删除尾随空格

Question

默认情况下，pyparsing只删除前导空格。在

所以分析这个

'between ( 1, map(  v7 , 2  ) )'

给了我

^{pr2}$

而不是

['between', [['1'], ['map', [['v7'], ['2']]]]]

但是如何使用pyparsing删除后面的空白呢？在

代码是：

from pyparsing import Forward, Word, alphas, alphanums, nums, Literal, Group, delimitedList, Optional
lparen = Literal("(").suppress()
rparen = Literal(")").suppress()
name = Optional(Word(alphanums + '_ ', alphanums + '_ '))
functor = Word(alphas, alphanums)
integer  = Word( nums )
expression = Forward()
arg =  Group(expression) |name | integer 
args = delimitedList(arg)
expression << ((functor + lparen + args + rparen) | name)

print( expression.parseString("between ( 1, map(  v7 , 2  ) )"))
>>> ['between', ['1'], ['map', ['v7 '], ['2  ']]]

pyparsing。版本=“2.0.1”和python3。在

此时，我在解析之前和之后使用额外的外部函数来处理和修复数据。在

Answer 1

发布的代码有两次出现alphanums + '_ '，将它们都改为alphanums + '_'（删除空格）可以实现您所描述的内容。在

from pyparsing import Forward, Word, alphas, alphanums, nums, Literal, Group, delimitedList, Optional

lparen = Literal("(").suppress()
rparen = Literal(")").suppress()
name = Optional(Word(alphanums + '_', alphanums + '_'))
functor = Word(alphas, alphanums)
integer = Word(nums)
expression = Forward()
arg = Group(expression) | name | integer 
args = delimitedList(arg)
expression << ((functor + lparen + args + rparen) | name)

print( expression.parseString("between ( 1, map(  v7 , 2  ) )"))

印刷品：

^{pr2}$

我不确定你是否能从评论中得出这个结论，但看起来Paul McGuire就是这样描述的。也就是说，pyparsing只在匹配的内容中包含空格，因为alphanums + '_ '在匹配的内容中包含一个空格。在