1就是这样，我们再也没有将任何人称为“ Person 0”。

希望这对您有所帮助，我很乐意为此工作！