在python中,如何减少timedelta邻域的日期时间列表?
如果我有
dates = [
dt.datetime(1970, 1, 1, 0, 2),
dt.datetime(1970, 1, 1, 0, 3),
dt.datetime(1970, 1, 1, 0, 7),
dt.datetime(1970, 1, 1, 0, 8)
]
和一个timedelta
delta = dt.timedelta(minutes=2)
我怎么能得到这个?
expected = [
dt.datetime(1970, 1, 1, 0, 2, 30),
dt.datetime(1970, 1, 1, 0, 7, 30)
]
编辑
数字示例(如果我有此数字列表)
numbers = [1,2,6,7]
delta = 1
我尝试将接近值分组,并获得该组的特征值(中心值)。增量是两个值之间的最大距离。
对于数字,特征值是
[1.5, 6.5]
因为这些值在[1,2]和[6,7]中分组,并计算出平均值。
最佳答案
import datetime as dt
dates = [
dt.datetime(1970, 1, 1, 0, 2),
dt.datetime(1970, 1, 1, 0, 3),
dt.datetime(1970, 1, 1, 0, 12),
dt.datetime(1970, 1, 1, 0, 7),
dt.datetime(1970, 1, 1, 0, 8),
dt.datetime(1970, 1, 1, 0, 9),
dt.datetime(1970, 1, 1, 0, 13)
]
def group_dates(dates, delta):
it = iter(dates)
prev = next(it)
grouped, total = [[prev]], delta.total_seconds()
for dte in it:
if (dte - prev).total_seconds() <= total:
grouped[-1].append(dte)
else:
grouped.append([dte])
prev = dte
return grouped
def td(l):
seconds = sum((d - dt.datetime(1970, 1, 1)).total_seconds() for d in l) / len(l)
return dt.datetime.utcfromtimestamp(seconds)
from pprint import pprint as pp
pp([td(sub) for sub in group_dates(dates,dt.timedelta(minutes=2))])
为避免不必要的函数调用,请检查len:
pp([td(sub) if len(sub) > 1 else sub[0] for sub in [datetime.datetime(1970, 1, 1, 0, 2, 30),
datetime.datetime(1970, 1, 1, 0, 12),
datetime.datetime(1970, 1, 1, 0, 8),
datetime.datetime(1970, 1, 1, 0, 13)]group_dates(dates,dt.timedelta(minutes=2))])
或随手产生值:
def group_dates(dates, delta):
it = iter(dates)
prev = next(it)
grouped, total = (prev,),delta.total_seconds()
for dte in it:
if (dte - prev).total_seconds() <= total:
grouped = grouped + (dte,)
else:
yield td(grouped)
grouped = (dte,)
prev = dte
yield td(grouped)
pp(list(group_dates(dates, delta=dt.timedelta(minutes=2))))
[datetime.datetime(1970, 1, 1, 0, 2, 30),
datetime.datetime(1970, 1, 1, 0, 12),
datetime.datetime(1970, 1, 1, 0, 8),
datetime.datetime(1970, 1, 1, 0, 13)]
一些时间:
In [28]: dates = [
dt.datetime(1970, 1, 1, 0, 2),
dt.datetime(1970, 1, 1, 0, 3),
dt.datetime(1970, 1, 1, 0, 4),
dt.datetime(1970, 1, 1, 0, 7),
dt.datetime(1970, 1, 1, 0, 8),
dt.datetime(1970, 1, 1, 0, 9),
dt.datetime(1970, 1, 1, 0, 15),
dt.datetime(1970, 1, 1, 0, 22),
dt.datetime(1970, 1, 1, 0, 24),
dt.datetime(1970, 1, 1, 0, 27)
]
In [41]: for i in range(10000):
dates.append(dates[-1]+dt.timedelta(minutes=choice([1,2,3,4])))
....:
In [42]: timeit [td(sub) if len(sub) > 1 else sub[0] for sub in group_dates(dates,dt.timedelta(minutes=2))]
100 loops, best of 3: 15.8 ms per loop
In [43]: timeit reduce_datetime_list_by_delta(dates, delta)
100 loops, best of 3: 16.9 ms per loop
In [44]: timeit timestamps = map(avgtm, groupby(dates, key=grouper(delta)))
10 loops, best of 3: 18.8 ms per loop
In [45]: timeit (list(group_dates_iter(dates, delta = dt.timedelta(minutes=2))))
10 loops, best of 3: 18.4 ms per loop