假设我有这些分配日期为所有员工工作:
01-JANUARY-2019 TO 31-JANUARY-2019
假设下面的员工已经分配了工作日期,可能是两个或两个以上的员工在同一块板上工作,比如员工A和员工D
Employee A : 01-JANUARY-2019 TO 04-JANUARY-2019
Employee B : 22-JANUARY-2019 TO 25-JANUARY-2019
Employee C : 10-JANUARY-2019 TO 20-JANUARY-2019
Employee D : 02-JANUARY-2019 TO 06-JANUARY-2019
我试着从范围中找出一个一个的日期,然后与slab进行比较,但这需要更多的时间
from datetime import timedelta, date
def daterange(date1, date2):
for n in range(int ((date2 - date1).days)+1):
yield date1 + timedelta(n)
因此,预期结果将是:
Total Allocation : 01 January to 31 January == 31 Days
Allocated Employee in Between Dates :
So Final Gap would expect as result = 31 Days - (6 + 11 + 4) = 10 DAYS
这正是你所期待的
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