将仅返回令牌的解析动作附加到解析器,该解析器在ParseResults列表中生成单个组,从而生成一个双重嵌套的字典(result3)。
我将此问题发布在pyparsing GitHub issues page上,将其视为不一致的地方。但是也许不是,我只是不理解ParseResults对象。在下面,我稍微扩展了我的观察范围。
from __future__ import print_function
from pyparsing import *
name = Word(alphas)('name')
score = Word(nums + '.')('score')
nameScore = Group(name + score)
line1 = nameScore('Rider')
line2 = nameScore('Rider') + nameScore('Bull')
result1 = line1.parseString('Mauney 46.5')
result2 = line2.parseString('Mauney 46.5 Asteroid 46')
print("### before parse action is added ###")
print("result1.dump():\n" + result1.dump() + "\n")
print("result2.dump():\n" + result2.dump() + "\n")
line1.setParseAction(lambda t: t)
line2.setParseAction(lambda t: t)
result3 = line1.parseString('Mauney 46.5')
result4 = line2.parseString('Mauney 46.5 Asteroid 46')
print("### after parse action was added ###")
print("result3.dump():\n" + result3.dump() + "\n")
print("result4.dump():\n" + result4.dump() + "\n")
执行上面的结果如下。
### before parse action is added ###
result1.dump():
[['Mauney', '46.5']]
- Rider: ['Mauney', '46.5']
- name: 'Mauney'
- score: '46.5'
result2.dump():
[['Mauney', '46.5'], ['Asteroid', '46']]
- Bull: ['Asteroid', '46']
- name: 'Asteroid'
- score: '46'
- Rider: ['Mauney', '46.5']
- name: 'Mauney'
- score: '46.5'
### after parse action was added ###
result3.dump():
[['Mauney', '46.5']]
- Rider: [['Mauney', '46.5']] <=================
- Rider: ['Mauney', '46.5']
- name: 'Mauney'
- score: '46.5'
result4.dump():
[['Mauney', '46.5'], ['Asteroid', '46']]
- Bull: ['Asteroid', '46']
- name: 'Asteroid'
- score: '46'
- Rider: ['Mauney', '46.5']
- name: 'Mauney'
- score: '46.5'
最外面的ParseResults对象似乎是其自己的字典的一部分。
这是预期的行为吗?如果是,为什么会发生?
一些进一步的观察结果。
如果“ ParseResults”列表包含两组,则不会出现双重嵌套(结果2)。
如果对line1
的解析操作更改为
line1.setParseAction(lambda t: [t[0]])
获得所需的结果:
result1.dump():
[['Mauney', '46.5']]
- Rider: ['Mauney', '46.5']
- name: 'Mauney'
- score: '46.5'