我有一个可能性列表和一个所需的输入内容：
SortedSet我想生成关闭列表。示例：
possibles = [20, 30, 40, 50, 60, 70, 80, 100]
desired = [20, 30, 40]
我当前的版本一次只改变一个元素，因此，当距离大于1时，我就缺少很多组合。
# Distance of 1 (i.e. 1 element changes to a close-by)
[30, 30, 40]
[20, 40, 40]
[20, 30, 30]
[20, 30, 50]
# Distance of 2:
[40, 30, 40]
[30, 30, 50]
[30, 40, 40]
...
我很确定应该有一个模块可以进行这种迭代（itertools？），您能指出我写函数吗？
谢谢。
编辑：
有关尝试的更新...
我试图生成一个所需大小相同的列表，其中每个元素对应于我必须移动所需元素的数量。
def generate_close_by(possibles, desired):
for k in range(1, 4):
for i, elt in enumerate(desired):
id = possibles.index(elt)
new = desired[:]
if id < len(possibles)-k-1:
new[i] = possibles[id+k]
yield (new)
if id > k:
new[i] = possibles[id-k]
yield (new)
# Output
[30, 30, 40]
[20, 40, 40]
[20, 30, 50]
[20, 30, 30]
[40, 30, 40]
[20, 50, 40]
[20, 30, 60]
[50, 30, 40]
[20, 60, 40]
[20, 30, 70]
然后计划尝试创建新列表，如果不能（超出范围），则继续。尚不可行，但这可能是一个好方法。
答案 0 :(得分：1)
是的，itertools在这里确实很有用。您想要的是找到possibles列表中具有所需长度的所有子集（包含重复项），并且该功能的集合为itertools.product
from itertools import product
possibles = [20, 30, 40, 50, 60, 70, 80, 100]
desired = [20, 30, 40]
def fake_hamming(cur, desired, possibles):
assert len(cur) == len(desired)
hamm = 0
for i in range(len(cur)):
assert cur[i] in possibles
assert desired[i] in possibles
hamm += abs(possibles.index(cur[i]) - possibles.index(desired[i]))
return hamm
def generate_close_by(desired, possibles, dist):
all_possible_lists = product(possibles, repeat=len(desired))
return [l for l in all_possible_lists if fake_hamming(l, desired, possibles) == dist]
print(generate_close_by(desired, possibles,1))
>>> [(20, 20, 40), (20, 30, 30), (20, 30, 50), (20, 40, 40), (30, 30, 40)]
编辑，您可以更改产品的组合（请参见下面的@tobias_k注释），这也是fake_hamming函数xD
同样，大列表的速度确实很慢，但这是最通用的方法
答案 1 :(得分：1)
def distribute(number, bucket):
if bucket == 1:
yield [number]
if number != 0:
yield [-1 * number]
elif number == 0:
yield [0]*bucket
else:
for i in range(number+1):
for j in distribute(number-i, 1):
for k in distribute(i, bucket-1):
yield j+k
def generate(possibles, desired, distance):
for index_distance_tuple in distribute(distance, len(desired)):
retval = desired[:]
for i, index in enumerate(index_distance_tuple):
if index + i < 0 or index + i >= len(possibles):
break
retval[i] = possibles[index + i]
else:
yield retval
对于距离1：
for i in generate(possibles, desired, 1):
print(i)
输出：
[30, 30, 40]
[20, 40, 40]
[20, 20, 40]
[20, 30, 50]
[20, 30, 30]
对于距离2：
for i in generate(possibles, desired, 2):
print(i)
输出：
[40, 30, 40]
[30, 40, 40]
[30, 20, 40]
[30, 30, 50]
[30, 30, 30]
[20, 50, 40]
[20, 40, 50]
[20, 40, 30]
[20, 20, 50]
[20, 20, 30]
[20, 30, 60]
[20, 30, 20]
答案 2 :(得分：1)
我想我将展示一个更漫长的方法，它可以更容易地推广。
我首先写下问题。
possible_pts = [20, 30, 40, 50, 60, 70, 80, 100]
starting_pt_in_idx = [0, 1, 2]
distance = 2
有3个轴可以“改变”。我首先找到轴变化的组合。
N = len(starting_pt_in_idx)
axis = list(range(N))
import itertools
axismoves = list(itertools.combinations_with_replacement(axis, distance))
print(axismoves)
接下来，我们将其装箱。例如，如果我看到轴0出现两次，则它变为[2,0,0]。
abs_movements = []
for combi in axismoves:
move_bin = [0] * N
for i in combi:
move_bin[i] += 1
abs_movements.append(move_bin)
print(abs_movements)
以上给出了绝对的动作。要找到实际的运动，我们必须考虑沿该轴的变化可以是正值或负值。
import copy
actual_movements = []
for movement in abs_movements:
actual_movements.append(movement)
for i, move in enumerate(movement):
if move != 0:
_movement = copy.deepcopy(movement)
_movement[i] = - move
actual_movements.append(_movement)
print(actual_movements)
最后一步是将索引转换为实际位置。因此，我们首先编写此辅助函数。
def translate_idx_to_pos(idx_vect, points):
idx_bounds = [0, len(points) - 1]
pos_point = [0] * len(idx_vect)
for i, idx_pos in enumerate(idx_vect):
if idx_pos < idx_bounds[0] or idx_pos > idx_bounds[1]:
return None
else:
pos_point[i] = points[idx_pos]
return pos_point
使用实际运动对起点索引进行操作，然后将其转换回位置。
from operator import add
final_pts = []
for movement in actual_movements:
final_pt_in_idx = list(map(add, starting_pt_in_idx, movement))
final_point = translate_idx_to_pos(final_pt_in_idx, possible_pts)
if final_point is not None:
final_pts.append(final_point)
print(final_pts)
这给出了
[40, 30, 40]
[30, 40, 40]
[30, 20, 40]
[30, 30, 50]
[30, 30, 30]
[20, 50, 40]
[20, 40, 50]
[20, 20, 50]
[20, 40, 30]
[20, 30, 60]
[20, 30, 20]
答案 3 :(得分：0)
您可以尝试一种递归方法：跟踪剩余距离并生成仅相关元素的组合。
def get_with_distance(poss, des, dist, k=0):
if k < len(des):
i = poss.index(des[k])
for n in range(-dist, dist+1):
if 0 <= i + n < len(poss):
for comb in get_with_distance(poss, des, dist - abs(n), k+1):
yield [poss[i + n]] + comb
elif dist == 0:
yield []
如果仍然还有dist，但des列表为空，这仍然会遇到一些“死角”，但是总的来说，与生成所有预先组合，然后检查它们的距离。
如果可能的元素列表较长，则可能要先创建一个dict映射到其索引的元素，因此不必每次都poss.index(first)。
示例：
possibles = [20, 30, 40, 50, 60, 70, 80, 100]
desired = [20, 30, 40]
for x in get_with_distance(possibles, desired, 2):
print(x)
输出：
[20, 20, 30]
[20, 20, 50]
[20, 30, 20]
[20, 30, 60]
[20, 40, 30]
[20, 40, 50]
[20, 50, 40]
[30, 20, 40]
[30, 30, 30]
[30, 30, 50]
[30, 40, 40]
[40, 30, 40]