我以前用过这个函数。安全、快捷、易于维护。

def deep_get(dictionary, keys, default=None):
    return reduce(lambda d, key: d.get(key, default) if isinstance(d, dict) else default, keys.split("."), dictionary)

例子:

>>> from functools import reduce
>>> def deep_get(dictionary, keys, default=None):
...     return reduce(lambda d, key: d.get(key, default) if isinstance(d, dict) else default, keys.split("."), dictionary)
...
>>> person = {'person':{'name':{'first':'John'}}}
>>> print (deep_get(person, "person.name.first"))
John
>>> print (deep_get(person, "person.name.lastname"))
None
>>> print (deep_get(person, "person.name.lastname", default="No lastname"))
No lastname
>>>

Defaultdict()是你的朋友!

对于一个二维字典，你可以这样做:

d = defaultdict(defaultdict)
d[1][2] = 3

更多维度你可以:

d = defaultdict(lambda :defaultdict(defaultdict))
d[1][2][3] = 4

由于您有一个星型模式设计，您可能希望它的结构更像一个关系表，而不是字典。

import collections

class Jobs( object ):
    def __init__( self, state, county, title, count ):
        self.state= state
        self.count= county
        self.title= title
        self.count= count

facts = [
    Jobs( 'new jersey', 'mercer county', 'plumbers', 3 ),
    ...

def groupBy( facts, name ):
    total= collections.defaultdict( int )
    for f in facts:
        key= getattr( f, name )
        total[key] += f.count

这类事情对于创建一个没有SQL开销的类似数据仓库的设计大有帮助。

可以对Collections.defaultdict进行子类化以生成嵌套的dict。然后向该类添加任何有用的迭代方法。

>>> from collections import defaultdict
>>> class nesteddict(defaultdict):
    def __init__(self):
        defaultdict.__init__(self, nesteddict)
    def walk(self):
        for key, value in self.iteritems():
            if isinstance(value, nesteddict):
                for tup in value.walk():
                    yield (key,) + tup
            else:
                yield key, value


>>> nd = nesteddict()
>>> nd['new jersey']['mercer county']['plumbers'] = 3
>>> nd['new jersey']['mercer county']['programmers'] = 81
>>> nd['new jersey']['middlesex county']['programmers'] = 81
>>> nd['new jersey']['middlesex county']['salesmen'] = 62
>>> nd['new york']['queens county']['plumbers'] = 9
>>> nd['new york']['queens county']['salesmen'] = 36
>>> for tup in nd.walk():
    print tup


('new jersey', 'mercer county', 'programmers', 81)
('new jersey', 'mercer county', 'plumbers', 3)
('new jersey', 'middlesex county', 'programmers', 81)
('new jersey', 'middlesex county', 'salesmen', 62)
('new york', 'queens county', 'salesmen', 36)
('new york', 'queens county', 'plumbers', 9)

这是一个返回任意深度的嵌套字典的函数:

from collections import defaultdict
def make_dict():
    return defaultdict(make_dict)

像这样使用它:

d=defaultdict(make_dict)
d["food"]["meat"]="beef"
d["food"]["veggie"]="corn"
d["food"]["sweets"]="ice cream"
d["animal"]["pet"]["dog"]="collie"
d["animal"]["pet"]["cat"]="tabby"
d["animal"]["farm animal"]="chicken"

迭代所有内容，如下所示:

def iter_all(d,depth=1):
    for k,v in d.iteritems():
        print "-"*depth,k
        if type(v) is defaultdict:
            iter_all(v,depth+1)
        else:
            print "-"*(depth+1),v

iter_all(d)

打印出来:

- food
-- sweets
--- ice cream
-- meat
--- beef
-- veggie
--- corn
- animal
-- pet
--- dog
---- labrador
--- cat
---- tabby
-- farm animal
--- chicken

你可能最终想要这样做，使新的项目不能添加到字典。递归地将所有这些defaultdict转换为普通dict很容易。

def dictify(d):
    for k,v in d.iteritems():
        if isinstance(v,defaultdict):
            d[k] = dictify(v)
    return dict(d)

我喜欢把它包装在一个类中，并实现__getitem__和__setitem__，这样它们就实现了一个简单的查询语言:

>>> d['new jersey/mercer county/plumbers'] = 3
>>> d['new jersey/mercer county/programmers'] = 81
>>> d['new jersey/mercer county/programmers']
81
>>> d['new jersey/mercer country']
<view which implicitly adds 'new jersey/mercer county' to queries/mutations>

如果你想要更花哨，你也可以实现如下内容:

>>> d['*/*/programmers']
<view which would contain 'programmers' entries>

但大多数情况下，我认为这样的事情执行起来真的很有趣:D