Question

根据它们的键来“修剪”字典的最快方法是什么？我的理解是，自Python 3.7起，字典现在可以保留顺序

我有一个字典，其中包含关键字（类型为datetime）：val（类型为float）。字典是按时间顺序排列的。

time_series_dict = 
{"2019-02-27 14:00:00": 95,
"2019-02-27 15:00:00": 98,
"2019-02-27 16:25:00: 80,
.............
"2019-03-01 12:15:00": 85
}

我想整理字典，删除开始日期和结束日期之外的所有内容。字典可以有1000个值。是否有比以下方法更快的方法：

for k in list(time_series_dict.keys()):
    if not start_date <= k <= end_date:
        del time_series_dict[k]

Answer 1

如果词典中有1000个键，并且您要从有序的时间戳序列的开头和结尾删除键，请考虑使用binary search在键的列表副本中查找截止点。 Python为此包含了bisect module：

from bisect import bisect_left, bisect_right

def trim_time_series_dict(tsd, start_date, end_date):
    ts = list(tsd)
    before = bisect_right(ts, start_date)  # insertion point at > start_date
    after = bisect_left(ts, end_date)      # insertion point is < end_date
    for i in range(before):                # up to == start_date
        del tsd[ts[i]]
    for i in range(after + 1, len(ts)):    # from >= end_date onwards
        del tsd[ts[i]]

我已经运行了time trials来看看这是否会与您的典型数据集有所不同；如预期的那样，当删除的键的数量显着低于输入字典的长度时，它会得到回报。

定时试用设置（导入，构建测试数据字典以及开始和结束日期，定义测试功能）

>>> import random
>>> from bisect import bisect_left, bisect_right
>>> from datetime import datetime, timedelta
>>> from itertools import islice
>>> from timeit import Timer
>>> def randomised_ordered_timestamps():
...     date = datetime.now().replace(second=0, microsecond=0)
...     while True:
...         date += timedelta(minutes=random.randint(15, 360))
...         yield date.strftime('%Y-%m-%d %H:%M:%S')
...
>>> test_data = {ts: random.randint(50, 500) for ts in islice(randomised_ordered_timestamps(), 10000)}
>>> start_date = next(islice(test_data, 25, None))                 # trim 25 from the start
>>> end_date = next(islice(test_data, len(test_data) - 25, None))  # trim 25 from the end
>>> def iteration(t, start_date, end_date):
...     time_series_dict = t.copy()  # avoid mutating test data
...     for k in list(time_series_dict.keys()):
...         if not start_date <= k <= end_date:
...             del time_series_dict[k]
...
>>> def bisection(t, start_date, end_date):
...     tsd = t.copy()  # avoid mutating test data
...     ts = list(tsd)
...     before = bisect_right(ts, start_date)  # insertion point at > start_date
...     after = bisect_left(ts, end_date)      # insertion point is < end_date
...     for i in range(before):                # up to == start_date
...         del tsd[ts[i]]
...     for i in range(after + 1, len(ts)):    # from >= end_date onwards
...         del tsd[ts[i]]
...

试验结果：

>>> count, total = Timer("t.copy()", "from __main__ import test_data as t").autorange()
>>> baseline = total / count
>>> for test in (iteration, bisection):
...     timer = Timer("test(t, s, e)", "from __main__ import test, test_data as t, start_date as s, end_date as e")
...     count, total = timer.autorange()
...     print(f"{test.__name__:>10}: {((total / count) - baseline) * 1000000:6.2f} microseconds")
...
 iteration: 671.33 microseconds
 bisection:  80.92 microseconds

（该测试会减去先制作字典副本的基准成本）。

但是，对于这类操作，可能会有更有效的数据结构。我签出了sortedcontainers project，因为它包含一个直接支持键二等分的SortedDict() type。不幸的是，尽管它的性能比您的迭代方法要好，但在这里我不能比对键列表的副本进行平分更好。

>>> from sortedcontainers import SortedDict
>>> test_data_sorteddict = SortedDict(test_data)
>>> def sorteddict(t, start_date, end_date):
...     tsd = t.copy()
...     # SortedDict supports slicing on the key view
...     keys = tsd.keys()
...     del keys[:tsd.bisect_right(start_date)]
...     del keys[tsd.bisect_left(end_date) + 1:]
...
>>> count, total = Timer("t.copy()", "from __main__ import test_data_sorteddict as t").autorange()
>>> baseline = total / count
>>> timer = Timer("test(t, s, e)", "from __main__ import sorteddict as test, test_data_sorteddict as t, start_date as s, end_date as e")
>>> count, total = timer.autorange()
>>> print(f"sorteddict: {((total / count) - baseline) * 1000000:6.2f} microseconds")
sorteddict: 249.46 microseconds

但是，我使用的项目可能不正确。从SortedDict对象中删除键是O（NlogN），所以我怀疑这就是问题所在。从其他9950个键值对创建一个新的SortedDict()对象仍然比较慢（超过2毫秒，这不是您要与其他方法进行比较的时间）。

但是，如果您要使用SortedDict.irange() method，则可以简单地忽略值，而不是删除它们，并遍历字典键的子集：

for ts in timeseries(start_date, end_date, inclusive=(False, False)):
    # iterates over all start_date > timestamp > end_date keys, in order.

无需删除任何内容。 irange()实现在后台使用二等分。

Answer 2

import time

import timeit

print(timeit.timeit(setup="""import datetime
time_series_dict = {}
for i in range(10000):
    t =datetime.datetime.now().strftime('%Y-%m-%d %H:%M:%S:%f')
    time_series_dict[t] = i
    if i ==100:
        start_time = t
    if i == 900:
        end_time = t
        """,
stmt="""
tmp = time_series_dict.copy()
for k in list(tmp.keys()):
    if not start_time <= k <= end_time:
        del tmp[k]

""",
number=10000
))
print(timeit.timeit(setup="""import datetime
time_series_dict = {}
for i in range(10000):
    t =datetime.datetime.now().strftime('%Y-%m-%d %H:%M:%S:%f')
    time_series_dict[t] = i
    if i ==100:
        start_time = t
    if i == 900:
        end_time = t
""",
stmt="""
tmp = time_series_dict.copy()
result = {}
for k in list(tmp.keys()):
    if start_time <= k <= end_time:
        result[k] = tmp[k]
""",
number=10000
))
print(timeit.timeit(setup="""
import datetime
from bisect import bisect_left, bisect_right

time_series_dict = {}
for i in range(10000):
    t =datetime.datetime.now().strftime('%Y-%m-%d %H:%M:%S:%f')
    time_series_dict[t] = i
    if i ==100:
        start_time = t
    if i == 900:
        end_time = t

""",
stmt="""
tmp = time_series_dict.copy()
def trim_time_series_dict(tsd, start_date, end_date):
    ts = list(tsd)
    before = bisect_right(ts, start_date)  # insertion point at > start_date
    after = bisect_left(ts, end_date)      # insertion point is < end_date
    for i in range(before):                # up to == start_date
        del tsd[ts[i]]
    for i in range(after + 1, len(ts)):    # from >= end_date onwards
        del tsd[ts[i]]

trim_time_series_dict(tmp, start_time, end_time)
""",
number=10000
))

测试结果

12.558672609
9.662761111
7.990544049

是否根据键修剪有序字典？

2 个答案: