按每个子列表中的某些值对子列表列表进行分组

Question

我有一本包含列表的字典。例如，

{1: [[sender11, receiver11, text11, address11]], 
 2: [[sender21, receiver21, text21, address21], [sender22, receiver22, text22, address22]], 
 3: [[sender31, receiver31, text31, address31], [sender32, receiver32, text32, address32], [sender33, receiver33, text33, address33]]
 4: [[sender41, receiver41, text41, address41], [sender42, receiver42, text42, address42], [sender43, receiver43, text43, address43], [sender44, receiver44, text44, address44]]}

我想要做的是，对于包含具有 2 个或更多元素的列表的字典元素（即本例中的 dict[2]、dict[3] 和 dict[4]），我对 sender, receiver, text 进行比较每个列表值。对于每组具有相同sender, receiver, text 的列表值，我会做一些事情。

例如，在dict[3] 中，如果sender31, receiver31, text31 与sender32, receiver32, text32 和sender33, receiver33, text33 的值相同，那么我将对所有3 个列表值进行处理。

在dict[4] 中说，如果sender41, receiver41, text41 与sender42, receiver42, text42 的值相同，而sender43, receiver43, text43 与sender44, receiver44, text44 的值相同，但与sender41, receiver41, text41 不同，那么我将处理这两个单独分组。

我编写了一个 Python 脚本，它几乎可以蛮力比较 sender21, receiver21, text21 和 sender22, receiver22, text22 的值，即

if sender21 == sender22 and receiver21 == receiver22 and text21 == text22:
   # Do something

这效率不高，因为它只适用于 2 个列表值，但我不知道我应该如何实现它以便它适用于任何数量大于 1 的列表值。

Answer 1

我认为defaultdict 是显而易见的方式：

from collections import defaultdict

def collate(seq):
    groups = defaultdict(list)
    for subseq in seq:
        groups[tuple(subseq[:3])].append(subseq[3])
    return groups

根据您的实际数据，您可以将上述函数中的tuple(subseq[:3]) 替换为例如(subseq[1], subseq[4], subseq[5])，或附加的 subseq[3] 和 subseq 本身...这取决于您对数据的处理方式。

但是，键必须是元组而不是列表，因为键必须是不可变的。

例子：

>>> data = [
...     ['S1', 'R1', 'T1', 'A3'],
...     ['S2', 'R2', 'T2', 'A4'],
...     ['S1', 'R1', 'T1', 'A5'],
...     ['S2', 'R2', 'T2', 'A6']
... ]

>>> collate(data)
defaultdict(<type 'list'>, {
    ('S2', 'R2', 'T2'): ['A4', 'A6'],
    ('S1', 'R1', 'T1'): ['A3', 'A5']
})

您可以像使用任何其他字典一样使用它，例如

>>> for (sender, receiver, text), addresses in collate(data).items():
...     print sender, receiver, text
...     print '|'.join(addresses)
...     print
... 
S2 R2 T2
A4|A6

S1 R1 T1
A3|A5