Python元組列表，其中包含要指定的鍵值

Question

我很難將包含鍵的元組列表轉換為字典。

我有一個元組列表：

[('season', '1', 'episode', '1', 'showkey', '1736'),
('season', '1', 'episode', '2', 'showkey', '1737'),
('season', '1', 'episode', '3', 'showkey', '1738'),
('season', '1', 'episode', '4', 'showkey', '1739'),
('season', '1', 'episode', '5', 'showkey', '1740'),
('season', '1', 'episode', '6', 'showkey', '1741'),
('season', '1', 'episode', '7', 'showkey', '1742'),
('season', '1', 'episode', '8', 'showkey', '1743'),
('season', '1', 'episode', '9', 'showkey', '1744'),
('season', '1', 'episode', '10', 'showkey', '1745'),
('season', '1', 'episode', '11', 'showkey', '1746'),
('season', '1', 'episode', '12', 'showkey', '1747'),
('season', '1', 'episode', '13', 'showkey', '1748')]

我該如何用鍵值對制作字典：

{'season':1, 'episode':2, 'showkey':1736}

謝謝你的幫助！

Answer 1

如果要為每個元組使用一本字典：

>>> i = ('season', '1', 'episode', '2', 'showkey', '1737')
>>> dict(zip(i[::2], i[1::2]))
{'season': '1', 'episode': '2', 'showkey': '1737'}

要將其擴展為您的列表：

new_list = [dict(zip(i[::2], i[1::2])) for i in list_of_data]

通過結合以下事實來起作用：

1. dict（）構造函數采用一對可迭代的對，可以將其轉換為鍵/值。
2. zip（）將返回傳遞給它的兩個可迭代對象的元組對。
3. 切片允許您添加起始偏移量和step參數。

結合以上內容：

1. 首先，通過跳過第一個元素i[1::2] ，通過奇數位置i[::2]剪切元組，然后通過偶數位置對元組進行切片，這給出了兩個代表鍵的列表，第二個代表值：

    >>> i[::2] ('season', 'episode', 'showkey') >>> i[1::2] ('1', '2', '1737') 

2. 接下來，我們將這兩個列表壓縮以提供給我們（鍵，值）對：

    >>> zip(i[::2], i[1::2]) [('season', '1'), ('episode', '2'), ('showkey', '1737')] 

3. 最后，我們將結果列表傳遞給dict()構造函數：

    >>> dict([('season', '1'), ('episode', '2'), ('showkey', '1737')]) {'season': '1', 'episode': '2', 'showkey': '1737'} 

Answer 2

我將使用列表推導，通過將元組中的每個2個連續元素從元組中創建字典。

代碼-

d = [dict((i[j],i[j+1]) for j in range(0,len(i),2)) for i in lt]

這里lt是元組的列表。

范例-

>>> lt = [('season', '1', 'episode', '1', 'showkey', '1736'),
... ('season', '1', 'episode', '2', 'showkey', '1737'),
... ('season', '1', 'episode', '3', 'showkey', '1738'),
... ('season', '1', 'episode', '4', 'showkey', '1739'),
... ('season', '1', 'episode', '5', 'showkey', '1740'),
... ('season', '1', 'episode', '6', 'showkey', '1741'),
... ('season', '1', 'episode', '7', 'showkey', '1742'),
... ('season', '1', 'episode', '8', 'showkey', '1743'),
... ('season', '1', 'episode', '9', 'showkey', '1744'),
... ('season', '1', 'episode', '10', 'showkey', '1745'),
... ('season', '1', 'episode', '11', 'showkey', '1746'),
... ('season', '1', 'episode', '12', 'showkey', '1747'),
... ('season', '1', 'episode', '13', 'showkey', '1748')]
>>>
>>> d= [dict((i[j],i[j+1]) for j in range(0,len(i),2)) for i in lt]
>>> d
[{'showkey': '1736', 'episode': '1', 'season': '1'}, {'showkey': '1737', 'episode': '2', 'season': '1'}, {'showkey': '1738', 'episode': '3', 'season': '1'}, {'showkey': '1739', 'episode': '4', 'season': '1'}, {'showkey': '1740', 'episode':
 'season': '1'}, {'showkey': '1744', 'episode': '9', 'season': '1'}, {'showkey': '1745', 'episode': '10', 'season': '1'}, {'showkey': '1746', 'episode': '11', 'season': '1'}, {'showkey': '1747', 'episode': '12', 'season': '1'}, {'showkey':

Answer 3

map(lambda datum: dict(zip(datum[::2], datum[1::2])), data)

在慢動作中：

results = []
for datum in data:
    # pick off every alternate item from the tuple, starting with the zeroth.
    keys = datum[::2]
    # pick off every alternate item from the tuple, starting with the oneth.
    values = datum[1::2]
    kv = zip(keys, values)
    new_dict = dict(kv)
    results.append(new_dict)

請原諒錯字，在我的手機上輸入。

Answer 4

x=[('season', '1', 'episode', '1', 'showkey', '1736'),
('season', '1', 'episode', '2', 'showkey', '1737'),
('season', '1', 'episode', '3', 'showkey', '1738'),
('season', '1', 'episode', '4', 'showkey', '1739'),
 ('season', '1', 'episode', '5', 'showkey', '1740'),
  ('season', '1', 'episode','6', 'showkey', '1741'),
 ('season', '1', 'episode', '7', 'showkey', '1742'),
('season', '1', 'episode', '8', 'showkey', '1743'),
('season', '1', 'episode', '9', 'showkey', '1744'),
('season', '1', 'episode', '10', 'showkey', '1745'),
('season', '1', 'episode', '11', 'showkey', '1746'),
('season', '1', 'episode', '12', 'showkey', '1747'),
('season', '1', 'episode', '13', 'showkey', '1748')]
d={}
m=0
for i in x:
    d1={}
    for j in i[::2]:

        d1[j]=i[i.index(j)+1]

    d[m]=d1
    m=m+1

print d

您可以創建字典的字典以保存所有數據。

Answer 5

inputd = [('season', '1', 'episode', '1', 'showkey', '1736'),
('season', '1', 'episode', '2', 'showkey', '1737'),
('season', '1', 'episode', '3', 'showkey', '1738'),
('season', '1', 'episode', '4', 'showkey', '1739'),
('season', '1', 'episode', '5', 'showkey', '1740'),
('season', '1', 'episode', '6', 'showkey', '1741'),
('season', '1', 'episode', '7', 'showkey', '1742'),
('season', '1', 'episode', '8', 'showkey', '1743'),
('season', '1', 'episode', '9', 'showkey', '1744'),
('season', '1', 'episode', '10', 'showkey', '1745'),
('season', '1', 'episode', '11', 'showkey', '1746'),
('season', '1', 'episode', '12', 'showkey', '1747'),
('season', '1', 'episode', '13', 'showkey', '1748')]


>>> map(dict, map(lambda x: zip(x[::2],x[1::2]), inputd))
[{'season': '1', 'episode': '1', 'showkey': '1736'}, {'season': '1', 'episode': '2', 'showkey': '1737'}, {'season': '1', 'episode': '3', 'showkey': '1738'}, {'season': '1', 'episode': '4', 'showkey': '1739'}, {'season': '1', 'episode': '5', 'showkey': '1740'}, {'season': '1', 'episode': '6', 'showkey': '1741'}, {'season': '1', 'episode': '7', 'showkey': '1742'}, {'season': '1', 'episode': '8', 'showkey': '1743'}, {'season': '1', 'episode': '9', 'showkey': '1744'}, {'season': '1', 'episode': '10', 'showkey': '1745'}, {'season': '1', 'episode': '11', 'showkey': '1746'}, {'season': '1', 'episode': '12', 'showkey': '1747'}, {'season': '1', 'episode': '13', 'showkey': '1748'}]

關於python slice

1. x[::2] -表示從頭到尾，每次跳過2個位置
2. x[1::2] -表示從索引1開始到結束，每次向前跳過2個位置