如何生成与所需列表保持固定距离的元素列表

我想我将展示一种更为冗长的方法,它更容易归纳。

我先把问题写下来。

possible_pts = [20, 30, 40, 50, 60, 70, 80, 100]
starting_pt_in_idx = [0, 1, 2]
distance = 2

有三个轴可以“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,我们把它放进垃圾箱。例如,如果我看到轴0出现两次,它就会变成[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)

上面给出了绝对运动。要找到实际的运动,我们必须考虑沿该轴的变化可以是正的也可以是负的。

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。所以我们首先编写这个助手函数。

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

使用实际的移动作用于起始点索引,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]

是的,你是对的, itertools 在这里会很有用的。您需要的是查找 possibles 列出重复项,执行此操作的函数是 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 现在,您可以更改产品的组合(请参见下面的@tobias_k comment),这里还有 fake_hamming 函数xd 当然,对于大名单来说,这会很慢,但这是最通用的方法。

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]

您可以尝试一种递归方法:跟踪剩余的距离并生成相关元素的组合。

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 []

如果还存在一些“死胡同”,这仍然会遇到一些“死胡同”。 dist 剩余,但 des 列表是空的,但总的来说,比起前面生成所有组合并在后面检查它们的距离,检查组合的次数要少得多。

如果可能元素的列表较长,则可能需要创建 dict 首先将元素映射到它们的索引,因此您不必这样做 poss.index(first) 每次。

例子:

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]