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

对于下面(从上面复制),是否有一种方法来实现append函数。我试图使用嵌套字典存储值作为数组。

class Vividict(dict):
    def __missing__(self, key):
        value = self[key] = type(self)() # retain local pointer to value
    return value  

我目前的实现如下:

totalGeneHash=Vividict()
        
for keys in GenHash:
    for second in GenHash[keys]:
        if keys in sampleHash:
            total_val = GenHash[keys][second]
                totalGeneHash[gene][keys].append(total_val)
This is the error I get: AttributeError: 'Vividict' object has no attribute 'append'

其他回答

只是因为我还没见过这么小的字典,这里有一个词典,你想怎么嵌套就怎么嵌套,毫不费力:

# yo dawg, i heard you liked dicts                                                                      
def yodict():
    return defaultdict(yodict)

您可以创建一个YAML文件,并使用PyYaml读取它。

第一步:创建一个YAML文件"employment.yml":

new jersey:
  mercer county:
    pumbers: 3
    programmers: 81
  middlesex county:
    salesmen: 62
    programmers: 81
new york:
  queens county:
    plumbers: 9
    salesmen: 36

第二步:用Python阅读

import yaml
file_handle = open("employment.yml")
my_shnazzy_dictionary = yaml.safe_load(file_handle)
file_handle.close()

现在my_shnazzy_dictionary有你所有的值。如果需要动态执行此操作,可以将YAML创建为字符串,并将其提供给YAML .safe_load(…)。

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)

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