递归列表拆分和索引

Question

我编写了以下代码以递归方式拆分列表。 它首先递归地拆分左侧，直到并且除非剩下一个元素。

代码：

def split(class_names):
 
  while len(class_names)>1:
    n=len(class_names)
    mid=n//2
    left=class_names[:mid]
    right=class_names[mid:]
    splits.append([left,right])
    class_names=left
    index=1
    split(left)
    class_names=right
  return splits
class_names=[1,2,3,4,5,6,7,8,9,10]
splits=[]
splits=split(class_names)

for ii in splits:
  print(ii)

输出：

[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
[[1, 2], [3, 4, 5]]
[[1], [2]]
[[3], [4, 5]]
[[4], [5]]
[[6, 7], [8, 9, 10]]
[[6], [7]]
[[8], [9, 10]]
[[9], [10]]

问题：我需要树及其索引的形式。 左侧应在末尾添加 0，右侧应在末尾添加 1。 例如：

                 [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]]
                            /\
                        0  /  \ 1
          [[1, 2], [3, 4, 5]]  [[6, 7], [8, 9, 10]]
                /\
            0  /  \ 1
       [[1], [2]]  [[3], [4, 5]]
Then the output should be like:
[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]] = 0
[[1, 2], [3, 4, 5]] = 00
[[6, 7], [8, 9, 10]] = 01
[[1], [2]] = 000

Answer 1

您已经降低了递归，但需要在左右遍历时跟踪索引。 这是一种带有附加参数的方法，并被重写为生成器：

def split(class_names, index='0'):
    if (n := len(class_names)) < 2:
        return
    mid = n // 2
    left, right = class_names[:mid], class_names[mid:]
    yield [left, right], index
    yield from split(left, index + '0')
    yield from split(right, index + '1')

class_names = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for node, index in split(class_names):
    print(f'{node} = {index}')

输出：

[[1, 2, 3, 4, 5], [6, 7, 8, 9, 10]] = 0
[[1, 2], [3, 4, 5]] = 00
[[1], [2]] = 000
[[3], [4, 5]] = 001
[[4], [5]] = 0011
[[6, 7], [8, 9, 10]] = 01
[[6], [7]] = 010
[[8], [9, 10]] = 011
[[9], [10]] = 0111