问题描述
这是我的任务:
这是我的实现:
#Todo - construct a CFG for simple arithmetic expressions
all_numbers = "'zero' | 'one' | 'two' | 'three' | 'four' | 'five' " \
+ "| 'six' | 'seven' | 'eight' | 'nine' | 'ten' | 'eleven' " \
+ "| 'twelve' | 'thirteen' | 'fourteen' | 'fifteen' " \
+ "| 'sixteen' | 'seventeen' | 'eighteen' | 'nineteen' | 'twenty'"
arithmetic_grammar = nltk.CFG.fromstring("""
S -> operand operator operand | S operator operand | S operator S
operand -> """+all_numbers+"""
operator -> 'plus' | 'minus' | 'times' | 'divided by'
""")
print(arithmetic_grammar)
#Todo
# "fifteen minus five"
tree2 = nltk.Tree.fromstring("(S (operand fifteen) (operator minus) (operand fifteen))")
# "four divided by two plus one"
tree3 = nltk.Tree.fromstring("(S (S (operand four) (operator divided by) (operand two)) (operator plus) (operand one))")
def validate(tree,grammar):
return functools.reduce(lambda accum,production:
accum and production in grammar.productions(),tree.productions())
使用它来指示我的 CFG 是否正确编写,我得到:
print(validate(tree2,arithmetic_grammar))
print(validate(tree3,arithmetic_grammar))
如何通过保留最初的“除以”表达式来表示除法来解决这个问题? 我可以将其更改为“divided_by”,但我希望它保持原样。 提前致谢。
解决方法
我发现一个有效的解决方案是:
#TODO - construct a CFG for simple arithmetic expressions
all_numbers = "'zero' | 'one' | 'two' | 'three' | 'four' | 'five' " \
+ "| 'six' | 'seven' | 'eight' | 'nine' | 'ten' | 'eleven' " \
+ "| 'twelve' | 'thirteen' | 'fourteen' | 'fifteen' " \
+ "| 'sixteen' | 'seventeen' | 'eighteen' | 'nineteen' | 'twenty'"
arithmetic_grammar = nltk.CFG.fromstring("""
S -> operand operator operand | S operator operand | S operator S
operand -> """+all_numbers+"""
operator -> 'plus' | 'minus' | 'times' | 'divided' b
b -> 'by'
""")
print(arithmetic_grammar)