将字符串的部分解析为字典？

Question

我不熟悉字符串解析库;并想从去：

'foo=5 z v xz er bar=" hel o" c z a == "hi" b = "who"' 

对此解析词典：

{'foo':5, 'bar': ' hel o', 'a': 'hi', b: 'who'} 

但我不知道从哪里开始。你能否给我一些处理这种转换的建议？

Answer 1

您可以使用正则表达式。见python's documentation on regex或tutorial's point tutorial。

像这样的东西可以工作：

import re 

regex = re.compile(r"(\w+ ?=+ ?\d+|\w+ ?=+ ?\"(?: *\w*)*\")") 

#your example string: 
s = 'foo=5 z v xz er bar=" hel o" c z a == "hi" b = "who"' 

matches = regex.findall(s) 

dict1 = {} 
for m in matches: 
    elems = m.split("=") 
    #elems[0] = key 
    #elems[len(elems)-1] = value, to account for the case of multiple ='s 

    try: 
     #see if the element is a number 
     dict1[str(elems[0])] = int(elems[len(elems) - 1]) 

    except: 
     #if type casting didn't work, just store it as a string 
     dict1[str(elems[0])] = elems[len(elems) - 1] 

这里的正则表达式分解：

(\w+ ?=+ ?\d+|\w+ ?=+ ?\"(?: *\w*)*\") 

\w+意味着一个或多个字母数字字符。

\d+表示一位或多位数字。

(?:regex)*表示匹配正则表达式的0个或更多副本而不分配组＃。

(regex1|regex2)表示查找与regex1匹配的字符串或与regex2匹配的字符串。

\"是引号的转义序列。

=+意味着匹配一个或多个“=”标志

_?意味着匹配0或1位（假装“_”为空格）

Answer 2

Pyparsing是一个解析库，让你建立你的一次匹配表达式。

from pyparsing import Word, alphas, alphanums, nums, oneOf, quotedString, removeQuotes 

identifier = Word(alphas, alphanums) 
integer = Word(nums).setParseAction(lambda t: int(t[0])) 
value = integer | quotedString.setParseAction(removeQuotes) 

# equals could be '==' or '=' 
# (suppress it so it does not get included in the resulting tokens) 
EQ = oneOf("= ==").suppress() 

# define the expression for an assignment 
assign = identifier + EQ + value 

下面是代码应用此解析器

# search sample string for matching assignments 
s = 'foo=5 z v xz er bar=" hel o" c z a == "hi" b = "who"' 
assignments = assign.searchString(s) 
dd = {} 
for k,v in assignments: 
    dd[k] = v 

# or more simply 
#dd = dict(assignments.asList()) 

print dd 

给出：

{'a': 'hi', 'b': 'who', 'foo': 5, 'bar': ' hel o'}