Question

请帮我解决以下问题：我有逗号分隔的字符串和与这些字符串相关的值。昏迷的数量是不可预测的，并且可以被限制为6.我必须将其转换为python字典。例如：

aa.bb.cc 6 =>  mydict['aa']['bb']['cc']=6
aa.bb.dd.ee 8 = mydict['aa']['bb']['dd']['ee']=8

我的python版本是2.7.9

Answer 1

一种方法是使用defaultdict

from collections import defaultdict as defd
dic = defd(dict)

def create_dict(astr):
    keys, val = astr.split(' ')
    keys = keys.split('.')
    val = int(val)
    prevdic = dic
    for j, k in enumerate(keys):
        if j == len(keys)-1:
            prevdic[k] = val
        elif k not in prevdic:
            prevdic[k] = defd(dict)
        prevdic = prevdic[k]

create_dict('aa.bb.cc 6')
create_dict('aa.bb.dd.ee 8')
dic['aa']['bb']['cc'] ## returns 6
dic['aa']['bb']['dd']['ee'] ## returns 8

Answer 2

我假设你想要一本带有多个键的字典（你说＆＃34; 一个字典＆＃34;）：

import re

s1 = 'a.bb.dd.ee 8'
s2 = 'aa.bb.cc 6'

def create_d(s):
    fields = re.split(r'[. ]', s)
    v = int(fields.pop())

    d = { field:v for field in fields }
    return d

mydict = create_d(s1)
print mydict

mydict.update(create_d(s2))
print mydict

给出：

{'a': 8, 'ee': 8, 'dd': 8, 'bb': 8}
{'a': 8, 'aa': 6, 'bb': 6, 'ee': 8, 'dd': 8, 'cc': 6}

Answer 3

您可以尝试使用简单的代码

str1 = 'aa.bb.cc 6'
str2 = 'aa.bb.cc.dd 8'
dictstr = {}
strdelim = str1.split(' ')
dictstr[tuple(strdelim[0].split('.'))] = strdelim[1]
strdelim = str2.split(' ')
dictstr[tuple(strdelim[0].split('.'))] = strdelim[1]
print dictstr

结果

{('aa', 'bb', 'cc'): '6', ('aa', 'bb', 'cc', 'dd'): '8'}

如何将昏迷分隔的字符串转换为python字典？

3 个答案: