我如何找到所有可能的方法来将一个数组放入 4 个槽中,并且数组的数量可能 more/less 多于 4 个?
How do I find all possible ways I can fit an array to 4 slots with the possibility of the array having more/less than 4 numbers?
假设我有 4 个桶和一个数字数组,例如 [1,2,3,4,5]
| || || || |
|__||__||__||__|
1 2 3 4
5 2 3 4
1 5 3 4
etc...
我也可以少于 4 个数字,例如 [1, 2, 3]
| || || || |
|__||__||__||__|
1 2 3
1 3 2
1 2 3
etc...
如何找到存储桶中数字的所有可能组合(如 [1,2,3,4]
如果长度 >= 4 或 [1,2,None,None]
如果长度 < 4)?
您可以使用递归生成器函数:
def combos(nums, buckets, c = []):
if len(c) == buckets:
yield c
else:
if len(c) + len(nums) < buckets:
yield from combos(nums, buckets, c+[None])
for i, a in enumerate(nums):
yield from combos(nums[:i]+nums[i+1:], buckets, c+[a])
print(list(combos([1, 2, 3, 4, 5], 4)))
print(list(combos([1, 2, 3], 4)))
输出:
[[1, 2, 3, 4], [1, 2, 3, 5], [1, 2, 4, 3], [1, 2, 4, 5], [1, 2, 5, 3], [1, 2, 5, 4], [1, 3, 2, 4], [1, 3, 2, 5], [1, 3, 4, 2], [1, 3, 4, 5], [1, 3, 5, 2], [1, 3, 5, 4], [1, 4, 2, 3], [1, 4, 2, 5], [1, 4, 3, 2], [1, 4, 3, 5], [1, 4, 5, 2], [1, 4, 5, 3], [1, 5, 2, 3], [1, 5, 2, 4], [1, 5, 3, 2], [1, 5, 3, 4], [1, 5, 4, 2], [1, 5, 4, 3], [2, 1, 3, 4], [2, 1, 3, 5], [2, 1, 4, 3], [2, 1, 4, 5], [2, 1, 5, 3], [2, 1, 5, 4], [2, 3, 1, 4], [2, 3, 1, 5], [2, 3, 4, 1], [2, 3, 4, 5], [2, 3, 5, 1], [2, 3, 5, 4], [2, 4, 1, 3], [2, 4, 1, 5], [2, 4, 3, 1], [2, 4, 3, 5], [2, 4, 5, 1], [2, 4, 5, 3], [2, 5, 1, 3], [2, 5, 1, 4], [2, 5, 3, 1], [2, 5, 3, 4], [2, 5, 4, 1], [2, 5, 4, 3], [3, 1, 2, 4], [3, 1, 2, 5], [3, 1, 4, 2], [3, 1, 4, 5], [3, 1, 5, 2], [3, 1, 5, 4], [3, 2, 1, 4], [3, 2, 1, 5], [3, 2, 4, 1], [3, 2, 4, 5], [3, 2, 5, 1], [3, 2, 5, 4], [3, 4, 1, 2], [3, 4, 1, 5], [3, 4, 2, 1], [3, 4, 2, 5], [3, 4, 5, 1], [3, 4, 5, 2], [3, 5, 1, 2], [3, 5, 1, 4], [3, 5, 2, 1], [3, 5, 2, 4], [3, 5, 4, 1], [3, 5, 4, 2], [4, 1, 2, 3], [4, 1, 2, 5], [4, 1, 3, 2], [4, 1, 3, 5], [4, 1, 5, 2], [4, 1, 5, 3], [4, 2, 1, 3], [4, 2, 1, 5], [4, 2, 3, 1], [4, 2, 3, 5], [4, 2, 5, 1], [4, 2, 5, 3], [4, 3, 1, 2], [4, 3, 1, 5], [4, 3, 2, 1], [4, 3, 2, 5], [4, 3, 5, 1], [4, 3, 5, 2], [4, 5, 1, 2], [4, 5, 1, 3], [4, 5, 2, 1], [4, 5, 2, 3], [4, 5, 3, 1], [4, 