如何使用元组访问深度嵌套的字典?
问题内容:
我想就扩大了例如自动激活
从前面的回答给nosklo允许通过元组字典的访问。
nosklo的解决方案如下所示:
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}}}
我有一种情况,我要给定一个任意下标元组的节点。如果我不知道元组有多少层,该如何设计一种方法来设置适当的节点?
我在想,也许我可以使用如下语法:
mytuple = (1,2,3)
a[mytuple] = 4
但是我很难提出一个可行的实施方案。
更新资料
我有一个基于@JCash答案的完整示例:
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)
使用扩展切片符号可以实现与上述相同的输出:
d = NestedDict()
d[1,2,3] = 4
d[1,3,3] = 5
d[1,2,'test'] = 6
问题答案:
这似乎有效
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)