我有一个[1,2,3]
数组
我想使用数组的所有元素进行所有可能的组合:
结果:
[[1], [2], [3]]
[[1,2], [3]]
[[1], [2,3]]
[[1,3], [2]]
[[1,2,3]]
最佳答案
既然这个好问题已经重生,那么这里就有一个新的答案。
该问题以递归方式解决:如果您已经有n-1个元素的分区,如何使用它来划分n个元素?将第n个元素放在现有子集中的一个子集中,或将其添加为新的单例子集。仅此而已;没有itertools
,没有集合,没有重复的输出,并且总共只有n次调用partition()
:
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 = list(range(1,5))
for n, p in enumerate(partition(something), 1):
print(n, sorted(p))
输出:
1 [[1, 2, 3, 4]]
2 [[1], [2, 3, 4]]
3 [[1, 2], [3, 4]]
4 [[1, 3, 4], [2]]
5 [[1], [2], [3, 4]]
6 [[1, 2, 3], [4]]
7 [[1, 4], [2, 3]]
8 [[1], [2, 3], [4]]
9 [[1, 3], [2, 4]]
10 [[1, 2, 4], [3]]
11 [[1], [2, 4], [3]]
12 [[1, 2], [3], [4]]
13 [[1, 3], [2], [4]]
14 [[1, 4], [2], [3]]
15 [[1], [2], [3], [4]]