Python：在某些條件下將元組列表轉換為字典

Question

我創建了一個像這樣的列表：

Book = [(24, '2008-10-30', 'Start'), (24, '2008-12-20', 'End','sold'), 
 (25, '2009-01-01', 'Start'), (25, '2009-11-14', 'End', 'returned'),
 (26, '2010-04-03', 'Start'), (26, '2010-10-11', 'End', 'sold'),...]

我想將其轉換成這樣的字典：

bookDict = { 24: {'Start': '2008-10-30', 'End': '2008-12-20','reason':'sold'},
  25: {'Start': '2009-01-01', 'End': '2009-11-14','reason':'returned'},
  26: {'Start': '2010-04-03', 'End': '2010-10-11','reason':'sold'},...}

對於作為Book列表中元組的第一個值的字典中的每個鍵（這是一個代碼），我希望有兩個元組作為每個鍵的值。 其中一個與該特定代碼的“開始”點有關，另一個與該特定代碼的“結束”點有關。

我還有一個問題。 對於某些代碼，有多個日期不同的“終點”。 我只想保留終點，並保留較晚的日期。 像這樣的事情：

Book = [(24, '2008-10-30', 'Start'), (24, '2008-12-20', 'End', 'sold'), 
 (24, '2009-02-04', 'End', 'sold'), (24, '2009-11-25', 'End', 'sold')]

對於上面的示例字典應保留以下內容：

bookDict = { 24: {'Start': '2008-10-30', 'End': '2009-11-25','reason':'sold'},

誰能幫我嗎？

Answer 1

您可以使用itertools.groupby ， min和max ：

import itertools
def quantity_key(d):
  return list(map(int, d[1].split('-')))

Book = [(24, '2008-10-30', 'Start'), (24, '2008-12-20', 'End','sold'), (25, '2009-01-01', 'Start'), (25, '2009-11-14', 'End', 'returned'), (26, '2010-04-03', 'Start'), (26, '2010-10-11', 'End', 'sold')]
new_books = {a:list(b) for a, b in itertools.groupby(Book, key=lambda x:x[0])}
final_books = {a:{'Start':min(b, key=quantity_key)[1], 'End':max(b, key=quantity_key)[1], 'reason':max(b, key=quantity_key)[-1]} for a, b in new_books.items()}

輸出：

{24: {'Start': '2008-10-30', 'End': '2008-12-20', 'reason': 'sold'}, 25: {'Start': '2009-01-01', 'End': '2009-11-14', 'reason': 'returned'}, 26: {'Start': '2010-04-03', 'End': '2010-10-11', 'reason': 'sold'}}

---

每個鍵有兩個以上的值：

Book = [(24, '2008-10-30', 'Start'), (24, '2008-12-20', 'End', 'sold'), (24, '2009-02-04', 'End', 'sold'), (24, '2009-11-25', 'End', 'sold')]
new_books = {a:list(b) for a, b in itertools.groupby(Book, key=lambda x:x[0])}
final_books = {a:{'Start':min(b, key=quantity_key)[1], 'End':max(b, key=quantity_key)[1], 'reason':max(b, key=quantity_key)[-1]} for a, b in new_books.items()}

輸出：

{24: {'Start': '2008-10-30', 'End': '2009-11-25', 'reason': 'sold'}}

Answer 2

這是一個滿足兩個條件的解決方案。

每當它收到新書ID時，它都會為其創建dict並在遇到list數據時將其填充。

對於多個結束條目，您的日期格式允許使用字符串比較來獲取最新日期。

books = [(24, '2008-10-30', 'Start'), (24, '2008-12-20', 'End','sold'),
 (25, '2009-01-01', 'Start'), (25, '2009-11-14', 'End', 'returned'),
 (26, '2010-04-03', 'Start'), (26, '2010-10-11', 'End', 'sold'),
 (26, '2011-10-11', 'End', 'returned')] # The latest 'End' entry should be picked

bookDict = {}

for info in books:
    id_ = info[0]
    type_ = info[2]

    book = bookDict.setdefault(id_, {})

    if type_ == 'Start':
        book[type_] = info[1]

    elif type_ == 'End' and info[1] > book.get(type_, ''):
        book[type_] = info[1]
        book['reason'] = info[3]

輸出：

bookDict
# {24: {'Start': '2008-10-30', 'End': '2008-12-20', 'reason': 'sold'},
#  25: {'Start': '2009-01-01', 'End': '2009-11-14', 'reason': 'returned'},
#  26: {'Start': '2010-04-03', 'End': '2010-10-11', 'reason': 'returned'}}

Answer 3

您可以執行以下操作：

for t in Book:
    index, date, marker, *rest = t
    entry = d.setdefault(index, {})
    end_date = entry.get("End", "1900-01-01")
    if marker == "Start" or date > end_date:
        entry[marker] = date
        if rest:
            entry["reason"] = rest[0]

Answer 4

盡管它可以適用於第二部分，但這僅回答了OP的第一部分問題。

您可以將collections.defaultdict用於O（n）解決方案：

book = [(24, '2008-10-30', 'Start'), (24, '2008-12-20', 'End','sold'), 
        (25, '2009-01-01', 'Start'), (25, '2009-11-14', 'End', 'returned'),
        (26, '2010-04-03', 'Start'), (26, '2010-10-11', 'End', 'sold')]

from collections import defaultdict

d = defaultdict(dict)

for key, date, *data in book:
    d[key][data[0]] = date
    if len(data) == 2:
        d[key]['reason'] = data[1]

---

另外，您可以捕獲IndexError而不是測試元IndexError度：

for key, date, *data in book:
    d[key][data[0]] = date
    try:
        d[key]['reason'] = data[1]
    except IndexError:
        continue