Question

我想用pyparsing解析一些单词和一些数字。简单吧。

from pyparsing import *

A = Word(nums).setResultsName('A')
B = Word(alphas).setResultsName('B')
expr = OneOrMore(A | B)

result = expr.parseString("123 abc 456 7 d")
print result

上面的代码打印['123', 'abc', '456', '7', 'd']。一切顺利。现在我想用这些解析的值做一些工作。对于此任务，我需要知道它们是否与A或B匹配。有没有办法区分这两者。

我在一些研究后发现的唯一内容是items类的ParseResults方法。但它只返回[('A', '7'), ('B', 'd')]，只返回最后两个匹配。

我的计划/目标如下：

for elem in result:
    if elem.is_of_type('A'):
        # do stuff
    elif elem.is_of_type('B'):
        # do something else

如何区分A和B？

Answer 1

使用getName（）做得很好。您还可以使用标记显式装饰返回的标记，指示所做的匹配：

def makeDecoratingParseAction(marker):
    def parse_action_impl(s,l,t):
        return (marker, t[0])
    return parse_action_impl

A = Word(nums).setParseAction(makeDecoratingParseAction("A"))
B = Word(alphas).setParseAction(makeDecoratingParseAction("B"))
expr = OneOrMore(A | B)

result = expr.parseString("123 abc 456 7 d")
print result.asList()

给出：

[('A', '123'), ('B', 'abc'), ('A', '456'), ('A', '7'), ('B', 'd')]

现在，您可以遍历返回的元组，每个元组都标有相应的标记。

您可以更进一步，使用类来捕获类型和特定于类型的解析后逻辑，然后将该类作为表达式的解析操作传递。这将在返回的ParseResults中创建类的实例，然后您可以使用某种exec或doIt方法直接执行：

class ResultsHandler(object):
    """Define base class to initialize location and tokens.
       Call subclass-specific post_init() if one is defined."""
    def __init__(self, s,locn,tokens):
        self.locn = locn
        self.tokens = tokens
        if hasattr(self, "post_init"):
            self.post_init()

class AHandler(ResultsHandler):
    """Handler for A expressions, which contain a numeric string."""
    def post_init(self):
        self.int_value = int(self.tokens[0])
        self.odd_even = ("EVEN","ODD")[self.int_value % 2]
    def doIt(self):
        print "An A-Type was found at %d with value %d, which is an %s number" % (
                self.locn, self.int_value, self.odd_even)

class BHandler(ResultsHandler):
    """Handler for B expressions, which contain an alphabetic string."""
    def post_init(self):
        self.string = self.tokens[0]
        self.vowels_count = sum(self.string.lower().count(c) for c in "aeiou")
    def doIt(self):
        print "A B-Type was found at %d with value %s, and contains %d vowels" % (
                self.locn, self.string, self.vowels_count)


# pass expression-specific handler classes as parse actions
A = Word(nums).setParseAction(AHandler)
B = Word(alphas).setParseAction(BHandler)
expr = OneOrMore(A | B)

# parse string and run handlers
result = expr.parseString("123 abc 456 7 d")
for handler in result:
    handler.doIt()

打印：

An A-Type was found at 0 with value 123, which is an ODD number
A B-Type was found at 4 with value abc, and contains 1 vowels
An A-Type was found at 8 with value 456, which is an EVEN number
An A-Type was found at 12 with value 7, which is an ODD number
A B-Type was found at 14 with value d, and contains 0 vowels

Answer 2

我不完全确定原因，但在您的.setResultsName()来电中，您需要指定listAllMatches=True（默认为False）。完成后，您可以循环遍历result并通过检查result的相应子事件中的成员身份来检查每个令牌是否与给定表达式匹配。

from pyparsing import *

#                                    ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
A = Word(nums  ).setResultsName('A', listAllMatches=True)
B = Word(alphas).setResultsName('B', listAllMatches=True)
expr = OneOrMore(A | B)

result = expr.parseString("123 abc 456 7 d")

for elem in result:
    if elem in list(result['A']):
        print(elem, 'is in A')
    elif elem in list(result['B']):
        print(elem, 'is in B')

打印：

123 is in A
abc is in B
456 is in A
7 is in A
d is in B

这是kludgey，我不确定这是否是规范正确的做法，但它似乎有效。

Answer 3

我自己找到了解决方案。 ParseResults类有getName()方法。但只是迭代result并打印getName()会引发错误，因为ParseResults对象会产生字符串。一种解决方案是将A和B放入Groups。不确定这是否是最佳方式，但它的效果非常好。

from pyparsing import *

A = Group(Word(nums)).setResultsName('A', listAllMatches=True)
B = Group(Word(alphas)).setResultsName('B', listAllMatches=True)
expr = OneOrMore(A | B)

result = expr.parseString("123 abc 456 7 d")

for i in  result:
    if i.getName() == 'A':
       print i[0], 'is a number'
    elif i.getName() == 'B':
       print i[0], 'is a string'

在pyparsing中区分匹配

3 个答案: