比较 Python 嵌套列表并计算重复项 [重复]

Question

我有两个带有字符串的嵌套列表（list_a 和 list_b），详情如下：

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')
]

我想在list_a 中从list_b 中找到相同的行，计算“重复”行并将list_a 与一个附加列（出现次数）合并为一个新列表，如下所示：

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)
]

有没有快速有效的方法来做到这一点？

Answer 1

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 上使用 dict((row, 0) for row in list_a) 而不是字典解析。

Answer 2

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]]`

Answer 3

这些不是嵌套列表，而是元组。这实际上是你的储蓄。请参阅 Most Efficient way to calculate Frequency of values in a Python list? ，它几乎可以立即工作。要获取重复项，请取两个字典的keys()，并计算它们的差异。