比较组合的开始和结束日期以确定排序

Question

对于总是包含 start 和 end 日期的 Python 字典列表，您将如何根据“组合”start 和 end 日期？

根据以下标准从上到下获得最终结果的最简单（最 Pythonic）方法是什么：

先按 end_date（降序）排序，然后按 start_date（降序）排序。
如果有两个对象具有相同的 end_date，则最新的 start_date 排在第一位，即：然后按 start_date 对这些项目进行排序。
如果 start_date 和 end_date 相同，则这些项目的顺序不是问题，可以忽略或保持原样。

import datetime


blah = [
    {"id": 1, "start_date": datetime.date(2021, 5, 1), "end_date": None},
    {"id": 2, "start_date": datetime.date(2013, 2, 1), "end_date": None},
    {"id": 3, "start_date": datetime.date(2017, 1, 1), "end_date": datetime.date(2018, 1, 1)},
    {"id": 4, "start_date": datetime.date(2016, 5, 1), "end_date": datetime.date(2019, 6, 1)},
    {"id": 5, "start_date": datetime.date(2012, 1, 1), "end_date": datetime.date(2015, 1, 1)},
    {"id": 6, "start_date": datetime.date(2008, 1, 1), "end_date": datetime.date(2011, 1, 1)},
    {"id": 7, "start_date": datetime.date(2006, 1, 1), "end_date": datetime.date(2008, 1, 1)},
    {"id": 8, "start_date": datetime.date(2005, 1, 15), "end_date": datetime.date(2010, 1, 15)},
    {"id": 9, "start_date": datetime.date(2002, 1, 15), "end_date": datetime.date(2002, 1, 15)},
    {"id": 10, "start_date": datetime.date(2002, 1, 1), "end_date": datetime.date(2006, 1, 1)},
    {"id": 11, "start_date": datetime.date(2002, 1, 1), "end_date": datetime.date(2006, 1, 1)},
    {"id": 12, "start_date": datetime.date(2001, 2, 1), "end_date": datetime.date(2003, 1, 1)},
    {"id": 13, "start_date": datetime.date(2001, 1, 15), "end_date": datetime.date(2003, 1, 15)},
    {"id": 14, "start_date": datetime.date(1998, 1, 1), "end_date": datetime.date(2001, 1, 1)},
    {"id": 15, "start_date": datetime.date(1997, 1, 15), "end_date": datetime.date(1997, 1, 15)}
]

# Do something here...and return `result`.

result = [
    {"id": 1, "start_date": datetime.date(2021, 5, 1), "end_date": None},
    {"id": 2, "start_date": datetime.date(2013, 2, 1), "end_date": None},
    {"id": 4, "start_date": datetime.date(2016, 5, 1), "end_date": datetime.date(2019, 6, 1)},
    {"id": 3, "start_date": datetime.date(2017, 1, 1), "end_date": datetime.date(2018, 1, 1)},
    {"id": 5, "start_date": datetime.date(2012, 1, 1), "end_date": datetime.date(2015, 1, 1)},
    {"id": 6, "start_date": datetime.date(2008, 1, 1), "end_date": datetime.date(2011, 1, 1)},
    {"id": 8, "start_date": datetime.date(2005, 1, 15), "end_date": datetime.date(2010, 1, 15)},
    {"id": 7, "start_date": datetime.date(2006, 1, 1), "end_date": datetime.date(2008, 1, 1)},
    {"id": 11, "start_date": datetime.date(2002, 1, 1), "end_date": datetime.date(2006, 1, 1)},
    {"id": 10, "start_date": datetime.date(2002, 1, 1), "end_date": datetime.date(2006, 1, 1)},
    {"id": 9, "start_date": datetime.date(2002, 1, 15), "end_date": datetime.date(2002, 1, 15)},
    {"id": 12, "start_date": datetime.date(2001, 2, 1), "end_date": datetime.date(2003, 1, 1)},
    {"id": 13, "start_date": datetime.date(2001, 1, 15), "end_date": datetime.date(2003, 1, 15)},
    {"id": 14, "start_date": datetime.date(1998, 1, 1), "end_date": datetime.date(2001, 1, 1)},
    {"id": 15, "start_date": datetime.date(1997, 1, 15), "end_date": datetime.date(1997, 1, 15)}
]

Answer 1

What would be the simplest (most Pythonic) way to obtain the end result ...

