将一起出现在单独列表中的项目组合在一起

Question

我有 2 个嵌套列表，我想编写一些代码来运行两个列表中的每个子列表，并将两个列表中一起出现的任何元素组合在一起。我正在做的项目实际上有巨大的嵌套列表，所以我创建了以下 2 个列表来稍微简化问题（我只有一年的 python 经验）。如果可以创建一个将这两个列表中的元素组合在一起的函数，那么我可以将该函数应用于实际项目。这个问题可能类似于：Find items that appear together on multiple lists, 但我无法理解该问题中编写的代码，正如我所说，我对 python 比较陌生。

my_list = [['a', 'd', 'l'], ['c', 'e', 't'], ['q', 'x'], ['p', 'f', 'd', 'k']

sec_list = [['f', 'd', 'w', 'a'], ['c', 'e', 'u', 'h'], ['q', 'x', 'd', 'z'], ['p', 'k']]

##The output should be something like:

[['a', 'd'], ['c', 'e'], ['q', 'x'], ['p', 'k'], ['f', 'd']]```

Thanks

Answer 1

您可以使用zip 遍历两个序列并找到具有集合交集的共同元素。请注意，您的代码在 my_list

中缺少结束符 ]

my_list = [['a', 'd', 'l'], ['c', 'e', 't'], ['q', 'x'], ['p', 'f', 'd', 'k']]
sec_list = [['f', 'd', 'w', 'a'], ['c', 'e', 'u', 'h'], ['q', 'x', 'd', 'z'], ['p', 'k']]

# each item of my_list and sec_list are lists
# zip allows parallel iteration so l1 and l2 are the pairs of inner lists
# sets are designed for tasks like finding common elements
# the & sign is python for set intersection 
matches = []
for l1, l2 in zip(my_list, sec_list):
    matches.append(list(set(l1) & set(l2)))

这可以合并为一个列表理解

my_list = [['a', 'd', 'l'], ['c', 'e', 't'], ['q', 'x'], ['p', 'f', 'd', 'k']]
sec_list = [['f', 'd', 'w', 'a'], ['c', 'e', 'u', 'h'], ['q', 'x', 'd', 'z'], ['p', 'k']]
matches = [list(set(l1) & set(l2)) for l1, l2 in zip(my_list, sec_list)]

Answer 2

基于 cmets：

my_list = [["a", "d", "l"], ["c", "e", "t"], ["q", "x"], ["p", "f", "d", "k"]]
sec_list = [
    ["f", "d", "w", "a"],
    ["c", "e", "u", "h"],
    ["q", "x", "d", "z"],
    ["p", "k"],
]

# to speed up, convert the sublists to sets
tmp1 = [set(i) for i in my_list]
tmp2 = [set(i) for i in sec_list]

out = []
for s1 in tmp1:
    for s2 in tmp2:
        m = s1 & s2
        if m:
            out.append(list(m))
print(out)

打印：

[['d', 'a'], ['d'], ['e', 'c'], ['x', 'q'], ['d', 'f'], ['d'], ['k', 'p']]

Answer 3

如果你想保留重复，你可以使用collections.Counter这个解决方案：

from collections import Counter

my_list = [['a', 'd', 'l'], ['c', 'e', 't'], ['q', 'x'], ['p', 'f', 'd', 'k', 'k']]

sec_list = [['f', 'd', 'w', 'a'], ['c', 'e', 'u', 'h'], ['q', 'x', 'd', 'z'], ['p', 'k', 'k']]

    
result = [list((Counter(a) & Counter(b)).elements()) for a in my_list for b in sec_list]
result = [x for x in result if len(x) > 0]
    
print(result)

输出：

[['a', 'd'], ['d'], ['c', 'e'], ['x', 'q'], ['f', 'd'], ['d'], ['k', 'k', 'p']]                                                                                                                                                                                                         

基于 cmets 更新。