Shorthand for defining PyParsing field names

Question

I have a few pyparsing tokens defined as follows:

field = Word(alphas + "_").setName("field")

Is there really no shorthand for this?

Furthermore, this does not seem to work, the dictionary returned by expression.parseString() is always an empty one.

Answer 1

You are confusing setName and setResultsName. setName assigns a name to the expression so that exception messages are more meaningful. Compare:

>>> integer1 = Word(nums)
>>> integer1.parseString('x')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "c:\python26\lib\site-packages\pyparsing-1.5.6-py2.6.egg\pyparsing.py", line 1032, in parseString
    raise exc
pyparsing.ParseException: Expected W:(0123...) (at char 0), (line:1, col:1)

and:

>>> integer2 = Word(nums).setName("integer")
>>> integer2.parseString('x')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "c:\python26\lib\site-packages\pyparsing-1.5.6-py2.6.egg\pyparsing.py", line 1032, in parseString
    raise exc
pyparsing.ParseException: Expected integer (at char 0), (line:1, col:1)

setName gives a name to the expression itself.

setResultsName on the other hand gives a name to the parsed data that is returned, like named fields in a regex.

>>> expr = integer.setResultsName('age') + integer.setResultsName('credits')
>>> data = expr.parseString('20 110')
>>> print data.dump()
['20', '110']
- age: 20
- credits: 110

And as @Kimvais has mentioned, there is a shortcut for setResultsName:

>>> expr = integer('age') + integer('credits')

Note also that setResultsName returns a copy of the expression - that is the only way that using the same expression multiple times with different names works.

Answer 2

field = Word(alphas + "_")("field")

seems to work.