我今晚刚开始使用pyparsing
,我已经构建了一个复杂的语法,它描述了我正在非常有效地工作的一些来源。它非常简单而且非常强大。但是,我在使用ParsedResults
时遇到了一些麻烦。我需要能够按照它们被找到的顺序迭代嵌套的标记,并且我发现它有点令人沮丧。我已将问题抽象为一个简单的案例:
import pyparsing as pp
word = pp.Word(pp.alphas + ',.')('word*')
direct_speech = pp.Suppress('“') + pp.Group(pp.OneOrMore(word))('direct_speech*') + pp.Suppress('”')
sentence = pp.Group(pp.OneOrMore(word | direct_speech))('sentence')
test_string = 'Lorem ipsum “dolor sit” amet, consectetur.'
r = sentence.parseString(test_string)
print r.asXML('div')
print ''
for name, item in r.sentence.items():
print name, item
print ''
for item in r.sentence:
print item.getName(), item.asList()
据我所知,这应该有用吗?这是输出:
<div>
<sentence>
<word>Lorem</word>
<word>ipsum</word>
<direct_speech>
<word>dolor</word>
<word>sit</word>
</direct_speech>
<word>amet,</word>
<word>consectetur.</word>
</sentence>
</div>
word ['Lorem', 'ipsum', 'amet,', 'consectetur.']
direct_speech [['dolor', 'sit']]
Traceback (most recent call last):
File "./test.py", line 27, in <module>
print item.getName(), item.asList()
AttributeError: 'str' object has no attribute 'getName'
XML输出似乎表明字符串的解析完全符合我的意愿,但我无法迭代句子,例如重构它。
有办法做我需要的吗?
谢谢!
修改
我一直在用这个:
for item in r.sentence:
if isinstance(item, basestring):
print item
else:
print item.getName(), item
但它对我没那么大帮助,因为我无法区分不同类型的字符串。这是一个稍微扩展的例子:
word = pp.Word(pp.alphas + ',.')('word*')
number = pp.Word(pp.nums + ',.')('number*')
direct_speech = pp.Suppress('“') + pp.Group(pp.OneOrMore(word | number))('direct_speech*') + pp.Suppress('”')
sentence = pp.Group(pp.OneOrMore(word | number | direct_speech))('sentence')
test_string = 'Lorem 14 ipsum “dolor 22 sit” amet, consectetur.'
r = sentence.parseString(test_string)
for i, item in enumerate(r.sentence):
if isinstance(item, basestring):
print i, item
else:
print i, item.getName(), item
输出是:
0 Lorem
1 14
2 ipsum
3 word ['dolor', '22', 'sit']
4 amet,
5 consectetur.
没太大帮助。我无法区分word
和number
,而direct_speech
元素标记为word
?!
我显然错过了一些东西。我想做的就是:
for item in r.sentence:
if (item is a number):
do something
elif (item is a word):
do something else
etc. ...
我应该以不同的方式接近这个吗?
答案 0 :(得分:5)
r.sentence
包含字符串和ParseResults的混合,只有ParseResults支持getName()
。您是否尝试过迭代r.sentence
?如果我使用asList()打印出来,我得到:
['Lorem', 'ipsum', ['dolor', 'sit'], 'amet,', 'consectetur.']
或者这个片段:
for item in r.sentence:
print type(item),item.asList() if isinstance(item,pp.ParseResults) else item
给出:
<type 'str'> Lorem
<type 'str'> ipsum
<class 'pyparsing.ParseResults'> ['dolor', 'sit']
<type 'str'> amet,
<type 'str'> consectetur.
我不确定我是否回答了你的问题,但这是否会说明下一步该怎么做?
(欢迎来到Pyparsing)
答案 1 :(得分:1)
.asXML()
并解析生成的XML。这是我的例子:
import pyparsing as pp
word = pp.Word(pp.alphas + ',.')('word*')
number = pp.Word(pp.nums + ',.')('number*')
direct_speech = pp.Suppress('“') + pp.Group(pp.OneOrMore(word | number))('direct_speech*') + pp.Suppress('”')
sentence = pp.Group(pp.OneOrMore(word | number | direct_speech))('sentence')
test_string = 'Lorem 14 ipsum “dolor 22 sit” amet, consectetur.'
r = sentence.parseString(test_string)
from lxml import etree
xml = etree.fromstring(r.sentence.asXML('sentence'))
for el in xml:
if len(el):
print el.tag
for sub_el in el:
print ' ', sub_el.tag, ':', sub_el.text
else:
print el.tag, ':', el.text
输出:
word : Lorem
number : 14
word : ipsum
direct_speech
word : dolor
number : 22
word : sit
word : amet,
word : consectetur.
似乎在房子周围很长,但似乎没有更好的方法。