遍历嵌套字典

Question

我正在尝试实现从字典继承的简单树类。 这是我的代码：

class tree(dict):
    def __init__(self, hsymbol="/"):
        self.hsymbol = hsymbol
    def __setitem__(self, key, value):
        if key[0] == self.hsymbol : key = key[1:]
        parts = key.split(self.hsymbol, 1)
        if len(parts) == 2:
            if parts[0] not in self: self[parts[0]] = tree(hsymbol = self.hsymbol)
            self[parts[0]].__setitem__(parts[1], value)
        else:
            super(tree, self).__setitem__(key, value)

    def __getitem__(self, key):
        if key[0] == self.hsymbol : key = key[1:]
        parts = key.split(self.hsymbol, 1)
        if len(parts) == 2:
            if parts[0] not in self: raise KeyError(parts[0])
            return self[parts[0]][parts[1]]
        else:
            if key not in self: raise KeyError(parts[0])
            return super(tree, self).__getitem__(key)
    def __contains__(self,key):
        if key[0] == self.hsymbol : key = key[1:]
        parts = key.split(self.hsymbol, 1)
        if len(parts) == 2:
            if not super(tree, self).__contains__(parts[0]): return False
            return parts[1] in self[parts[0]]
        else:
            if not super(tree, self).__contains__(key): return False
            return True
    def __delitem__(self, key):
        if key[0] == self.hsymbol : key = key[1:]
        parts = key.split(self.hsymbol, 1)
        if len(parts) == 2:
            if parts[0] not in self: raise KeyError(parts[0])
            self[parts[0]].__delitem__(parts[1])
        else:
            if key not in list(self): raise KeyError(parts[0])
            super(tree,self).__delitem__(key)
    def keys(self,parent=""):
        #if parent is None: parent = self.hsymbol
        names = []
        for name in super(tree, self).keys():
            if isinstance(self[name], tree):
                names += self[name].keys(parent=parent+self.hsymbol+name)
            else:
                names.append(parent+self.hsymbol+name)
        return names

所以一切都很好，虽然我不确定按键功能的实现：

>>> t=tree()
>>> t['/user/Johnson/incoming'] = 2200
>>> t['/user/Johnson/family'] = 4
>>> t['/user/Johnson/play'] = False
>>> t['/user/Smith/incoming'] = 12000
>>> t['/user/Smith/family'] = 1
>>> t['/user/Smith/play'] = True
>>> t['user/Smith/incoming']
12000    
>>> print t
{'user': {'Smith': {'play': True, 'incoming': 12000, 'family': 1}, 'Johnson': {'play': False, 'incoming': 2200, 'family': 4}}}
>>> print t.keys()
['/user/Smith/play', '/user/Smith/incoming', '/user/Smith/family', '/user/Johnson/play', '/user/Johnson/incoming', '/user/Johnson/family']
>>> t
{'user': {'Smith': {'play': True, 'incoming': 12000, 'family': 1}, 'Johnson': {'play': False, 'incoming': 2200, 'family': 4}}}

...但不是通过它的迭代：

>>> for k in t:
...  print k
... 
user
>>> 

我怎样才能得到这样的东西？

/user/Smith/play
/user/Smith/incoming
/user/Smith/family
/user/Johnson/play
/user/Johnson/incoming
/user/Johnson/family

很确定一定是树类的__iter__和next属性，不过还没想好怎么写。

我搜索过 Stack Overflow 没有运气：

• python 递归迭代嵌套字典

• 通过嵌套字典从字典迭代继承的python类

• 通过嵌套字典进行python迭代

Answer 1

是的，您需要__iter__（迭代器将自动拥有next()）。

遵循您现有的逻辑：

def __iter__(self, parent=""):
    for name in super(tree, self).keys():
        if isinstance(self[name], tree):
            for item in self[name].__iter__(parent=parent+self.hsymbol+name):
                yield item
        else:
            yield parent+self.hsymbol+name

与您当前的 keys() 实现不同，这仅根据需要遍历树：如果客户端只需要前两个键，则它只调用 next() 两次，因此迭代器仅继续经过两个 @ 987654326@s.

（我可能会建议将keys 简单地实现为return list(iter(self))——这样您就可以为那些想要避免不必要地遍历整棵树的低效率的人使用惰性方法，否则使用非惰性方法）。