从平面列表中查找所有列表和子列表组合
Finding all list and sublist combination from flat list
我正在尝试找到一种优雅的方式来从初始平面列表生成所有可能的组合。
例如:
[在:]
l = ["aaa", "bbb", "ccc"]
[输出:]
[["aaa"], ["bbb"], ["ccc"]]
[["aaa", "bbb"], ["ccc"]]
[["aaa", "ccc"], ["bbb"]]
[["bbb", "ccc"], ["aaa"]]
[["aaa", "bbb", "ccc"]]
如您所见,顺序对我来说并不重要。所以我会避免获得这种情况:
[["aaa"], ["bbb"], ["ccc"]]
[["bbb"], ["aaa"], ["ccc"]]
...
[["ccc"], ["aaa"], ["bbb"]]
此外,我的输出列表的每个子列表都必须包含初始列表的每个元素。
我没有找到任何明显的 itertools.combination()
解决方案
谢谢
问题比较模糊,我想你要找的是set-partitions:
def partition(collection):
if len(collection) == 1:
yield [ collection ]
return
first = collection[0]
for smaller in partition(collection[1:]):
# insert `first` in each of the subpartition's subsets
for n, subset in enumerate(smaller):
yield smaller[:n] + [[ first ] + subset] + smaller[n+1:]
# put `first` in its own subset
yield [ [ first ] ] + smaller
something = ['a', 'b', 'c']
for n, p in enumerate(partition(something), 1):
print(n, sorted(p))
输出:
1 [['a', 'b', 'c']]
2 [['a'], ['b', 'c']]
3 [['a', 'b'], ['c']]
4 [['a', 'c'], ['b']]
5 [['a'], ['b'], ['c']]
我正在尝试找到一种优雅的方式来从初始平面列表生成所有可能的组合。
例如:
[在:]
l = ["aaa", "bbb", "ccc"]
[输出:]
[["aaa"], ["bbb"], ["ccc"]]
[["aaa", "bbb"], ["ccc"]]
[["aaa", "ccc"], ["bbb"]]
[["bbb", "ccc"], ["aaa"]]
[["aaa", "bbb", "ccc"]]
如您所见,顺序对我来说并不重要。所以我会避免获得这种情况:
[["aaa"], ["bbb"], ["ccc"]]
[["bbb"], ["aaa"], ["ccc"]]
...
[["ccc"], ["aaa"], ["bbb"]]
此外,我的输出列表的每个子列表都必须包含初始列表的每个元素。
我没有找到任何明显的 itertools.combination()
谢谢
问题比较模糊,我想你要找的是set-partitions:
def partition(collection):
if len(collection) == 1:
yield [ collection ]
return
first = collection[0]
for smaller in partition(collection[1:]):
# insert `first` in each of the subpartition's subsets
for n, subset in enumerate(smaller):
yield smaller[:n] + [[ first ] + subset] + smaller[n+1:]
# put `first` in its own subset
yield [ [ first ] ] + smaller
something = ['a', 'b', 'c']
for n, p in enumerate(partition(something), 1):
print(n, sorted(p))
输出:
1 [['a', 'b', 'c']]
2 [['a'], ['b', 'c']]
3 [['a', 'b'], ['c']]
4 [['a', 'c'], ['b']]
5 [['a'], ['b'], ['c']]