在Python中，将一个数字串的键拆分为单个数字键

我想翻一下字典：

dictionary = {
    4388464: ['getting']
    827862 : ['Taruma', 'Varuna']
    ...
}

进入：

这将允许我使用字典，比如：dictionary[8][2][7][8][6][2][words]，而不是：dictionary[827862]

印刷品：

{4: {3: {8: {8: {1: {'words': ['got']},
                 4: {6: {4: {'words': ['getting']}}}}}}},
 8: {2: {7: {8: {6: {2: {'words': ['Taruma', 'Varuna']}}}}}}}

---

生成词典的简短解决方案：

def num2dict( n, d ):
  if n < 10:
    return { n: d }
  else:
    q, r = divmod( n, 10 )
    return num2dict( q, { r: d } )

print( num2dict( 4388464, { 'words': [ 'getting' ] } ) )

---

您可以尝试使用递归defaultdict：

from collections import defaultdict

# define a hierarchical defaultdict (of defaultdicts (of defaultdicts...))
class recursivedefaultdict(defaultdict):
    def __init__(self):
        self.default_factory = type(self)

# add an iterator recursively to create entries, sub-entries, etc.
def addToTree(it, v, accum):
    try:
        addToTree(it, v, accum[it.next()])
    except StopIteration:
        accum["words"] = v

# test it out
dictionary = { 
    4388464: ['getting'], 
    43881: ['got'], 
    827862 : ['Taruma', 'Varuna'], 
} 

d2 = recursivedefaultdict()
for k,v in dictionary.iteritems():
    addToTree(iter(str(k)), v, d2)


# use recursion again to view the results
def dumpDict(d,indent=""):
    for k,v in d.iteritems():
        if k == "words":
            print "%s- %s : %s" % (indent, k, v)
        else:
            print "%s- %s:" % (indent, k)
            dumpDict(v, indent+"  ")

dumpDict(d2)

给出：

- 8:
  - 2:
    - 7:
      - 8:
        - 6:
          - 2:
            - words : ['Taruma', 'Varuna']
- 4:
  - 3:
    - 8:
      - 8:
        - 1:
          - words : ['got']
        - 4:
          - 6:
            - 4:
              - words : ['getting']

---

我认为递归defaultdict是创建这些长度不可预测的嵌套dict的一种很好的方法。但请注意，如果我们添加的下一个值使用43884作为键，则会出现问题，因为d2[4][3][8][4]已经存在条目.

你为什么要这样做？我使用的字典非常大，我很想看看这种方法是否能更快地在字典中找到值。告诉我你在开玩笑。哦，这太尴尬了。打开包含字典的pickle文件需要很长时间！我仍然很好奇如何做到这一点。@THC4k仅在一般情况下，使用一个好的哈希函数。在最坏的情况下，会打开哈希表中的查找。当然，如果在新字典中添加另一个以相同数字开头的条目，则会导致查找失败。是的，我在弄清楚字典的用途之前写了这篇文章。

from collections import defaultdict

# define a hierarchical defaultdict (of defaultdicts (of defaultdicts...))
class recursivedefaultdict(defaultdict):
    def __init__(self):
        self.default_factory = type(self)

# add an iterator recursively to create entries, sub-entries, etc.
def addToTree(it, v, accum):
    try:
        addToTree(it, v, accum[it.next()])
    except StopIteration:
        accum["words"] = v

# test it out
dictionary = { 
    4388464: ['getting'], 
    43881: ['got'], 
    827862 : ['Taruma', 'Varuna'], 
} 

d2 = recursivedefaultdict()
for k,v in dictionary.iteritems():
    addToTree(iter(str(k)), v, d2)


# use recursion again to view the results
def dumpDict(d,indent=""):
    for k,v in d.iteritems():
        if k == "words":
            print "%s- %s : %s" % (indent, k, v)
        else:
            print "%s- %s:" % (indent, k)
            dumpDict(v, indent+"  ")

dumpDict(d2)

- 8:
  - 2:
    - 7:
      - 8:
        - 6:
          - 2:
            - words : ['Taruma', 'Varuna']
- 4:
  - 3:
    - 8:
      - 8:
        - 1:
          - words : ['got']
        - 4:
          - 6:
            - 4:
              - words : ['getting']