5, 3, 2], [5, 1, 2, 3], [5, 1, 2, 4], [5, 1, 3, 2], [5, 1, 3, 4], [5, 1, 4, 2], [5, 1, 4, 3], [5, 2, 1, 3], [5, 2, 1, 4], [5, 2, 3, 1], [5, 2, 3, 4], [5, 2, 4, 1], [5, 2, 4, 3], [5, 3, 1, 2], [5, 3, 1, 4], [5, 3, 2, 1], [5, 3, 2, 4], [5, 3, 4, 1], [5, 3, 4, 2], [5, 4, 1, 2], [5, 4, 1, 3], [5, 4, 2, 1], [5, 4, 2, 3], [5, 4, 3, 1], [5, 4, 3, 2]]
[[None, 1, 2, 3], [None, 1, 3, 2], [None, 2, 1, 3], [None, 2, 3, 1], [None, 3, 1, 2], [None, 3, 2, 1], [1, None, 2, 3], [1, None, 3, 2], [1, 2, None, 3], [1, 2, 3, None], [1, 3, None, 2], [1, 3, 2, None], [2, None, 1, 3], [2, None, 3, 1], [2, 1, None, 3], [2, 1, 3, None], [2, 3, None, 1], [2, 3, 1, None], [3, None, 1, 2], [3, None, 2, 1], [3, 1, None, 2], [3, 1, 2, None], [3, 2, None, 1], [3, 2, 1, None]]
在每次递归调用时,如果组合尚未形成,代码会做两件事:
- 检查指定的桶数是否大于输入的数字列表。如果是这样,则输出用
None
. 填充
- 迭代剩余的数字列表,并将每个迭代值添加到运行结果,并进行递归。
使用内置函数itertools.permutations
。
import itertools
def func(arr: list, slot_count=4):
if (v := slot_count - len(arr)) > 0:
arr.extend([None for _ in range(v)])
return itertools.permutations(arr, slot_count)
print(list(func([1, 2, 3])))
print(list(func([1, 2, 3, 4, 5])))
[(1, 2, 3, None), (1, 2, None, 3), (1, 3, 2, None), (1, 3, None, 2), (1, None, 2, 3), (1, None, 3, 2), (2, 1, 3, None), (2, 1, None, 3), (2, 3, 1, None), (2, 3, None, 1), (2, None, 1, 3), (2, None, 3, 1), (3, 1, 2, None), (3, 1, None, 2), (3, 2, 1, None), (3, 2, None, 1), (3, None, 1, 2), (3, None, 2, 1), (None, 1, 2, 3), (None, 1, 3, 2), (None, 2, 1, 3), (None, 2, 3, 1), (None, 3, 1, 2), (None, 3, 2, 1)]
[(1, 2, 3, 4), (1, 2, 3, 5), (1, 2, 4, 3), (1, 2, 4, 5), (1, 2, 5, 3), (1, 2, 5, 4), (1, 3, 2, 4), (1, 3, 2, 5), (1, 3, 4, 2), (1, 3, 4, 5), (1, 3, 5, 2), (1, 3, 5, 4), (1, 4, 2, 3), (1, 4, 2, 5), (1, 4, 3, 2), (1, 4, 3, 5), (1, 4, 5, 2), (1, 4, 5, 3), (1, 5, 2, 3), (1, 5, 2, 4), (1, 5, 3, 2), (1, 5, 3, 4), (1, 5, 4, 2), (1, 5, 4, 3), (2, 1, 3, 4), (2, 1, 3, 5), (2, 1, 4, 3), (2, 1, 4, 5), (2, 1, 5, 3), (2, 1, 5, 4), (2, 3, 1, 4), (2, 3, 1, 5), (2, 3, 4, 1), (2, 3, 4, 5), (2, 3, 5, 1), (2, 3, 5, 4), (2, 4, 1, 3), (2, 4, 1, 5), (2, 4, 3, 1), (2, 4, 3, 5), (2, 4, 5, 1), (2, 4, 5, 3), (2, 5, 1, 3), (2, 5, 1, 4), (2, 5, 3, 1), (2, 5, 3, 4), (2, 5, 4, 1), (2, 5, 4, 3), (3, 1, 2, 4), (3, 1, 2, 5), (3, 1, 4, 2), (3, 1, 4, 5), (3, 1, 5, 2), (3, 1, 5, 4), (3, 2, 1, 4), (3, 2, 1, 5), (3, 2, 4, 1), (3, 2, 4, 5), (3, 2, 5, 1), (3, 2, 5, 4), (3, 4, 1, 2), (3, 4, 1, 5), (3, 4, 2, 1), (3, 4, 2, 5), (3, 4, 5, 1), (3, 4, 5, 2), (3, 5, 1, 2), (3, 5, 1, 4), (3, 5, 2, 1), (3, 5, 2, 4), (3, 5, 4, 1), (3, 5, 4, 2), (4, 1, 2, 3), (4, 1, 2, 5), (4, 1, 3, 2), (4, 1, 3, 5), (4, 1, 5, 2), (4, 1, 5, 3), (4, 2, 1, 3), (4, 2, 1, 5), (4, 2, 3, 1), (4, 2, 3, 5), (4, 2, 5, 1), (4, 2, 5, 3), (4, 3, 1, 2), (4, 3, 1, 5), (4, 3, 2, 1), (4, 3, 2, 5), (4, 3, 5, 1), (4, 3, 5, 2), (4, 5, 1, 2), (4, 5, 1, 3), (4, 5, 2, 1), (4, 5, 2, 3), (4, 5, 3, 1), (4, 5, 3, 2), (5, 1, 2, 3), (5, 1, 2, 4), (5, 1, 3, 2), (5, 1, 3, 4), (5, 1, 4, 2), (5, 1, 4, 3), (5, 2, 1, 3), (5, 2, 1, 4), (5, 2, 3, 1), (5, 2, 3, 4), (5, 2, 4, 1), (5, 2, 4, 3), (5, 3, 1, 2), (5, 3, 1, 4), (5, 3, 2, 1), (5, 3, 2, 4), (5, 3, 4, 1), (5, 3, 4, 2), (5, 4, 1, 2), (5, 4, 1, 3), (5, 4, 2, 1), (5, 4, 2, 3), (5, 4, 3, 1), (5, 4, 3, 2)]
假设我有 4 个桶和一个数字数组,例如 [1,2,3,4,5]
| || || || |
|__||__||__||__|
1 2 3 4
5 2 3 4
1 5 3 4
etc...
我也可以少于 4 个数字,例如 [1, 2, 3]
| || || || |
|__||__||__||__|
1 2 3
1 3 2
1 2 3
etc...
如何找到存储桶中数字的所有可能组合(如 [1,2,3,4]
如果长度 >= 4 或 [1,2,None,None]
如果长度 < 4)?
您可以使用递归生成器函数:
def combos(nums, buckets, c = []):
if len(c) == buckets:
yield c
else:
if len(c) + len(nums) < buckets:
yield from combos(nums, buckets, c+[None])
for i, a in enumerate(nums):
yield from combos(nums[:i]+nums[i+1:], buckets, c+[a])
print(list(combos([1, 2, 3, 4, 5], 4)))
print(list(combos([1, 2, 3], 4)))
输出:
[[1, 2, 3, 4], [1, 2, 3, 5], [1, 2, 4, 3], [1, 2, 4, 5], [1, 2, 5, 3], [1, 2, 5, 4], [1, 3, 2, 4], [1, 3, 2, 5], [1, 3, 4, 2], [1, 3, 4, 5], [1, 3, 5, 2], [1, 3, 5, 4], [1, 4, 2, 3], [1, 4, 2, 5], [1, 4, 3, 2], [1, 4, 3, 5], [1, 4, 5, 2], [1, 4, 5, 3], [1, 5, 2, 3], [1, 5, 2, 4], [1, 5, 3, 2], [1, 5, 3, 4], [1, 5, 4, 2], [1, 5, 4, 3], [2, 1, 3, 4], [2, 1, 3, 5], [2, 1, 4, 3], [2, 1, 4, 5], [2, 1, 5, 3], [2, 1, 5, 4], [2, 3, 1, 4], [2, 3, 1, 5], [2, 3, 4, 1], [2, 3, 4, 5], [2, 3, 5, 1], [2, 3, 5, 4], [2, 4, 1, 3], [2, 4, 1, 5], [2, 4, 3, 1], [2, 4, 3, 5], [2, 4, 5, 1], [2, 4, 5, 3], [2, 5, 1, 3], [2, 5, 1, 4], [2, 5, 3, 1], [2, 5, 3, 4], [2, 5, 4, 1], [2, 5, 4, 3], [3, 1, 2, 4], [3, 1, 2, 5], [3, 1, 4, 2], [3, 1, 4, 5], [3, 1, 5, 2], [3, 1, 5, 4], [3, 2, 1, 4], [3, 2, 1, 5], [3, 2, 4, 1], [3, 2, 4, 5], [3, 2, 5, 1], [3, 2, 5, 4], [3, 4, 1, 2], [3, 4, 1, 5], [3, 4, 2, 1], [3, 4, 2, 5], [3, 4, 5, 1], [3, 4, 5, 2], [3, 5, 1, 2], [3, 5, 1, 4], [3, 5, 2, 1], [3, 5, 2, 4], [3, 5, 4, 1], [3, 5, 4, 2], [4, 1, 2, 3], [4, 1, 2, 5], [4, 1, 3, 2], [4, 1, 3, 5], [4, 1, 5, 2], [4, 1, 5, 3], [4, 2, 1, 3], [4, 2, 1, 5], [4, 2, 3, 1], [4, 2, 3, 5], [4, 2, 5, 1], [4, 2, 5, 3], [4, 3, 1, 2], [4, 3, 1, 5], [4, 3, 2, 1], [4, 3, 2, 5], [4, 3, 5, 1], [4, 3, 5, 2], [4, 5, 1, 2], [4, 5, 1, 3], [4, 5, 2, 1], [4, 5, 2, 3], [4, 5, 3, 1], [4, 5, 3, 2], [5, 1, 2, 3], [5, 1, 2, 4], [5, 1, 3, 2], [5, 1, 3, 4], [5, 1, 4, 2], [5, 1, 4, 3], [5, 2, 1, 3], [5, 2, 1, 4], [5, 2, 3, 1], [5, 2, 3, 4], [5, 2, 4, 1], [5, 2, 4, 3], [5, 3, 1, 2], [5, 3, 1, 4], [5, 3, 2, 1], [5, 3, 2, 4], [5, 3, 4, 1], [5, 3, 4, 2], [5, 4, 1, 2], [5, 4, 1, 3], [5, 4, 2, 1], [5, 4, 2, 3], [5, 4, 3, 1], [5, 4, 3, 2]]
[[None, 1, 2, 3], [None, 1, 3, 2], [None, 2, 1, 3], [None, 2, 3, 1], [None, 3, 1, 2], [None, 3, 2, 1], [1, None, 2, 3], [1, None, 3, 2], [1, 2, None, 3], [1, 2, 3, None], [1, 3, None, 2], [1, 3, 2, None], [2, None, 1, 3], [2, None, 3, 1], [2, 1, None, 3], [2, 1, 3, None], [2, 3, None, 1], [2, 3, 1, None], [3, None, 1, 2], [3, None, 2, 1], [3, 1, None, 2], [3, 1, 2, None], [3, 2, None, 1], [3, 2, 1, None]]
在每次递归调用时,如果组合尚未形成,代码会做两件事:
- 检查指定的桶数是否大于输入的数字列表。如果是这样,则输出用
None
. 填充
- 迭代剩余的数字列表,并将每个迭代值添加到运行结果,并进行递归。
使用内置函数itertools.permutations
。
import itertools
def func(arr: list, slot_count=4):
if (v := slot_count - len(arr)) > 0:
arr.extend([None for _ in range(v)])
return itertools.permutations(arr, slot_count)
print(list(func([1, 2, 3])))
print(list(func([1, 2, 3, 4, 5])))
