Question

我创建了一个这样的列表：

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'),...]

我想将其转换为这样的字典：

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'},...}

对于字典中的每个键，它是Book列表中元组的第一个值（它是一个代码），我希望有两个元组作为每个键的值。其中一个与“开始”有关。点和另一个相关的＆＃39;结束＆＃39;该特定代码的要点。

我还有另一个问题。对于某些代码，有一个以上的结尾＆＃39;点不同的日期。我想只保留较晚日期的结束点。这样的事情：

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

对于上面的例子，字典应该保留这个：

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

有人可以帮我吗？

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()}

输出：

{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'}}

每个键有两个以上的值：

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()}

输出：

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

Answer 2

这是一个满足这两个标准的解决方案。

每当它为新书ID添加时，它会为其创建dict并在遇到list中的数据时将其填入。

对于多个 End 条目，您的日期格式允许使用字符串比较来获取最新日期。

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]

输出：

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

你可以这样做：

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

这只回答了OP问题的第一部分，虽然它可以适用于第二部分。

您可以将collections.defaultdict用于O（n）解决方案：

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]

或者，您可以捕获IndexError而不是测试元组长度：

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

Python：在某些条件下将元组列表转换为字典

4 个答案: