[問題] 窮舉"將List中元素分群"的所有可能方法

[sharkbay (Shark Bay)]

目前已知這個方法能滿足這個題目, 但是空間或運算量太大, 想請問有沒有別的做法?

import itertools

powerset = lambda iterable: itertools.chain.from_iterable(
        itertools.combinations(list(iterable), r)
        for r in range(len(list(iterable)) + 1))
flatten = lambda list2d: [item for sublist in list2d for item in sublist]

x = list("abcd")
xx = [list(val) for val in list(powerset(x)) if 0 != len(val)]
xxx = [list(val) for val in list(powerset(xx)) if 0 != len(val)]
xxxx = [list(val) for val in xxx if x == list(sorted(flatten(val)))]

xxxx =
[[['a', 'b', 'c', 'd']],
 [['a'], ['b', 'c', 'd']],
 [['b'], ['a', 'c', 'd']],
 [['c'], ['a', 'b', 'd']],
 [['d'], ['a', 'b', 'c']],
 [['a', 'b'], ['c', 'd']],
 [['a', 'c'], ['b', 'd']],
 [['a', 'd'], ['b', 'c']],
 [['a'], ['b'], ['c', 'd']],
 [['a'], ['c'], ['b', 'd']],
 [['a'], ['d'], ['b', 'c']],
 [['b'], ['c'], ['a', 'd']],
 [['b'], ['d'], ['a', 'c']],
 [['c'], ['d'], ['a', 'b']],
 [['a'], ['b'], ['c'], ['d']]]

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推 LP9527: tuple+ set幾行就搞定了
→ LP9527: 資料少的話就這樣吧  https://i.imgur.com/0CbjrkT.jpg
→ LP9527: 喔看錯了
推 sherees: itertools.combinations
→ ro9956882: 這樣試試看 https://imgur.com/bbYlZR8
→ ro9956882: [['a','b'],['c']]我用['ab','c']表示 這樣過程中不用
→ ro9956882: 用到deepcopy 要轉換格式做完再一次轉
→ ro9956882: 如果有重複元素(多個'a')就要再改一下 0.0
elements = list("abcdefghijl")
combinations = [[]]
for element in elements :
    for i in range(len(combinations)):
        for j in range(len(combinations[i])):
            new = combinations[i][:]
            new[j] += element
            combinations.append(new)
        combinations[i].append(element)

combinations = [list(sorted([list(c) for c in combination]))
                for combination in combinations]
combinations = list(sorted(combinations))

我重抄一份在此
※ 編輯: sharkbay (106.1.116.121 臺灣), 01/23/2021 00:50:44


def partition(collection):
    global counter
    if len(collection) == 1:
        yield [collection]
        return
    first = collection[0]
    for smaller in partition(collection[1:]):
        for n, subset in enumerate(smaller):
            yield smaller[:n] + [[first] + subset] + smaller[n + 1:]
        yield [[first]] + smaller

google到一份新的code
※ 編輯: sharkbay (106.1.116.121 臺灣), 01/23/2021 06:20:52