我有一个数据结构,本质上相当于一个嵌套的字典。假设它是这样的:

{'new jersey': {'mercer county': {'plumbers': 3,
                                  'programmers': 81},
                'middlesex county': {'programmers': 81,
                                     'salesmen': 62}},
 'new york': {'queens county': {'plumbers': 9,
                                'salesmen': 36}}}

现在,维护和创造这个是相当痛苦的;每次我有一个新的州/县/职业,我都必须通过讨厌的try/catch块创建下层字典。此外,如果我想遍历所有值,就必须创建恼人的嵌套迭代器。

我也可以使用元组作为键,像这样:

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

这使得遍历值非常简单和自然,但在语法上更痛苦的事情,如聚合和查看字典的子集(例如,如果我只想逐个状态查看)。

基本上,有时我想把一个嵌套字典看作一个平面字典,有时我想把它看作一个复杂的层次结构。我可以把这些都打包到一个类中,但似乎有人已经这样做了。或者,似乎有一些非常优雅的语法结构可以做到这一点。

我怎样才能做得更好呢?

附录:我知道setdefault(),但它并不是真正的干净语法。此外,您创建的每个子字典仍然需要手动设置setdefault()。


当前回答

至于“讨厌的尝试/捕捉障碍”:

d = {}
d.setdefault('key',{}).setdefault('inner key',{})['inner inner key'] = 'value'
print d

收益率

{'key': {'inner key': {'inner inner key': 'value'}}}

你可以使用它从你的平面字典格式转换为结构化格式:

fd = {('new jersey', 'mercer county', 'plumbers'): 3,
 ('new jersey', 'mercer county', 'programmers'): 81,
 ('new jersey', 'middlesex county', 'programmers'): 81,
 ('new jersey', 'middlesex county', 'salesmen'): 62,
 ('new york', 'queens county', 'plumbers'): 9,
 ('new york', 'queens county', 'salesmen'): 36}

for (k1,k2,k3), v in fd.iteritems():
    d.setdefault(k1, {}).setdefault(k2, {})[k3] = v

其他回答

正如其他人所建议的,关系数据库可能对您更有用。您可以使用内存中的sqlite3数据库作为数据结构来创建表,然后查询它们。

import sqlite3

c = sqlite3.Connection(':memory:')
c.execute('CREATE TABLE jobs (state, county, title, count)')

c.executemany('insert into jobs values (?, ?, ?, ?)', [
    ('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),
])

# some example queries
print list(c.execute('SELECT * FROM jobs WHERE county = "Queens County"'))
print list(c.execute('SELECT SUM(count) FROM jobs WHERE title = "Programmers"'))

这只是一个简单的例子。您可以为州、县和职称定义单独的表。

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。

我也有类似的事情。我有很多这样的案例:

thedict = {}
for item in ('foo', 'bar', 'baz'):
  mydict = thedict.get(item, {})
  mydict = get_value_for(item)
  thedict[item] = mydict

但要深入很多层次。这是“。”Get (item,{})",这是一个键,因为如果已经没有字典,它将创建另一个字典。与此同时,我一直在想办法对付 这个更好。现在,有很多

value = mydict.get('foo', {}).get('bar', {}).get('baz', 0)

所以,我做了:

def dictgetter(thedict, default, *args):
  totalargs = len(args)
  for i,arg in enumerate(args):
    if i+1 == totalargs:
      thedict = thedict.get(arg, default)
    else:
      thedict = thedict.get(arg, {})
  return thedict

如果你这样做,效果是一样的:

value = dictgetter(mydict, 0, 'foo', 'bar', 'baz')

更好吗?我想是的。

我发现setdefault非常有用;它检查一个键是否存在,如果不存在就添加它:

d = {}
d.setdefault('new jersey', {}).setdefault('mercer county', {})['plumbers'] = 3

Setdefault总是返回相关的键,所以你实际上是在原地更新'd'的值。

说到迭代,我相信你可以很容易地编写一个生成器,如果Python中还没有这样的生成器:

def iterateStates(d):
    # Let's count up the total number of "plumbers" / "dentists" / etc.
    # across all counties and states
    job_totals = {}

    # I guess this is the annoying nested stuff you were talking about?
    for (state, counties) in d.iteritems():
        for (county, jobs) in counties.iteritems():
            for (job, num) in jobs.iteritems():
                # If job isn't already in job_totals, default it to zero
                job_totals[job] = job_totals.get(job, 0) + num

    # Now return an iterator of (job, number) tuples
    return job_totals.iteritems()

# Display all jobs
for (job, num) in iterateStates(d):
    print "There are %d %s in total" % (job, num)
class JobDb(object):
    def __init__(self):
        self.data = []
        self.all = set()
        self.free = []
        self.index1 = {}
        self.index2 = {}
        self.index3 = {}

    def _indices(self,(key1,key2,key3)):
        indices = self.all.copy()
        wild = False
        for index,key in ((self.index1,key1),(self.index2,key2),
                                             (self.index3,key3)):
            if key is not None:
                indices &= index.setdefault(key,set())
            else:
                wild = True
        return indices, wild

    def __getitem__(self,key):
        indices, wild = self._indices(key)
        if wild:
            return dict(self.data[i] for i in indices)
        else:
            values = [self.data[i][-1] for i in indices]
            if values:
                return values[0]

    def __setitem__(self,key,value):
        indices, wild = self._indices(key)
        if indices:
            for i in indices:
                self.data[i] = key,value
        elif wild:
            raise KeyError(k)
        else:
            if self.free:
                index = self.free.pop(0)
                self.data[index] = key,value
            else:
                index = len(self.data)
                self.data.append((key,value))
                self.all.add(index)
            self.index1.setdefault(key[0],set()).add(index)
            self.index2.setdefault(key[1],set()).add(index)
            self.index3.setdefault(key[2],set()).add(index)

    def __delitem__(self,key):
        indices,wild = self._indices(key)
        if not indices:
            raise KeyError
        self.index1[key[0]] -= indices
        self.index2[key[1]] -= indices
        self.index3[key[2]] -= indices
        self.all -= indices
        for i in indices:
            self.data[i] = None
        self.free.extend(indices)

    def __len__(self):
        return len(self.all)

    def __iter__(self):
        for key,value in self.data:
            yield key

例子:

>>> db = JobDb()
>>> db['new jersey', 'mercer county', 'plumbers'] = 3
>>> db['new jersey', 'mercer county', 'programmers'] = 81
>>> db['new jersey', 'middlesex county', 'programmers'] = 81
>>> db['new jersey', 'middlesex county', 'salesmen'] = 62
>>> db['new york', 'queens county', 'plumbers'] = 9
>>> db['new york', 'queens county', 'salesmen'] = 36

>>> db['new york', None, None]
{('new york', 'queens county', 'plumbers'): 9,
 ('new york', 'queens county', 'salesmen'): 36}

>>> db[None, None, 'plumbers']
{('new jersey', 'mercer county', 'plumbers'): 3,
 ('new york', 'queens county', 'plumbers'): 9}

>>> db['new jersey', 'mercer county', None]
{('new jersey', 'mercer county', 'plumbers'): 3,
 ('new jersey', 'mercer county', 'programmers'): 81}

>>> db['new jersey', 'middlesex county', 'programmers']
81

>>>

编辑:现在使用通配符(None)查询时返回字典,否则返回单值。