我能想到的最简单（最 Pythonic）的方法是使用 pandas。

演示：

import datetime
import pandas as pd

blah = [
    {"id": 1, "start_date": datetime.date(2021, 5, 1), "end_date": None},
    {"id": 2, "start_date": datetime.date(2013, 2, 1), "end_date": None},
    {"id": 3, "start_date": datetime.date(2017, 1, 1), "end_date": datetime.date(2018, 1, 1)},
    {"id": 4, "start_date": datetime.date(2016, 5, 1), "end_date": datetime.date(2019, 6, 1)},
    {"id": 5, "start_date": datetime.date(2012, 1, 1), "end_date": datetime.date(2015, 1, 1)},
    {"id": 6, "start_date": datetime.date(2008, 1, 1), "end_date": datetime.date(2011, 1, 1)},
    {"id": 7, "start_date": datetime.date(2006, 1, 1), "end_date": datetime.date(2008, 1, 1)},
    {"id": 8, "start_date": datetime.date(2005, 1, 15), "end_date": datetime.date(2010, 1, 15)},
    {"id": 9, "start_date": datetime.date(2002, 1, 15), "end_date": datetime.date(2002, 1, 15)},
    {"id": 10, "start_date": datetime.date(2002, 1, 1), "end_date": datetime.date(2006, 1, 1)},
    {"id": 11, "start_date": datetime.date(2002, 1, 1), "end_date": datetime.date(2006, 1, 1)},
    {"id": 12, "start_date": datetime.date(2001, 2, 1), "end_date": datetime.date(2003, 1, 1)},
    {"id": 13, "start_date": datetime.date(2001, 1, 15), "end_date": datetime.date(2003, 1, 15)},
    {"id": 14, "start_date": datetime.date(1998, 1, 1), "end_date": datetime.date(2001, 1, 1)},
    {"id": 15, "start_date": datetime.date(1997, 1, 15), "end_date": datetime.date(1997, 1, 15)}
]

df = pd.DataFrame(blah)

result = df.sort_values(['end_date', 'start_date'], ascending=(False, False), na_position='first').to_dict('records')

for e in result:
    print(e)

输出：

{'id': 1, 'start_date': datetime.date(2021, 5, 1), 'end_date': None}
{'id': 2, 'start_date': datetime.date(2013, 2, 1), 'end_date': None}
{'id': 4, 'start_date': datetime.date(2016, 5, 1), 'end_date': datetime.date(2019, 6, 1)}
{'id': 3, 'start_date': datetime.date(2017, 1, 1), 'end_date': datetime.date(2018, 1, 1)}
{'id': 5, 'start_date': datetime.date(2012, 1, 1), 'end_date': datetime.date(2015, 1, 1)}
{'id': 6, 'start_date': datetime.date(2008, 1, 1), 'end_date': datetime.date(2011, 1, 1)}
{'id': 8, 'start_date': datetime.date(2005, 1, 15), 'end_date': datetime.date(2010, 1, 15)}
{'id': 7, 'start_date': datetime.date(2006, 1, 1), 'end_date': datetime.date(2008, 1, 1)}
{'id': 10, 'start_date': datetime.date(2002, 1, 1), 'end_date': datetime.date(2006, 1, 1)}
{'id': 11, 'start_date': datetime.date(2002, 1, 1), 'end_date': datetime.date(2006, 1, 1)}
{'id': 13, 'start_date': datetime.date(2001, 1, 15), 'end_date': datetime.date(2003, 1, 15)}
{'id': 12, 'start_date': datetime.date(2001, 2, 1), 'end_date': datetime.date(2003, 1, 1)}
{'id': 9, 'start_date': datetime.date(2002, 1, 15), 'end_date': datetime.date(2002, 1, 15)}
{'id': 14, 'start_date': datetime.date(1998, 1, 1), 'end_date': datetime.date(2001, 1, 1)}
{'id': 15, 'start_date': datetime.date(1997, 1, 15), 'end_date': datetime.date(1997, 1, 15)}

