pyparsing:词典列表的语法(erlang) [英] pyparsing: grammar for list of Dictionaries (erlang)

问题描述

我正在尝试构建语法来分析带有Erlang标签的元组列表，并将其映射到pyparsing中的Dict.我有字典列表时遇到问题.如果Dict仅包含一个元素，则该语法有效，但是当我添加第二个元素时，现在无法解决该问题.

I'm trying to build a grammar to parse an Erlang tagged tuple list, and map this to a Dict in pyparsing. I'm having problems when I have a list of Dicts. The grammar works if the Dict has just one element, but when I add a second can't work out now to get it to parse.

最新(简化的语法代码(在这种情况下，我删除了不必要的语言部分):

Current (simplified grammar code (I removed the bits of the language not necessary in this case):

#!/usr/bin/env python2.7

from pyparsing import *

# Erlang config file definition:
erlangAtom = Word( alphas + '_')
erlangString = dblQuotedString.setParseAction( removeQuotes )

erlangValue = Forward()
erlangList = Forward()

erlangElements = delimitedList( erlangValue )
erlangCSList = Suppress('[') + erlangElements + Suppress(']')
erlangList <<= Group( erlangCSList )
erlangTaggedTuple = Group( Suppress('{') + erlangAtom + Suppress(',') +
                           erlangValue + Suppress('}') )
erlangDict = Dict( Suppress('[') + delimitedList( erlangTaggedTuple ) + 
                   Suppress(']') )

erlangValue <<= ( erlangAtom | erlangString |
                  erlangTaggedTuple |
                  erlangDict | erlangList )

if __name__ == "__main__":
    working = """
[{foo,"bar"}, {baz, "bar2"}]
"""

    broken = """
[
    [{foo,"bar"}, {baz, "bar2"}],
    [{foo,"bob"}, {baz, "fez"}]
]
"""
    w = erlangValue.parseString(working)
    print w.dump()

    b = erlangValue.parseString(broken)
    print "b[0]:", b[0].dump()
    print "b[1]:", b[1].dump()

这给出了:

[['foo', 'bar'], ['baz', 'bar2']]
- baz: bar2
- foo: bar

b[0]: [['foo', 'bar'], ['baz', 'bar2'], ['foo', 'bob'], ['baz', 'fez']]
- baz: fez
- foo: bob

b[1]:
Traceback (most recent call last):
  File "./erl_testcase.py", line 39, in <module>
    print "b[1]:", b[1].dump()
  File "/Library/Python/2.7/site-packages/pyparsing.py", line 317, in __getitem__
    return self.__toklist[i]
IndexError: list index out of range

即working可以，但是broken不能解析为两个列表.

i.e. working works, but broken doesn't parse as two lists.

有什么想法吗?

编辑:调整了测试用例以更加明确地显示预期的输出.

Tweaked testcase to be more explicit about expected output.

推荐答案

好，所以我以前从未使用过pyparsing，因此，如果我的解决方案没有意义，请原谅.我们开始:

Ok, so I have never worked with pyparsing before, so excuse me if my solution does not make sense. Here we go:

据我了解，您需要的是三个主要结构.您最常犯的错误是对delimitedLists进行分组.它们已经被分组，因此您遇到了双重分组的问题.这是我的定义:

As far as I understand what you need is three main structures. The most common mistake you made was grouping delimitedLists. They are already grouped, so you have an issue of double grouping. Here are my definitions:

对于{a，"b"}:

erlangTaggedTuple = Dict(Group(Suppress('{') + erlangAtom + Suppress(',') + erlangValue + Suppress('}') ))

对于[{a，"b"}，{c，"d"}]:

for [{a,"b"}, {c,"d"}]:

erlangDict = Suppress('[') + delimitedList( erlangTaggedTuple ) + Suppress(']')

其余:

erlangList <<= Suppress('[') + delimitedList( Group(erlangDict|erlangList) ) + Suppress(']')

所以我对您代码的修正是:

So my fix for your code is:

#!/usr/bin/env python2.7

from pyparsing import *

# Erlang config file definition:
erlangAtom = Word( alphas + '_')
erlangString = dblQuotedString.setParseAction( removeQuotes )

erlangValue = Forward()
erlangList = Forward()

erlangTaggedTuple = Dict(Group(Suppress('{') + erlangAtom + Suppress(',') +
                           erlangValue + Suppress('}') ))
erlangDict = Suppress('[') + delimitedList( erlangTaggedTuple ) + Suppress(']') 
erlangList <<= Suppress('[') + delimitedList( Group(erlangDict|erlangList) ) + Suppress(']')

erlangValue <<= ( erlangAtom | erlangString |
                  erlangTaggedTuple |
                  erlangDict| erlangList )

if __name__ == "__main__":
    working = """
[{foo,"bar"}, {baz, "bar2"}]
"""

    broken = """
[
    [{foo,"bar"}, {baz, "bar2"}],
    [{foo,"bob"}, {baz, "fez"}]
]
"""
    w = erlangValue.parseString(working)
    print w.dump()

    b = erlangValue.parseString(broken)
    print "b[0]:", b[0].dump()
    print "b[1]:", b[1].dump()

哪个给出输出:

[['foo', 'bar'], ['baz', 'bar2']]
- baz: bar2
- foo: bar
b[0]: [['foo', 'bar'], ['baz', 'bar2']]
- baz: bar2
- foo: bar
b[1]: [['foo', 'bob'], ['baz', 'fez']]
- baz: fez
- foo: bob

希望有帮助，加油！

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