比較Python的嵌套列表和計數複製

我有兩個嵌套列表與字符串（list_a和list_b），詳細信息如下：比較Python的嵌套列表和計數複製

list_a = [ 
('shop1', 'stand1', 'shelf1', 'fruit1'), 
('shop1', 'stand1', 'shelf2', 'fruit2'), 
('shop1', 'stand1', 'shelf3', 'fruit3'), 
('shop1', 'stand2', 'shelf1', 'fruit1'), 
('shop1', 'stand2', 'shelf2', 'fruit2'), 
('shop1', 'stand2', 'shelf3', 'fruit3'), 
('shop2', 'stand3', 'shelf1', 'fruit1'), 
('shop2', 'stand3', 'shelf2', 'fruit2'), 
('shop2', 'stand3', 'shelf3', 'fruit3') 
] 
list_b = [ 
('shop1', 'stand1', 'shelf1', 'fruit1'), 
('shop1', 'stand1', 'shelf2', 'fruit2'), 
('shop1', 'stand1', 'shelf2', 'fruit2'), 
('shop1', 'stand1', 'shelf3', 'fruit3'), 
('shop1', 'stand1', 'shelf3', 'fruit3'), 
('shop1', 'stand1', 'shelf3', 'fruit3'), 
('shop1', 'stand2', 'shelf1', 'fruit1'), 
('shop1', 'stand2', 'shelf1', 'fruit1'), 
('shop1', 'stand2', 'shelf1', 'fruit1'), 
('shop1', 'stand2', 'shelf2', 'fruit2'), 
('shop1', 'stand2', 'shelf2', 'fruit2'), 
('shop1', 'stand2', 'shelf2', 'fruit2'), 
('shop1', 'stand2', 'shelf3', 'fruit3'), 
('shop2', 'stand3', 'shelf1', 'fruit1'), 
('shop2', 'stand3', 'shelf1', 'fruit1'), 
('shop2', 'stand3', 'shelf2', 'fruit2'), 
('shop2', 'stand3', 'shelf3', 'fruit3'), 
('shop2', 'stand3', 'shelf3', 'fruit3'), 
('shop2', 'stand3', 'shelf3', 'fruit3') 
]

，我想找到list_b相同的行中list_a，算上「複製」行和合並使用一個額外的列（發生次數）list_a作爲一個新的列表，這樣下面：

result_list = [ 
('shop1', 'stand1', 'shelf1', 'fruit1', 1), 
('shop1', 'stand1', 'shelf2', 'fruit2', 2), 
('shop1', 'stand1', 'shelf3', 'fruit3', 3), 
('shop1', 'stand2', 'shelf1', 'fruit1', 3), 
('shop1', 'stand2', 'shelf2', 'fruit2', 3), 
('shop1', 'stand2', 'shelf3', 'fruit3', 1), 
('shop2', 'stand3', 'shelf1', 'fruit1', 2), 
('shop2', 'stand3', 'shelf2', 'fruit2', 1), 
('shop2', 'stand3', 'shelf3', 'fruit3', 3) 
]

有什麼快速有效的方式做到這一點？

來源

2012-09-25 jusef

聽起來就像你想要一個數據庫。 – phg

可能的重複 - http://stackoverflow.com/questions/642763/python-intersection-of-two-lists或http://stackoverflow.com/questions/2029795/comparing-python-nested-lists –

WEll，didn沒有注意到你想要的頻率也..然後這些鏈接不包含你想要的.. –

dict_a = {row: 0 for row in list_a} 
for row in list_b: 
    if row in dict_a: 
     dict_a[row] += 1 

result = [row + (dict_a[row],) for row in list_a]

在Python 2.6中使用dict((row, 0) for row in list_a)而不是字典解析。

來源

2012-09-25 18:03:26

工程很漂亮，但我忘了提及我的Python版本，它是2.6，所以我改變了一下。非常感謝你！ – jusef

這些不是嵌套列表，而是元組。這實際上是你的儲蓄。請參閱 Most Efficient way to calculate Frequency of values in a Python list? 幾乎馬上就能工作。要獲取重複項，請使用這兩個詞典的keys()，並計算它們的差異。

來源

2012-09-25 17:59:48

從你的鏈接工程美麗。非常感謝你！ – jusef

使用Counter()：

>>> from collections import Counter 
    >>> count=Counter(list_b) 
    >>> [list(x)+[count[x]] for x in list_a] 

    [['shop1', 'stand1', 'shelf1', 'fruit1', 1], 
    ['shop1', 'stand1', 'shelf2', 'fruit2', 2], 
    ['shop1', 'stand1', 'shelf3', 'fruit3', 3], 
    ['shop1', 'stand2', 'shelf1', 'fruit1', 3], 
    ['shop1', 'stand2', 'shelf2', 'fruit2', 3], 
    ['shop1', 'stand2', 'shelf3', 'fruit3', 1], 
    ['shop2', 'stand3', 'shelf1', 'fruit1', 2], 
    ['shop2', 'stand3', 'shelf2', 'fruit2', 1], 
    ['shop2', 'stand3', 'shelf3', 'fruit3', 3]]`

來源

2012-09-25 18:02:30

工程很漂亮，但我忘了提及我的Python版本，它的2.6，所以我已經改變了一點。非常感謝你！ – jusef

你可以在pythom 2.6中使用'defaultdict'。 –

比較Python的嵌套列表和計數複製

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