[(1, 2, 3, None), (1, 2, None, 3), (1, 3, 2, None), (1, 3, None, 2), (1, None, 2, 3), (1, None, 3, 2), (2, 1, 3, None), (2, 1, None, 3), (2, 3, 1, None), (2, 3, None, 1), (2, None, 1, 3), (2, None, 3, 1), (3, 1, 2, None), (3, 1, None, 2), (3, 2, 1, None), (3, 2, None, 1), (3, None, 1, 2), (3, None, 2, 1), (None, 1, 2, 3), (None, 1, 3, 2), (None, 2, 1, 3), (None, 2, 3, 1), (None, 3, 1, 2), (None, 3, 2, 1)]
[(1, 2, 3, 4), (1, 2, 3, 5), (1, 2, 4, 3), (1, 2, 4, 5), (1, 2, 5, 3), (1, 2, 5, 4), (1, 3, 2, 4), (1, 3, 2, 5), (1, 3, 4, 2), (1, 3, 4, 5), (1, 3, 5, 2), (1, 3, 5, 4), (1, 4, 2, 3), (1, 4, 2, 5), (1, 4, 3, 2), (1, 4, 3, 5), (1, 4, 5, 2), (1, 4, 5, 3), (1, 5, 2, 3), (1, 5, 2, 4), (1, 5, 3, 2), (1, 5, 3, 4), (1, 5, 4, 2), (1, 5, 4, 3), (2, 1, 3, 4), (2, 1, 3, 5), (2, 1, 4, 3), (2, 1, 4, 5), (2, 1, 5, 3), (2, 1, 5, 4), (2, 3, 1, 4), (2, 3, 1, 5), (2, 3, 4, 1), (2, 3, 4, 5), (2, 3, 5, 1), (2, 3, 5, 4), (2, 4, 1, 3), (2, 4, 1, 5), (2, 4, 3, 1), (2, 4, 3, 5), (2, 4, 5, 1), (2, 4, 5, 3), (2, 5, 1, 3), (2, 5, 1, 4), (2, 5, 3, 1), (2, 5, 3, 4), (2, 5, 4, 1), (2, 5, 4, 3), (3, 1, 2, 4), (3, 1, 2, 5), (3, 1, 4, 2), (3, 1, 4, 5), (3, 1, 5, 2), (3, 1, 5, 4), (3, 2, 1, 4), (3, 2, 1, 5), (3, 2, 4, 1), (3, 2, 4, 5), (3, 2, 5, 1), (3, 2, 5, 4), (3, 4, 1, 2), (3, 4, 1, 5), (3, 4, 2, 1), (3, 4, 2, 5), (3, 4, 5, 1), (3, 4, 5, 2), (3, 5, 1, 2), (3, 5, 1, 4), (3, 5, 2, 1), (3, 5, 2, 4), (3, 5, 4, 1), (3, 5, 4, 2), (4, 1, 2, 3), (4, 1, 2, 5), (4, 1, 3, 2), (4, 1, 3, 5), (4, 1, 5, 2), (4, 1, 5, 3), (4, 2, 1, 3), (4, 2, 1, 5), (4, 2, 3, 1), (4, 2, 3, 5), (4, 2, 5, 1), (4, 2, 5, 3), (4, 3, 1, 2), (4, 3, 1, 5), (4, 3, 2, 1), (4, 3, 2, 5), (4, 3, 5, 1), (4, 3, 5, 2), (4, 5, 1, 2), (4, 5, 1, 3), (4, 5, 2, 1), (4, 5, 2, 3), (4, 5, 3, 1), (4, 5, 3, 2), (5, 1, 2, 3), (5, 1, 2, 4), (5, 1, 3, 2), (5, 1, 3, 4), (5, 1, 4, 2), (5, 1, 4, 3), (5, 2, 1, 3), (5, 2, 1, 4), (5, 2, 3, 1), (5, 2, 3, 4), (5, 2, 4, 1), (5, 2, 4, 3), (5, 3, 1, 2), (5, 3, 1, 4), (5, 3, 2, 1), (5, 3, 2, 4), (5, 3, 4, 1), (5, 3, 4, 2), (5, 4, 1, 2), (5, 4, 1, 3), (5, 4, 2, 1), (5, 4, 2, 3), (5, 4, 3, 1), (5, 4, 3, 2)]