从嵌套字典值（树）创建列表列表

Question

I have used hierarchical clustering to create a tree of clusters.

This results in:

dic = {6: {2: 2, 5: {3: 3, 4: {0: 0, 1: 1}}}}

lists in list according to clusters:

[ [[6]],
  [[2], [5]],
  [[2], [3], [4]],
  [[2], [3], [0], [1]] ]

lists in list according to values:

[ [[2, 3, 0, 1]],
  [[2], [3, 0, 1]],
  [[2], [3], [0, 1]],
  [[2], [3], [0], [1]] ]

What I want to end up with is 'lists in list according to values'.

Thanks

Answer 1

I just spent way too much time for this:

def list_of_list(d):
    if type(d)!=dict:
        return [[[d]]]
    results=[]
    for k,value in d.items():
        results.append(list_of_list(value))
    L=len(max(results,key=len))
    for i in range(len(results)):
        j=len(results[i]) 
        results[i].extend([results[i][j-1]]*(L-j))
    outputtop=[v  for result in results for v in result[0][0]]
    output=[[outputtop]]
    for l in range(L):
        output.append([ val  for result in results for  val in result[l]])
    return output

Answer 2

flat flattens input array
walk recursively walks tree and unpacks values

final list comprehension that generates array same length as number of values extracted
each array element is an array with x single value arrays from l ( l[i:i+1] ) concatenated with array slice of rest of elements in l ( l[x:] )

dic = {6: {2: 2, 5: {3: 3, 4: {0: 0, 1: 1}}}}
flat = lambda l: [item for sublist in l for item in sublist]
walk = lambda node: flat([walk(v) if type(v) is dict else [v] for v in node.values()])

l = walk(dic)
print([[l[i:i+1] for i in range(x)]+[l[x:]] for x in range(len(l))])