Python:搜索并删除嵌套的字典列表

Question

我有一个嵌套列表和字典树,我需要递归遍历并删除符合特定条件的整个字典.例如,我需要删除所有没有子项的"类型""文件夹"的词典(或一个空的子列表).

我仍然是一个初学者Pythonist所以请原谅蛮力.

这是一个格式化的示例字典,便于复制和粘贴.

{'children': [{'children': [{'key': 'group-1',
                         'name': 'PRD',
                         'parent': 'dc-1',
                         'type': 'Folder'},
                        {'children': [{'key': 'group-11',
                                       'name': 'App1',
                                       'parent': 'group-2',
                                       'type': 'Folder'}],
                         'key': 'group-2',
                         'name': 'QA',
                         'parent': 'dc-1',
                         'type': 'Folder'},
                        {'key': 'group-3',
                         'name': 'Keep',
                         'parent': 'dc-1',
                         'type': 'Host'}],
           'key': 'dc-1',
           'name': 'ABC',
           'parent': 'root',
           'type': 'Datacenter'}],
'key': 'root',
'name': 'Datacenters',
'parent': None,
'type': 'Folder'}

在这本字典中,唯一应该保留的树是/ root/dc-1/group-3.应首先删除group-11文件夹,然后删除其父级(因为孩子不再在那里),等等.

我已经尝试了许多不同的递归方法,但似乎无法让它正常工作.任何帮助将不胜感激.

def cleanup(tree):
    def inner(tree):
        if isinstance(tree, dict):
            if 'type' in tree and tree['type'] == 'Folder':
                if 'children' not in tree or not tree['children']:
                    print 'Deleting tree: ' + str(tree['name'])
                    if str(tree['key']) not in del_nodes:
                        del_nodes.append(str(tree['key']))
                else:
                    for item in tree.values():
                        inner(item)
                        # Delete empty folders here
                        if del_nodes:
                            print 'Perform delete here'
                            if 'children' in tree and isinstance(tree['children'], (list, tuple)):
                                getvals = operator.itemgetter('key')
                                tree['children'].sort(key=getvals)
                                result = []
                                # groupby is the wrong method.  I need a list of tree['children'] that doesn't contain keys in del_nodes
                                for k, g in itertools.groupby(tree['children'], getvals):
                                    result.append(g.next())

                                    tree['children'][:] = result

                            del_nodes = []
            else:
                for item in tree.values():
                    inner(item)
        elif isinstance(tree, (list, tuple)):
            for item in tree:
                inner(item)

                if isinstance(item, dict):
                    if 'type' in item and item['type'] == 'Folder':
                        if 'children' not in item or not item['children']:
                            print 'Delete ' + str(item['name'])
                            if str(item['key']) not in del_nodes:
                                del_nodes.append(str(item['key']))
                elif isinstance(item, (list, tuple)):
                    if not item:
                        print 'Delete ' + str(item['name'])
                        if str(item['key']) not in del_nodes:
                            del_nodes.append(str(item['key']))

    inner(tree)

Answer 1

我建议你编写一个函数来遍历数据结构并在每个节点上调用一个函数.

更新以避免"从迭代序列​​中删除项目"错误

例如

def walk(node,parent=None,func=None):
  for child in list(node.get('children',[])):
    walk(child,parent=node,func=func)
  if func is not None:
    func(node,parent=parent)

def removeEmptyFolders(node,parent):
  if node.get('type') == 'Folder' and len(node.get('children',[])) == 0:
    parent['children'].remove(node)

d = {'children': [{'children': [{'key': 'group-1',
                         'name': 'PRD',
                         'parent': 'dc-1',
                         'type': 'Folder'},
                        {'children': [{'key': 'group-11',
                                       'name': 'App1',
                                       'parent': 'group-2',
                                       'type': 'Folder'}],
                         'key': 'group-2',
                         'name': 'QA',
                         'parent': 'dc-1',
                         'type': 'Folder'},
                        {'key': 'group-3',
                         'name': 'Keep',
                         'parent': 'dc-1',
                         'type': 'Host'}],
           'key': 'dc-1',
           'name': 'ABC',
           'parent': 'root',
           'type': 'Datacenter'}],
'key': 'root',
'name': 'Datacenters',
'parent': None,
'type': 'Folder'}

笔记

• Walk函数使用三个参数,子节点,父节点和工作函数.
• 该步行函数调用的工作访问的子节点之后的功能.
• 该工作函数需要的参数,以便修剪孩子都子和父节点是一样简单parent['children'].remove(child)
• 更新:正如注释中所注意到的,如果在迭代时从序列中删除,它将跳过元素.for child in list(node.get('children',[]))在walk函数中复制子项列表,允许从父项的键中删除条目而不跳过.

然后:

>>> walk(d,func=removeEmptyFolders)
>>> from pprint import pprint
>>> pprint(d)
{'children': [{'children': [{'key': 'group-3',
                             'name': 'Keep',
                             'parent': 'dc-1',
                             'type': 'Host'}],
               'key': 'dc-1',
               'name': 'ABC',
               'parent': 'root',
               'type': 'Datacenter'}],
 'key': 'root',
 'name': 'Datacenters',
 'parent': None,
 'type': 'Folder'}