我有一个datetime.datetime实例、d和datetime.timedelta实例td，我正试图编写一个函数，将范围(d, d+td)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)
我想要一份的清单
[(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))]
有人能提出一种很好的、Python式的方法来实现这一点吗？
使用dateutil，您可以使用rrule:生成列表
import dateutil.rrule as rrule
import datetime
def hours_aligned(start, end, inc = True):
if inc: yield start
rule = rrule.rrule(rrule.HOURLY, byminute = 0, bysecond = 0, dtstart=start)
for x in rule.between(start, end, inc = False):
yield x
if inc: yield end
d = datetime.datetime(2012, 9, 8, 18, 53, 34)
td = datetime.timedelta(hours=2, minutes=34, seconds=5)
for x in hours_aligned(d,d+td):
print(x)
产生
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
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) )
这里的主要假设是，您最初指定的时间增量不是负数。如果你这样做，你会得到一个单独的区块列表[(x,y)]，其中y < x。