python - 如何在python中划分列表？

Question

我有这个列表示例 list1=[['p1', 'p2', 'p3', 'p4'], ['p5', 'p6', 'p7']]，如果任何变量与其他变量相同，它们会将其拆分为一组。假设 p1 和 p4 相同，p5 和 p6 相同。所以我想要一个看起来像 list2=[['p1', 'p4'], ['p2', 'p3'], ['p5', 'p6'], 'p7']. 那么我需要如何划分它们，请帮助。我正在使用最新版本的python。

好的，更具体地说，我的程序中的“相同”，我使用 p1 和 p4，如果它们对某个字符给出相同的结果，那么我将它们合并到一个组中。例子

if dictionary.get(p1, character) is dictionary.get(p4, character)

如果您有更多问题，请问我。

Answer 1

以下将为您提供该结果：

list1=[[1, 2, 2, 1], [3, 4, 3]]

print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]

[OUTPUT]
[[1, 1], [2, 2], [3, 3], [4]]

示例 1

list1=[['hello', 'hello', 'hello', 'what'], ['i', 'am', 'i']]

print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]

[OUTPUT]
[['what'], ['hello', 'hello', 'hello'], ['i', 'i'], ['am']]

示例 2

list1=[[1,2,3,2,1],[9,8,7,8,9],[5,4,6,4,5]]

print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]

[OUTPUT]
[[1, 1], [2, 2], [3], [8, 8], [9, 9], [7], [4, 4], [5, 5], [6]]

Answer 2

使用unique_everseen配方和collections.Counters：

from collections import Counter

def solve(lst):
    counters = map(Counter, lst)
    return [ [uniq]*c[uniq] for seq, c in zip(lst, counters)
                                                for uniq in unique_everseen(seq)]

演示：

>>> print(solve([[1, 2, 2, 1], [3, 4, 3]]))
[[1, 1], [2, 2], [3, 3], [4]]
>>> print(solve([['hello', 'hello', 'hello', 'what'], ['i', 'am', 'i']]))
[['hello', 'hello', 'hello'], ['what'], ['i', 'i'], ['am']]
>>> print(solve([[1,2,3,2,1],[9,8,7,8,9],[5,4,6,4,5]]))
[[1, 1], [2, 2], [3], [9, 9], [8, 8], [7], [5, 5], [4, 4], [6]]

如您所见，这也保留了项目的顺序。

---

unique_everseen配方代码：

from itertools import filterfalse

def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in filterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element