在python列表中计算日期的最佳/最快方法

Question

我有一个日期列表，目标是计算每个日期的出现次数，同时保持它们出现在原始列表中的顺序。请考虑以下示例：

列表only_dates如下所示：

[datetime.date(2017, 3, 9), datetime.date(2017, 3, 10), datetime.date(2017, 3, 10), datetime.date(2017, 3, 11)]

我正在尝试使用groupby：

import itertools
day_wise_counts = [(k, len(list(g))) for k, g in itertools.groupby(only_dates)]
print(str(day_wise_counts))

打印

[(datetime.date(2017, 3, 10), 1), (datetime.date(2017, 3, 9), 1), (datetime.date(2017, 3, 10), 1), (datetime.date(2017, 3, 11), 1)]

我理解这种情况正在发生，因为最终每个日期对象在分组时都被视为不同的日期对象。

我期待输出为：

[(datetime.date(2017, 3, 9), 1), (datetime.date(2017, 3, 10), 2), (datetime.date(2017, 3, 11), 1)]

我不一定在寻找元组列表。只要保持原始日期顺序，字典输出也就足够了。 （OrderedDict也许）。

我怎样才能做到这一点？

更新：有可能建议多种方法都能正常运行。但我应该提到我将为大量数据执行此操作。因此，如果您的解决方案在运行时间方面是最佳的，那就太好了。如果可以，请相应地编辑您的答案/评论。

更新2：数据大小可以达到100万行。

Answer 1

确实，您可以使用OrderedDict：

from collections import OrderedDict
import datetime

inp = [datetime.date(2017, 3, 9), datetime.date(2017, 3, 10),
       datetime.date(2017, 3, 10), datetime.date(2017, 3, 11)]

odct = OrderedDict()
for item in inp:
    try:
        odct[item] += 1
    except KeyError:
        odct[item] = 1

print(odct)

打印：

OrderedDict([(datetime.date(2017, 3, 9), 1),
             (datetime.date(2017, 3, 10), 2),
             (datetime.date(2017, 3, 11), 1)])

你还要求时间安排，所以这里是：

from collections import OrderedDict, Counter
import datetime
import random

# Functions

def ordereddict(inp):
    odct = OrderedDict()
    for item in inp:
        try:
            odct[item] += 1
        except KeyError:
            odct[item] = 1
    return odct


def dawg(inp):
    cnts=Counter(inp)
    seen=set()
    return [(e, cnts[e]) for e in inp if not (e in seen or seen.add(e))]


def chris1(inp):
    return [(item, inp.count(item)) for item in list(OrderedDict.fromkeys(inp))]


def chris2(inp):
    c = Counter(inp)
    return [(item,c[item]) for item in list(OrderedDict.fromkeys(inp))]


# Taken from answer: https://stackoverflow.com/a/23747652/5393381
class OrderedCounter(Counter, OrderedDict):  
    'Counter that remembers the order elements are first encountered'

    def __repr__(self):
        return '%s(%r)' % (self.__class__.__name__, OrderedDict(self))

    def __reduce__(self):
        return self.__class__, (OrderedDict(self),)


# Timing setup
timings = {ordereddict: [], dawg: [], chris1: [], chris2: [], OrderedCounter: []}
sizes = [2**i for i in range(1, 20)]

# Timing
for size in sizes:
    func_input = [datetime.date(2017, random.randint(1, 12), random.randint(1, 28)) for _ in range(size)]
    for func in timings:
        res = %timeit -o func(func_input)   # if you use IPython, otherwise use the "timeit" module
        timings[func].append(res)

并绘制：

%matplotlib notebook

import matplotlib.pyplot as plt
import numpy as np

fig = plt.figure(1)
ax = plt.subplot(111)

for func in timings:
    ax.plot([2**i for i in range(1, 20)], 
            [time.best for time in timings[func]], 
            label=str(func.__name__))
ax.set_xscale('log')
ax.set_yscale('log')
ax.set_xlabel('size')
ax.set_ylabel('time [seconds]')
ax.grid(which='both')
ax.legend()
plt.tight_layout()

我在Python-3.5上计时。在python-2.x上使用Counter的方法可能会慢一些（Counter针对python-3.x进行了优化）。此外chris2和dawg方法相互重叠（因为它们之间几乎没有时间差异）。

除了@Chris_Rands和OrderedCounter的第一种方法外，这些方法的表现非常相似，主要取决于列表中重复的数量。

它主要是1.5-2的差异因素。我无法在3＆＃34;快速＆＃34;中找到100万件物品的任何实时时差。方法

Answer 2

您可以使用list.count()一个列表推导，迭代从OrderedDict个唯一有序日期派生的列表：

import datetime
from collections import OrderedDict

lst = [datetime.date(2017, 3, 9), datetime.date(2017, 3, 10), datetime.date(2017, 3, 10), datetime.date(2017, 3, 11)]

[(item,lst.count(item)) for item in list(OrderedDict.fromkeys(lst))]
# [(datetime.date(2017, 3, 9), 1), (datetime.date(2017, 3, 10), 2), (datetime.date(2017, 3, 11), 1)]

或类似地使用collections.Counter代替list.count：

from collections import Counter

c = Counter(lst)

[(item,c[item]) for item in list(OrderedDict.fromkeys(lst))]
# [(datetime.date(2017, 3, 9), 1), (datetime.date(2017, 3, 10), 2), (datetime.date(2017, 3, 11), 1)]

或使用OrderedCounter。

编辑：通过@MSeifert查看优秀基准。

Answer 3

您可以使用计数器计算原始列表uniqify，以便在添加计数时维持顺序。

假设：

>>> dates=[datetime.date(2017, 3, 9), datetime.date(2017, 3, 10), datetime.date(2017, 3, 10), datetime.date(2017, 3, 11)]  

你可以这样做：

from collections import Counter

cnts=Counter(dates)
seen=set()
>>> [(e, cnts[e]) for e in dates if not (e in seen or seen.add(e))]
[(datetime.date(2017, 3, 9), 1), (datetime.date(2017, 3, 10), 2), (datetime.date(2017, 3, 11), 1)]

更新

您还可以使用键功能将计数器排序回原始列表的顺序，以获取该列表中第一个日期（X）条目的索引：

sorted([(k,v) for k,v in Counter(dates).items()], key=lambda t: dates.index(t[0])) 

（此速度与您的列表的排序或无序方式相关......）

有人说时间！

以下是一些较大的例子（400,000个日期）：

from __future__ import print_function
import datetime
from collections import Counter
from collections import OrderedDict

def dawg1(dates):
    seen=set()
    cnts=Counter(dates)
    return [(e, cnts[e]) for e in dates if not (e in seen or seen.add(e))]

def od_(dates):    
    odct = OrderedDict()
    for item in dates:
        try:
            odct[item] += 1
        except KeyError:
            odct[item] = 1
    return odct

def lc_(lst):
    return [(item,lst.count(item)) for item in list(OrderedDict.fromkeys(lst))]    

def dawg2(dates):
    return sorted([(k,v) for k,v in Counter(dates).items()], key=lambda t: dates.index(t[0]))    

if __name__=='__main__':
    import timeit  
    dates=[datetime.date(2017, 3, 9), datetime.date(2017, 3, 10), datetime.date(2017, 3, 10), datetime.date(2017, 3, 11)]*100000
    for f in (dawg, od_, lc_,sort_):
        print("   {:^10s}{:.4f} secs {}".format(f.__name__, timeit.timeit("f(dates)", setup="from __main__ import f, dates", number=100),f(dates))) 

打印（在Python 2.7上）：

 dawg1   10.7253 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]
  od_    21.8186 secs OrderedDict([(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)])
  lc_    17.0879 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]
 dawg2   8.6058 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]0000)]

PyPy：

 dawg1   7.1483 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]
  od_    4.7551 secs OrderedDict([(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)])
  lc_    27.8438 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]
 dawg2   4.7673 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]

Python 3.6：

 dawg1   3.4944 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]
  od_    4.6541 secs OrderedDict([(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)])
  lc_    2.7440 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]
 dawg2   2.1330 secs [(datetime.date(2017, 3, 9), 100000), (datetime.date(2017, 3, 10), 200000), (datetime.date(2017, 3, 11), 100000)]

最佳。