Answer 2

为了订购 end_date 我引入了一个“假日期”来使数据一致。此选择是任意的，但应避免与其他值发生冲突。内置函数 sorted、reversed 需要一个具有同质数据的迭代器，所以没有 None.

sorted return 一个列表，reversed 一个生成器。

# In order to make sense the question 3. I modify the start date for 2006, see comment
blah = [
    {"id": 1, "start_date": datetime.date(2021, 5, 1), "end_date": None},
    {"id": 2, "start_date": datetime.date(2013, 2, 1), "end_date": None},
    {"id": 3, "start_date": datetime.date(2017, 1, 1), "end_date": datetime.date(2018, 1, 1)},
    {"id": 4, "start_date": datetime.date(2016, 5, 1), "end_date": datetime.date(2019, 6, 1)},
    {"id": 5, "start_date": datetime.date(2012, 1, 1), "end_date": datetime.date(2015, 1, 1)},
    {"id": 6, "start_date": datetime.date(2008, 1, 1), "end_date": datetime.date(2011, 1, 1)},
    {"id": 7, "start_date": datetime.date(2006, 1, 1), "end_date": datetime.date(2008, 1, 1)},
    {"id": 8, "start_date": datetime.date(2005, 1, 15), "end_date": datetime.date(2010, 1, 15)},
    {"id": 9, "start_date": datetime.date(2002, 1, 15), "end_date": datetime.date(2002, 1, 15)},
    {"id": 10, "start_date": datetime.date(2002, 1, 2), "end_date": datetime.date(2006, 1, 1)}, # <---- modified start_date!
    {"id": 11, "start_date": datetime.date(2002, 1, 1), "end_date": datetime.date(2006, 1, 1)}, 
    {"id": 12, "start_date": datetime.date(2001, 2, 1), "end_date": datetime.date(2003, 1, 1)},
    {"id": 13, "start_date": datetime.date(2001, 1, 15), "end_date": datetime.date(2003, 1, 15)},
    {"id": 14, "start_date": datetime.date(1998, 1, 1), "end_date": datetime.date(2001, 1, 1)},
    {"id": 15, "start_date": datetime.date(1997, 1, 15), "end_date": datetime.date(1997, 1, 15)}
]

这里是代码。

import itertools as it
import datetime

FAKE_DATE = datetime.date(2999, 9, 9) # or any non-interfering date 

# 1
print(sorted(blah, key=lambda p: p['start_date']))

print(sorted(blah, reverse=True, key=lambda p: p['start_date'])) # reverse, A
print(list(reversed(sorted(blah, key=lambda p: p['start_date'])))) # reverse, B

# 2
order_2 = reversed(sorted(blah, key=lambda p: p['end_date'] if p['end_date'] is not None else FAKE_DATE))

print(list(order_2))

# 3
grp_by_end_dates = it.groupby(sorted(blah, key=lambda p: p['end_date'] if p['end_date'] is not None else FAKE_DATE), key=lambda p: p['end_date'])
order_3 = it.chain(*(sorted(list(i), reverse=True, key=lambda p: p['start_date']) for _, i in grp_by_end_dates))

print(list(order_3)

Answer 3

您可以简单地使用适当的关键字对数据进行排序以满足第 1-3 点；第 4 点自动满足，因为 Python 中的排序保证稳定：

result = sorted(blah,
                reverse = True,
                key=lambda d:(
                    d["end_date"] if d["end_date"] is not None else datetime.date(2999,12,31),
                    d["start_date"])
                )

比较组合的开始和结束日期以确定排序

Comparing a combined start and end date to determine ordering

python

sorting

dictionary

list-comprehension

date