将Python datetime拆分为小时对齐的块
我有一个将Python datetime拆分为小时对齐的块,python,datetime,split,Python,Datetime,Split,我有一个datetime.datetime实例,d和一个datetime.timedelta实例td,我试图编写一个函数,将(d,d+td)范围分解为[(d,x1),(x1,x2),…,(xn,d+td)],其中xn变量都与小时对齐 例如,如果 d = datetime.datetime(2012, 9, 8, 18, 53, 34) td = datetime.timedelta(hours=2, minutes=34, seconds=5) 我想要一份清单 [(datetime.datet
datetime.datetime
实例,d
和一个datetime.timedelta
实例td
,我试图编写一个函数,将(d,d+td)
范围分解为[(d,x1),(x1,x2),…,(xn,d+td)]
,其中xn
变量都与小时对齐
例如,如果
d = datetime.datetime(2012, 9, 8, 18, 53, 34)
td = datetime.timedelta(hours=2, minutes=34, seconds=5)
我想要一份清单
[(datetime.datetime(..., 18, 53, 34), datetime.datetime(..., 19, 0, 0)),
(datetime.datetime(..., 19, 0, 0), datetime.datetime(..., 20, 0, 0)),
(datetime.datetime(..., 20, 0, 0), datetime.datetime(..., 21, 0, 0)),
(datetime.datetime(..., 21, 0, 0), datetime.datetime(..., 21, 27, 39))]
有人能提出一个很好的、类似于蟒蛇的方法来实现这一点吗?chunks=[]
chunks = []
end = d + td
current = d
# Set next_current to the next hour-aligned datetime
next_current = (d + datetime.timedelta(hours=1)).replace(minute=0, second=0)
# Grab the start block (that ends on an hour alignment)
# and then any full-hour blocks
while next_current < end:
chunks.append( (current, next_current) )
# Advance both current and next_current to the following hour-aligned spots
current = next_current
next_current += datetime.timedelta(hours=1)
# Grab any remainder as the last segment
chunks.append( (current, end) )
结束=d+td
电流=d
#将next_current设置为下一小时对齐日期时间
next_current=(d+datetime.timedelta(小时=1)).replace(分钟=0,秒=0)
#抓住起始块(在一小时对齐时结束)
#然后是整小时的街区
当下一个_电流<结束时:
chunks.append((当前,下一个\当前))
#将当前和下一个_电流提前到以下时间点
电流=下一个\u电流
next_current+=datetime.timedelta(小时=1)
#抓取任何余数作为最后一段
chunks.append((当前,结束))
这里的主要假设是,初始指定的timedelta不是负值。如果您这样做,您将得到一个单块列表
[(x,y)]
其中y
。使用,您可以使用:
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2012-09-08 18:53:34
2012-09-08 19:00:00
2012-09-08 20:00:00
2012-09-08 21:00:00
2012-09-08 21:27:39
2012-09-08 18:53:34
2012-09-08 19:00:00
2012-09-08 20:00:00
2012-09-08 21:00:00
2012-09-08 21:27:39