Question

我是python的新手，正在寻找一种优雅的方式来完成以下工作。

我有一个字符串说：

s = u'(鞋子="AAA", last = "BBB", abcd)'

我正在考虑一个函数，它可以解析上面的字符串并以下列格式给出输出。

arg, kwarg = foo(s)

def foo():
    # the implementation I dont know.

我如何在python中执行此操作？

Answer 1

解析遵循某些语法规则的字符串的一种很好的方法是第三方pyparsing库。这是非常通用的，缺少允许用户输入的正式语法定义：

#coding:utf8
from pyparsing import *

# Names for symbols
_lparen = Suppress('(')
_rparen = Suppress(')')
_quote = Suppress('"')
_eq = Suppress('=')

# Parsing grammar definition
data = (_lparen +                        # left parenthesis
        delimitedList(                   # Zero or more comma-separated items
            Group(                       #   Group the contained unsuppressed tokens in a list
                Regex(u'[^=,)\s]+') +    #     Grab everything up to an equal, comma, endparen or whitespace as a token
                Optional(                #     Optionally...
                    _eq +                #       match an = 
                    _quote +             #       a quote
                    Regex(u'[^"]*') +    #       Grab everything up to another quote as a token
                    _quote)              #       a quote
                )                        #   EndGroup - will have one or two items.
            ) +                          # EndList
        _rparen)                         # right parenthesis

def process(s):
    items = data.parseString(s).asList()
    args = [i[0] for i in items if len(i) == 1]
    kwargs = {i[0]:i[1] for i in items if len(i) == 2}
    return args,kwargs

s = u'(鞋子="AAA", last = "BBB", abcd)'
args,kwargs = process(s)
for a in args:
    print a
for k,v in kwargs.items():
    print k,v

输出：

abcd
鞋子 AAA
last BBB

Answer 2

def foo(s):
    data={"kwargs":[],"args":[]}
    for item in s:
      if "=" in item: data['kwargs'].append(item)
      else: data['args'].append(item)
    return data

s = u'(鞋子="AAA", last = "BBB", abcd)'
s = s[1:-1] # get rid of the parenthesis
print foo(s)

解析字符串以识别kwargs和args

2 个答案: