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Question

在python中，如何通过timedelta邻域减少日期时间列表？

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)
    ]

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带数字的示例，如果我有这个数字列表

numbers = [1,2,6,7]
delta = 1

我尝试对近似值进行分组并获得该组的特征值（中心值）。 delta是值之间的最大距离。

对于数字，特征值是

[1.5, 6.5]

因为这些值在[1,2]和[6,7]中分组并计算了平均值。

Answer 1

问题描述已经解决了：你想使用groupby()中的itertools函数

所需要的只是一个稍微聪明的key函数，只要连续的时间戳比{{1}更接近，就会记住最后一个状态并继续给出相同的key值。 }。

分组后，将找到的组转换为平均时间，处理单个时间戳（包括示例）。

delta

产量输出：

import datetime as dt
from itertools import groupby

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),
        dt.datetime(1970, 1, 1, 0, 13)
    ]
delta = dt.timedelta(minutes=2)

class grouper:
    def __init__(self, delta):
        self.delta= delta
        self.last = None

    def __call__(self, tm):
        # we keep on returning the same key as long as successive time
        # stamps are within the last time stamp + delta
        self.last = tm if (self.last is None) or (tm - self.last)>self.delta \
                       else self.last
        return self.last

# transform the result of groupby into average times
def avgtm(item):
    (key, tms) = item
    tms = list(tms) # transform generator into list so we can index it
    return tms[0] + (tms[-1]-tms[0])/2 if len(tms)>1 else tms[0]

timestamps = map(avgtm, groupby(dates, key=grouper(delta)))
print "Time stamps: ",timestamps

Answer 2

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))])

要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))])

或者随时产生值：

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

Answer 3

import datetime as dt

def datetime_to_epoch(dtime):
    return (dtime - dt.datetime(1970,1,1)).total_seconds()

def datetime_sublists(datetime_list, time_delta = dt.timedelta(days=1)):
    sublists = []

    temp = [datetime_list[0]]
    for i in range(len(datetime_list)-1):
        prev_date = datetime_list[i]
        current_date = datetime_list[i+1]

        if current_date - prev_date <= time_delta:
            temp.append(current_date)
        else:
            sublists.append(temp)
            temp = [current_date]
    sublists.append(temp)

    return sublists

def reduce_datetime_list_by_delta(date_list, delta):
    sublist = datetime_sublists(date_list, delta)

    reduced = []
    for dates in sublist:
        epochs = [ datetime_to_epoch(date) for date in dates]
        epoch_average = sum(epochs)/len(epochs)
        reduced.append(dt.datetime.utcfromtimestamp(epoch_average))

    return reduced


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),
    dt.datetime(1970, 1, 1, 0, 12)
]

delta = dt.timedelta(minutes=2)

print reduce_datetime_list_by_delta(dates, delta)

按时间缩短日期时间列表

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3 个答案: