Question

我想翻开以下词典：

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

成：

dictionary = {
    4: {3: {8: {8: {4: {6: {4: {'words': ['getting']}}}}}}}
    8: {2: {7: {8: {6: {2: {'words': ['Taruma', 'Varuna']}}}}}}
    ...
}

这样我就可以使用以下词典：dictionary[8][2][7][8][6][2]['words']而不是：dictionary[827862]。

Answer 1

import pprint

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

d2 = {}

def add_it(d, k, words):
    knum = int(k[0])
    if len(k) == 1:
        d[knum] = {'words': words}
    else:
        dsub = d.setdefault(knum, {})
        add_it(dsub, k[1:], words)

for k, words in dictionary.items():
    add_it(d2, list(str(k)), words)


pprint.pprint(d2)

打印：

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

Answer 2

制作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' ] } ) )

Answer 3

您可以尝试使用递归的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是一种创建这些不可预测长度的嵌套dicts的漂亮方法。（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，因为已存在d2[4][3][8][8][4]的条目。）

在Python中将一串数字键分成单个数字键

3 个答案: