Question

我有一个格式为：

的python字符串

str = "name: srek age :24 description: blah blah"

有没有办法将其转换为类似

的字典

{'name': 'srek', 'age': '24', 'description': 'blah blah'}

其中每个条目都是从字符串中获取的（键，值）对。我尝试将字符串拆分为

列表

str.split()

然后手动删除:，检查每个标记名称，添加到字典中。这种方法的缺点是：这种方法很讨厌，我必须为每一对手动删除:，如果字符串中有多个单词'value'（例如，blah blah为description }），每个单词将是列表中的单独条目，这是不可取的。是否有任何Pythonic方式获取字典（使用python 2.7）？

Answer 1

>>> r = "name: srek age :24 description: blah blah"
>>> import re
>>> regex = re.compile(r"\b(\w+)\s*:\s*([^:]*)(?=\s+\w+\s*:|$)")
>>> d = dict(regex.findall(r))
>>> d
{'age': '24', 'name': 'srek', 'description': 'blah blah'}

<强>解释

\b           # Start at a word boundary
(\w+)        # Match and capture a single word (1+ alnum characters)
\s*:\s*      # Match a colon, optionally surrounded by whitespace
([^:]*)      # Match any number of non-colon characters
(?=          # Make sure that we stop when the following can be matched:
 \s+\w+\s*:  #  the next dictionary key
|            # or
 $           #  the end of the string
)            # End of lookahead

Answer 2

没有re：

r = "name: srek age :24 description: blah blah cat: dog stack:overflow"
lis=r.split(':')
dic={}
try :
 for i,x in enumerate(reversed(lis)):
    i+=1
    slast=lis[-(i+1)]
    slast=slast.split()
    dic[slast[-1]]=x

    lis[-(i+1)]=" ".join(slast[:-1])
except IndexError:pass    
print(dic)

{'age': '24', 'description': 'blah blah', 'stack': 'overflow', 'name': 'srek', 'cat': 'dog'}

Answer 3

以原始顺序显示字典的Aswini程序的其他变体

import os
import shutil
mystr = "name: srek age :24 description: blah blah cat: dog stack:overflow"
mlist = mystr.split(':')
dict = {}
list1 = []
list2 = []
try:
 for i,x in enumerate(reversed(mlist)):
    i = i + 1
    slast = mlist[-(i+1)]
    cut = slast.split()
    cut2 = cut[-1]
    list1.insert(i,cut2)
    list2.insert(i,x)
    dict.update({cut2:x})
    mlist[-(i+1)] = " ".join(cut[0:-1])
except:
 pass   

rlist1 = list1[::-1]
rlist2= list2[::-1]

print zip(rlist1, rlist2)

输出

[('name', 'srek'), ('age', '24'), ('description', 'blah blah'), ('cat', 'dog'), ('stack', 'overflow')]

从包含键值对的字符串中获取python字典

3 个答案: