序言其實詞法分析器在Python

Question

我感興趣的是能夠閱讀和使用python

爲了便於操縱序言「事實」，怎麼可能用Python語言編寫一個詞法分析器，可以讀出一組事實一個文本文件？

事實的集合可能看起來像：

track(1, 2.0, 4000, 3, 300). 
track(2, 1.0, 9000, 5, 500). 
track(3, 7.0, 9000, 2, 200). 
... 

Answer 1

我喜歡用pyparsing模塊創建這種性質在Python的解析器。 Pyparsing是一個解析器組合器庫;您通過將其他解析器組合在一起來構造解析器。下面是使用pyparsing，它將分析輸入數據爲例解析器（這將無法解析任何序言的事實，但我認爲這將是一個很好的起點）：

import pyparsing as pp 
#relationship will refer to 'track' in all of your examples 
relationship = pp.Word(pp.alphas).setResultsName('relationship') 

number = pp.Word(pp.nums + '.') 
variable = pp.Word(pp.alphas) 
# an argument to a relationship can be either a number or a variable 
argument = number | variable 

# arguments are a delimited list of 'argument' surrounded by parenthesis 
arguments= (pp.Suppress('(') + pp.delimitedList(argument) + 
      pp.Suppress(')')).setResultsName('arguments') 

# a fact is composed of a relationship and it's arguments 
# (I'm aware it's actually more complicated than this 
# it's just a simplifying assumption) 
fact = (relationship + arguments).setResultsName('facts', listAllMatches=True) 

# a sentence is a fact plus a period 
sentence = fact + pp.Suppress('.') 

# self explanatory 
prolog_sentences = pp.OneOrMore(sentence) 

現在我們有一個解析器您在prolog_sentences變量中的輸入。這裏是解析器的一個測試：

test="""track(1, 2.0, 4000, 3, 300). 
track(2, 1.0, 9000, 5, 500). 
track(3, 7.0, 9000, 2, 200).""" 

result = prolog_sentences.parseString(test) 

print result['facts'][0]['arguments'].asList() 
# outputs ['1', '2.0', '4000', '3', '300'] 

print result['facts'][1]['relationship'] 
# outputs 'track' 

print result['facts'][2]['arguments'][1] 
# outputs ['7.0']