如何使用python在一个月内平均选择休假？

实际上,必须有1天休息.

将问题改为每7天随机说2天(或将月份分成四段时间).然后,您将获得均匀分布.使用random.sample()作为Martijn Pieters建议.

您可以从第一周开始使用此技术生成两个值,然后按顺序生成它们,如果您需要逐个生成它们.

编辑：

正如tcaswell所观察到的,仍有一些情况下你最终连续十天值班.要解决这个问题,您可以每三天分配一天,创建十个列表,并从不影响7连续日标准的天数子集中随机删除两天.

或者,您可以使用原始算法生成列表,直到符合条件,因为您很可能无论如何都会获得有效的解决方案.你必须编写某种验证函数,但这样做很容易,因为你只计算最长的连续日期.

码：

第二种选择的实现.

import random
from itertools import chain
from itertools import count

def candidate(m):
    ''' Returns 2 days per week,in m days,where m is the length of the month. '''
    weeks = weeksmaker(m)
    return sorted(list(chain(*[random.sample(week,2) for week in weeks])))

def weeksmaker(m):
    ''' Divides a month up into four weeks,randomly assigning extra days to weeks. '''
    weeks = [range(i,i+7) for i in xrange(1,29,7)]
    for i in range(m - 28):
        weeks[random.randint(1,len(weeks))-1].append(i)
    c = count(1)
    return [[c.next() for day in week] for week in weeks]

def valid(days,c):
    ''' Validity check. Cant work more than c consecutive days. '''
    for i in xrange(1,len(days)):
        if days[i] - days[i-1] > c:
            return False
    else:
        return True

def daysoff(m,n,c):
    ''' In month length m,need n days off,cant work more than c consecutive days. '''
    while True:
        days = candidate(n)
        if valid(days,c):
            return days

>>> for i in range(28,32):
...     daysoff(i,7)
... 
[6,7,10,20,28]
[4,13,19,21,23,24]
[2,9,15,25,27]
[1,12,28]