python - 从 parseAction 返回未更改的标记会生成双重嵌套字典

Question

将只返回标记的解析操作附加到解析器，解析器会在 ParseResults 列表中生成单个组，从而生成一个双重嵌套字典 (result3)。

我在pyparsing GitHub 问题页面上发布了这个问题，将其视为不一致。但也许不是，我只是不理解 ParseResults 对象。下面我稍微扩展了我的观察。

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")

执行上述结果如下。

### 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'

看起来最外面的 ParseResults 对象是它自己的字典的一部分。

这是预期的行为，如果是，为什么会发生？

一些进一步的观察。

如果 ParseResults 列表包含两个组，则没有双重嵌套 (result2)。

如果解析操作line1更改为

line1.setParseAction(lambda t: [t[0]])

得到了想要的结果：

result1.dump():
[['Mauney', '46.5']]
- Rider: ['Mauney', '46.5']
  - name: 'Mauney'
  - score: '46.5'

标签： pythonparsingpyparsing