如何根据值删除列表的嵌套字典中的键

Question

I am working on a file representing a tree-like structure very similar to flare.json which is known for D3.js community. What's the best way to delete all the leaves of the tree in python? In other words, I want to remove all the keys that don't have a 'children' key in their value.

example :

{
 "name": "flare",
 "children": [
  {
   "name": "analytics",
   "children": [
    {
     "name": "cluster",
     "children": [
      {"name": "AgglomerativeCluster", "size": 3938},
      {"name": "CommunityStructure", "size": 3812},
      {"name": "HierarchicalCluster", "size": 6714},
      {"name": "MergeEdge", "size": 743}
     ]
    },
    {
     "name": "graph",
     "children": [
      {"name": "BetweennessCentrality", "size": 3534},
      {"name": "LinkDistance", "size": 5731},
      {"name": "MaxFlowMinCut", "size": 7840},
      {"name": "ShortestPaths", "size": 5914},
      {"name": "SpanningTree", "size": 3416}
     ]
    },
    {
     "name": "optimization",
     "children": [
      {"name": "AspectRatioBanker", "size": 7074}
     ] ...

which should become:

{
 "name": "flare",
 "children": [
  {
   "name": "analytics",
   "children": [
    {
     "name": "cluster",
    },
    {
     "name": "graph",
    },
    {
     "name": "optimization",

     ] ...

In other words, I'm just cutting the leaves of the tree. In a children list is empty, it should be removed.

I tried this only to remove the keys and it did not work:

def deleteLeaves(pTree):
    if pTree.has_key('children'):
        for child in pTree['children']:
            deleteLeaves(child)
    else:
        del pTree

Answer 1

This seems to approximate what you want:

def pruneLeaves(obj):

  if isinstance(obj, dict):
    isLeaf = True
    for key in obj.keys():
      if key == 'children': isLeaf = False
      if pruneLeaves(obj[key]): del obj[key]
    return isLeaf

  elif isinstance(obj, list):
    leaves = []
    for (index, element) in enumerate(obj):
      if pruneLeaves(element): leaves.append(index)
    leaves.reverse()
    for index in leaves: obj.pop(index)
    return not bool(obj)

  else:  # String values look like attributes in your dict, so never prune them
    return False

Tested with a truncated sample of your data:

data = {
 "name": "flare",
 "children": [
  {
   "name": "analytics",
   "children": [
    {
     "name": "cluster",
     "children": [
      {"name": "AgglomerativeCluster", "size": 3938},
      {"name": "CommunityStructure", "size": 3812},
      {"name": "HierarchicalCluster", "size": 6714},
      {"name": "MergeEdge", "size": 743}
     ]
    },
    {
     "name": "graph",
     "children": [
      {"name": "BetweennessCentrality", "size": 3534},
      {"name": "LinkDistance", "size": 5731},
      {"name": "MaxFlowMinCut", "size": 7840},
      {"name": "ShortestPaths", "size": 5914},
      {"name": "SpanningTree", "size": 3416}
     ]
    }
   ]
  }
  ]
}

pruneLeaves(data)
print data

And got these results:

{'name': 'flare', 'children': [{'name': 'analytics', 'children': [{'name': 'cluster'}, {'name': 'graph'}]}]}

Answer 2

I just edited the answer of @rchang to fix deletion of lists other than children .

def pruneLeaves(self,obj):
    if isinstance(obj, dict):
        isLeaf = True
        for key in obj.keys():
            if key=='children': 
                isLeaf = False
                if self.pruneLeaves(obj[key]): del obj[key]
        return isLeaf

    elif isinstance(obj, list) :
        leaves = []
        for (index, element) in enumerate(obj):
          if self.pruneLeaves(element): leaves.append(index)
        leaves.reverse()
        for index in leaves: obj.pop(index)
        return not bool(obj)
    else:  # String values look like attributes in your dict, so never prune them
        return False