我如何使Python字典成员访问通过点“。”?
例如,我想写mydict.val而不是mydict['val']。
我还想以这种方式访问嵌套字典。例如
mydict.mydict2.val
会提到
mydict = { 'mydict2': { 'val': ... } }
我如何使Python字典成员访问通过点“。”?
例如,我想写mydict.val而不是mydict['val']。
我还想以这种方式访问嵌套字典。例如
mydict.mydict2.val
会提到
mydict = { 'mydict2': { 'val': ... } }
当前回答
我的观点:出于我自己的目的,我开发了minydra,一个简单的命令行解析器,包括一个自定义类MinyDict(灵感来自addict):
In [1]: from minydra import MinyDict
In [2]: args = MinyDict({"foo": "bar", "yes.no.maybe": "idontknow"}).pretty_print(); args
╭──────────────────────────────╮
│ foo : bar │
│ yes.no.maybe : idontknow │
╰──────────────────────────────╯
Out[2]: {'foo': 'bar', 'yes.no.maybe': 'idontknow'}
In [3]: args.resolve().pretty_print(); args
╭──────────────────────────╮
│ foo : bar │
│ yes │
│ │no │
│ │ │maybe : idontknow │
╰──────────────────────────╯
Out[3]: {'foo': 'bar', 'yes': {'no': {'maybe': 'idontknow'}}}
In [4]: args.yes.no.maybe
Out[4]: "idontknow"
In [5]: "foo" in args
Out[5]: True
In [6]: "rick" in args
Out[6]: False
In [7]: args.morty is None
Out[7]: True
In [8]: args.items()
Out[8]: dict_items([('foo', 'bar'), ('yes', {'no': {'maybe': 'idontknow'}})])
它通过向json yaml和pickle添加转储/加载方法来上瘾,并且在MinyDict.update()中也有一个严格的模式来防止创建新键(这对于防止命令行中的错字很有用)
其他回答
基于epool的答案,这个版本允许你通过点操作符访问任何字典:
foo = {
"bar" : {
"baz" : [ {"boo" : "hoo"} , {"baba" : "loo"} ]
}
}
例如,foo.bar.baz[1]。爸爸回答“loo”。
class Map(dict):
def __init__(self, *args, **kwargs):
super(Map, self).__init__(*args, **kwargs)
for arg in args:
if isinstance(arg, dict):
for k, v in arg.items():
if isinstance(v, dict):
v = Map(v)
if isinstance(v, list):
self.__convert(v)
self[k] = v
if kwargs:
for k, v in kwargs.items():
if isinstance(v, dict):
v = Map(v)
elif isinstance(v, list):
self.__convert(v)
self[k] = v
def __convert(self, v):
for elem in range(0, len(v)):
if isinstance(v[elem], dict):
v[elem] = Map(v[elem])
elif isinstance(v[elem], list):
self.__convert(v[elem])
def __getattr__(self, attr):
return self.get(attr)
def __setattr__(self, key, value):
self.__setitem__(key, value)
def __setitem__(self, key, value):
super(Map, self).__setitem__(key, value)
self.__dict__.update({key: value})
def __delattr__(self, item):
self.__delitem__(item)
def __delitem__(self, key):
super(Map, self).__delitem__(key)
del self.__dict__[key]
Fabric有一个非常好的、最小的实现。将其扩展为允许嵌套访问,我们可以使用defaultdict,结果看起来像这样:
from collections import defaultdict
class AttributeDict(defaultdict):
def __init__(self):
super(AttributeDict, self).__init__(AttributeDict)
def __getattr__(self, key):
try:
return self[key]
except KeyError:
raise AttributeError(key)
def __setattr__(self, key, value):
self[key] = value
可以这样使用它:
keys = AttributeDict()
keys.abc.xyz.x = 123
keys.abc.xyz.a.b.c = 234
这详细阐述了Kugel的回答“从dict和派生并实现__getattr__和__setattr__”。现在你知道怎么做了!
用于无限级别的字典、列表、字典的列表和列表的字典的嵌套。
它还支持酸洗
这是这个答案的延伸。
class DotDict(dict):
# https://stackoverflow.com/a/70665030/913098
"""
Example:
m = Map({'first_name': 'Eduardo'}, last_name='Pool', age=24, sports=['Soccer'])
Iterable are assumed to have a constructor taking list as input.
"""
def __init__(self, *args, **kwargs):
super(DotDict, self).__init__(*args, **kwargs)
args_with_kwargs = []
for arg in args:
args_with_kwargs.append(arg)
args_with_kwargs.append(kwargs)
args = args_with_kwargs
for arg in args:
if isinstance(arg, dict):
for k, v in arg.items():
self[k] = v
if isinstance(v, dict):
self[k] = DotDict(v)
elif isinstance(v, str) or isinstance(v, bytes):
self[k] = v
elif isinstance(v, Iterable):
klass = type(v)
map_value: List[Any] = []
for e in v:
map_e = DotDict(e) if isinstance(e, dict) else e
map_value.append(map_e)
self[k] = klass(map_value)
def __getattr__(self, attr):
return self.get(attr)
def __setattr__(self, key, value):
self.__setitem__(key, value)
def __setitem__(self, key, value):
super(DotDict, self).__setitem__(key, value)
self.__dict__.update({key: value})
def __delattr__(self, item):
self.__delitem__(item)
def __delitem__(self, key):
super(DotDict, self).__delitem__(key)
del self.__dict__[key]
def __getstate__(self):
return self.__dict__
def __setstate__(self, d):
self.__dict__.update(d)
if __name__ == "__main__":
import pickle
def test_map():
d = {
"a": 1,
"b": {
"c": "d",
"e": 2,
"f": None
},
"g": [],
"h": [1, "i"],
"j": [1, "k", {}],
"l":
[
1,
"m",
{
"n": [3],
"o": "p",
"q": {
"r": "s",
"t": ["u", 5, {"v": "w"}, ],
"x": ("z", 1)
}
}
],
}
map_d = DotDict(d)
w = map_d.l[2].q.t[2].v
assert w == "w"
pickled = pickle.dumps(map_d)
unpickled = pickle.loads(pickled)
assert unpickled == map_d
kwargs_check = DotDict(a=1, b=[dict(c=2, d="3"), 5])
assert kwargs_check.b[0].d == "3"
kwargs_and_args_check = DotDict(d, a=1, b=[dict(c=2, d="3"), 5])
assert kwargs_and_args_check.l[2].q.t[2].v == "w"
assert kwargs_and_args_check.b[0].d == "3"
test_map()
通过pip安装dotmap
pip install dotmap
它能做你想让它做的所有事情,并继承dict的子类,所以它的操作就像一个普通的字典:
from dotmap import DotMap
m = DotMap()
m.hello = 'world'
m.hello
m.hello += '!'
# m.hello and m['hello'] now both return 'world!'
m.val = 5
m.val2 = 'Sam'
最重要的是,你可以将它转换为dict对象:
d = m.toDict()
m = DotMap(d) # automatic conversion in constructor
这意味着如果你想访问的东西已经是字典形式的,你可以把它转换成DotMap来方便访问:
import json
jsonDict = json.loads(text)
data = DotMap(jsonDict)
print data.location.city
最后,它会自动创建新的子DotMap实例,你可以这样做:
m = DotMap()
m.people.steve.age = 31
与Bunch的比较
完全公开:我是DotMap的创造者。我创建它是因为Bunch缺少这些功能
记住添加的顺序项并按此顺序迭代 自动创建子DotMap,当你有很多层次结构时,这节省了时间,并使代码更干净 从字典构造并递归地将所有子字典实例转换为DotMap
使用__getattr__,非常简单,适用于 Python 3.4.3
class myDict(dict):
def __getattr__(self,val):
return self[val]
blockBody=myDict()
blockBody['item1']=10000
blockBody['item2']="StackOverflow"
print(blockBody.item1)
print(blockBody.item2)
输出:
10000
StackOverflow