Question

给定python中的列表（或者java，我试图用两种语言来做），我怎样才能获得将列表拆分成不同分组的所有不同方法，因为每个分组必须至少具有一定的大小？我认为获得分割位置的组合是最好的方法。

列表和最小尺寸的示例输入是

[1,2,3], 2

，相应的输出应为

[[1,2], [1,3], [2,3], [1,2,3]]

Answer 1

在Python中，你可以递归地执行此操作：

def partition(lst, minsize=1):
    yield [lst]
    for n in range(minsize, len(lst)-minsize+1):
        for p in partition(lst[n:], minsize):
            yield [lst[:n]] + [l for l in p]

例如：

>>> lst = [1, 2, 3, 4, 5, 6, 7]
>>> partition(lst, 3)
[[[1, 2, 3, 4, 5, 6, 7]], 
 [[1, 2, 3], [4, 5, 6, 7]], 
 [[1, 2, 3, 4], [5, 6, 7]]]
>>> list(partition(lst, 2))
[[[1, 2, 3, 4, 5, 6, 7]], [[1, 2], [3, 4, 5, 6, 7]], 
 [[1, 2], [3, 4], [5, 6, 7]], [[1, 2], [3, 4, 5], [6, 7]], 
 [[1, 2, 3], [4, 5, 6, 7]], [[1, 2, 3], [4, 5], [6, 7]], 
 [[1, 2, 3, 4], [5, 6, 7]], [[1, 2, 3, 4, 5], [6, 7]]]

Answer 2

修改

根据OP的新输入/输出要求，它只是几行：

def groupings(lst, min_size): results = [tuple(lst)] for i in range(min_size, len(lst)): results.extend(combinations(lst, i)) return results

<强> ORIGINAL

（基于错误的假设，即OP希望给定最小分区大小的所有可能分区。）

所以itertools.combinations（）应该是一个起点。例如，

>>> list(combinations('ABCD', 2)) [('A', 'B'), ('A', 'C'), ('A', 'D'), ('B', 'C'), ('B', 'D'), ('C', 'D')]

所以这给了你一个答案。您的分组输出＆＃39; ABCD＆＃39;最小尺寸设置为2是：

[['A', 'B', 'C', 'D']] [['A', 'D'], ['B', 'C']] [['A', 'C'], ['B', 'D']] [['A', 'B'], ['C', 'D']]

所以高级流程应该是粗略的：

results = [] remaining = [([], list)] # Start without any groups and the full list while remaining not empty: groups, list = remaining.pop() if len(list) >= min_size: results.append(groups + list) for size in min_size to len(list) - 1: for combo in combinations: new <- (groups + combo, list - combo) if new will not cause duplicates: remaining.append(new)

以下是一些似乎有用的代码。为了避免重复，并处理原始列表可能有重复的情况，我修改了itertools.combinations中的代码，而不是仅使用该方法。

def groupings(lst, min_size): lst = list(lst) # List for storing our final groupings results = [] # Unfinished grouping, tuple storing the group and remaining sub-list # Initialize with the empty group and original list remaining = [([], lst)] # Continue as long as we have unfinished groupings while len(remaining): # Get an unfinished grouping current_group, current_lst = remaining.pop() n = len(current_lst) # If the last part of the list is big enough, # then record the grouping if n >= min_size: results.append(current_group + [current_lst]) # Otherwise, discard it else: continue # Helper set for getting remainder below all_indices = set(range(n)) # Iterate the group length from min_size to the length of our current list for r in range(min_size, n - 1): # This is a modified version of itertools.combinations() # http://docs.python.org/3.3/library/itertools.html#itertools.combinations # Add the first combination to our remaining list indices = list(range(r)) remainder = current_lst[r:] group = current_group + [current_lst[:r]] remaining.append((group, remainder)) while True: for i in reversed(range(r)): if indices[i] != i + n - r: break else: break indices[i] += 1 for j in range(i+1, r): indices[j] = indices[j-1] + 1 # If one of the remaining indexes is less than the minimum used index, # then a different iteration will handle it, so discard this one min_index = min(indices) remainder = [] for i in all_indices.difference(indices): remainder.append(current_lst[i]) if i < min_index: break else: # Add this combination to our remaining list group = current_group + [[current_lst[i] for i in indices]] remaining.append((group, remainder)) return results

结果：

>>> groupings('ABCDE', 2) [['A', 'B', 'C', 'D', 'E']] [['A', 'D', 'E'], ['B', 'C']] [['A', 'C', 'E'], ['B', 'D']] [['A', 'C', 'D'], ['B', 'E']] [['A', 'B', 'E'], ['C', 'D']] [['A', 'B', 'D'], ['C', 'E']] [['A', 'B', 'C'], ['D', 'E']] [['A', 'E'], ['B', 'C', 'D']] [['A', 'D'], ['B', 'C', 'E']] [['A', 'C'], ['B', 'D', 'E']] [['A', 'B'], ['C', 'D', 'E']]

如何获得至少一定大小的列表分组的所有组合？

2 个答案: