如何通过元组访问深层嵌套字典?

5 投票
1 回答
2459 浏览
提问于 2025-04-17 17:06

我想进一步解释一下之前一个回答中的自动生成字典的例子,这个回答是由nosklo提供的,目的是让我们可以通过元组来访问字典。

nosklo的解决方案看起来是这样的:


class AutoVivification(dict):
    """Implementation of perl's autovivification feature."""
    def __getitem__(self, item):
        try:
            return dict.__getitem__(self, item)
        except KeyError:
            value = self[item] = type(self)()
            return value

测试:

a = AutoVivification()

a[1][2][3] = 4
a[1][3][3] = 5
a[1][2]['test'] = 6

print a

输出:

{1: {2: {'test': 6, 3: 4}, 3: {3: 5}}}

我有一个情况,我想根据一些任意的下标元组来设置一个节点。如果我不知道这个元组会有多少层嵌套,我该如何设计一个方法来设置合适的节点呢?

我在想,也许我可以使用下面这样的语法:

mytuple = (1,2,3)
a[mytuple] = 4

但是我在想出一个可行的实现方法时遇到了困难。


更新

我有一个完全可用的例子,基于@JCash的回答:

class NestedDict(dict):
    """                                                                       
    Nested dictionary of arbitrary depth with autovivification.               

    Allows data access via extended slice notation.                           
    """
    def __getitem__(self, keys):
        # Let's assume *keys* is a list or tuple.                             
        if not isinstance(keys, basestring):
            try:
                node = self
                for key in keys:
                    node = dict.__getitem__(node, key)
                return node
            except TypeError:
            # *keys* is not a list or tuple.                              
                pass
        try:
            return dict.__getitem__(self, keys)
        except KeyError:
            raise KeyError(keys)
    def __setitem__(self, keys, value):
        # Let's assume *keys* is a list or tuple.                             
        if not isinstance(keys, basestring):
            try:
                node = self
                for key in keys[:-1]:
                    try:
                        node = dict.__getitem__(node, key)
                    except KeyError:
                        node[key] = type(self)()
                        node = node[key]
                return dict.__setitem__(node, keys[-1], value)
            except TypeError:
                # *keys* is not a list or tuple.                              
                pass
        dict.__setitem__(self, keys, value)

这个例子可以使用扩展切片语法,达到和上面相同的输出效果:

d = NestedDict()
d[1,2,3] = 4
d[1,3,3] = 5
d[1,2,'test'] = 6

1 个回答

5

这看起来有效

def __setitem__(self, key, value):
    if isinstance(key, tuple):
        node = self
        for i in key[:-1]:
            try:
                node = dict.__getitem__(node, i)
            except KeyError:
                node = node[i] = type(self)()
        return dict.__setitem__(node, i, value)
    return dict.__setitem__(self, key, value)

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