在一个较大的列表中，确定其元素构成确切元素的列表分类

Question

Python 3。你知道吗

序言

所以我有一个元素列表，可以这样排序（或任何其他方式）：

List = [c, d, a, c, b] 

为了让事情变得更简单：

List = [a, b, c, c, d]

所以列表可能有重复项，长度在1到10之间。你知道吗

用户将选择一个范围号，例如：

r = 3

范围号是逻辑的一部分，该逻辑产生的列表数包含长度为1到r的“列表”子序列的所有可能组合。每个组合可以以任何方式排列。考虑到本例中的'List'和'r'，用户将得到以下列表选择：

%Lists of length 1.  
ListSelection1 = [[a], [b], [c], [c], [d]]

%Lists of length 2.
ListSelection2 = [[a, b], [a, c], [a, c], [a, d], [b, c], [b, c], [b, d], [c, c], [c, d], [c, d]]

%Lists of length 3.
ListSelection3 = [[a, b, c], [a, b, c], [a, b, d], [a, c, c], [a, c, d], [a, c, d], [b, c, c], [b, c, d], [b, c, d], [c, c, d]]

这些列表将排列在另一个列表中，如下所示：

ListSelections = [ListSelection1, ListSelection2, ListSelection3]

问题

如何识别“ListSelections”中包含与“List”相同的元素选择的（子）列表的所有可能组合？元素的顺序无关紧要。关于上述示例，可接受的列表组合将是[b，c]和[a，c，d]，或[b]+[c]+[d]+[a，c]。你明白了。你知道吗

我需要确定所有正确的列表组合以及这些列表在“ListSelections”中的位置。你知道吗

Answer 1

我想你想找到的是multiset的分区集。为此，可以使用sympy如下所示：

from sympy.utilities.iterables import multiset_partitions
res=[p for p in multiset_partitions(["a","b","c","c","d"])]

输出：

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

Answer 2

您的说明有点混乱，但我怀疑您想要的是“排列”或“组合”：

https://docs.python.org/2/library/itertools.html

Python有一个库可以为您实现这一点。你知道吗

output = [itertools.permutations(input, r_) for r_ in range(r)]