如何通过元组访问深层嵌套字典?
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)