Question

我有开始和结束时间。如下所示

def overlap(start1, end1, start2, end2):
    pass

因此，两个时间安排都会安排会议，我想确保没有会议相互重叠。因此，如果两次会议重叠，则会返回True其他False。

我是用JavaScript完成的，我想用Python做，因为我是Python的新手，我不知道应该使用什么模块以及如何使用它们。这是我在JavaScript上的实现。

// time of first timespan
var x = new Date('01/01/2001 8:30:00').getTime();
var y = new Date('01/01/2001 9:30:00').getTime();

// time of second timespan
var a = new Date('01/01/2001 8:54:00').getTime();
var b = new Date('01/01/2001 9:00:00').getTime();

if (Math.min(x, y) <= Math.max(a, b) && Math.max(x, y) >= Math.min(a, b)) {
    // between
}

Answer 1

这个怎么样：

from dateutil import parser

def date_overlap(date1, date2):
    obj_1 = map(parser.parse, date1)
    obj_2 = map(parser.parse, date2)

    # is latest start before earliest end? If so, overlap!
    if max(obj_1[0], obj_2[0]) < min(obj_1[1], obj_2[1]):
        return True
    else:
        return False

date_1 = ('01/01/2001 8:30:00', '01/01/2001 9:30:00')
date_2 = ('01/01/2001 8:54:00', '01/01/2001 9:00:00')

print date_overlap(date_1, date_2)

dateutil.parser将字符串转换为日期。如果我们最新的开始是在我们最早的结束之前，那就有重叠。

注意：如果您使用的是Python 3.x，则需要将两个map输出包装为如下所示的列表：

from dateutil import parser

def date_overlap(date1, date2):
    obj_1 = list(map(parser.parse, date1))
    obj_2 = list(map(parser.parse, date2))

    # is latest start before earliest end? If so, overlap!
    if max(obj_1[0], obj_2[0]) < min(obj_1[1], obj_2[1]):
        return True
    else:
        return False

date_1 = ('01/01/2001 8:30:00', '01/01/2001 9:30:00')
date_2 = ('01/01/2001 8:54:00', '01/01/2001 9:00:00')

print date_overlap(date_1, date_2)

Answer 2

import datetime

def overlap(start1, end1, start2, end2):
    if start1 < start2 and end1 > start2 or \
       start1 > start2 and end1 < end2 or \
       start1 > start2 and start1 < end2: 
        return True
    else:
        return False

start_date1 = datetime.datetime(2017,1,1)
end_date1 = datetime.datetime(2017,1,3)
start_date2 = datetime.datetime(2017,1,2)
end_date2 = datetime.datetime(2017,1,4)

print(overlap(start_date1, end_date1, start_date2, end_date2))

你也可以使用时间。

Answer 3

这是我在短时间内可以做的JS代码的最接近的解释：

from datetime import datetime

#assuming time format comes in a string like this: '8:30:00'
#assumming provided time has sense, i.e. end1 > start1 and end2 > start2

TIME_FORMAT = '%H:%M:%S'

def overlap(start1, end1, start2, end2):
    #transform time        
    start1_time = datetime.strptime(start1, TIME_FORMAT )
    end1_time = datetime.strptime(end1, TIME_FORMAT )
    start2_time = datetime.strptime(start2, TIME_FORMAT )
    end2_time = datetime.strptime(end2, TIME_FORMAT )

    #checking conditions
    if (min(start1_time, end1_time) <= max(start2_time, end2_time)) \
    and \
    (max(start1_time, end1_time) >= min(start2_time, end2_time)):
        return True
    else:
        return False

>>>overlap('8:30:00','9:30:00','8:54:00','9:00:00')
True

>>>overlap('8:30:00','9:30:00','9:54:00','10:00:00')
False

PS：唯一的模块是datetime，因为Python在其核心提供了min和max。

EDIT1：仅使用时间而非日期。

EDIT1：缩进更新。

Answer 4

def overlap(start1, end1, start2, end2):
    return (start1 <= start2 <= end1) or (start2 <= start1 <= end2)

如何在Python中获得特定的时间范围

4 个答案: