比較組合的開始和結束日期以確定排序-有解無憂

使用始終包含start和end日期的 Python 字典串列，您將如何根據“組合”start和end日期對串列進行排序？

使用以下標準獲得最終結果的最簡單（最 Pythonic）方法是什么 - 從上到下：

先按end_date（降序）排序，然后按start_date（降序）排序。
如果有兩個具有相同的物件end_date，則最新的start_date先出現，即：然后按start_date這些專案的排序。
如果 thestart_date和 theend_date相同，那么這些專案的順序不是問題，可以忽略或保持原樣。

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)}
]

uj5u.com熱心網友回復：

獲得最終結果的最簡單（最 Pythonic）方法是什么？

我能想到的最簡單（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)

輸出：

{'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)}

uj5u.com熱心網友回復：

您可以簡單地使用適當的鍵對資料進行排序以滿足第 1-3 點；第 4 點自動滿足，因為 Python 中的排序保證是穩定的：

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"])
                )

uj5u.com熱心網友回復：

為了訂購wrt，end_date我引入了一個“假日期”以使資料保持一致。此選擇是任意的，但應避免與其他值發生沖突。內置的功能sorted，reversed需要均勻的資料迭代器，所以沒有None。

sorted回傳一個串列，reversed一個生成器。

# 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)}
]

這里是代碼。

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)

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標籤：Python 日期排序字典列表理解

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