Re: [Pyparsing] Reading n words

[Paul M. <pt...@au...>]

Andreas -

Pyparsing includes a built-in helper method, called countedArray.  Here is
how you use it:

from pyparsing import *

counted_list_of_words = countedArray(Word(alphas))
tests = """\
1 foo
2 foo baz
3 foo bar baz""".splitlines()

for t in tests:
    print counted_list_of_words.parseString(t)[0]

Prints:
['foo']
['foo', 'baz']
['foo', 'bar', 'baz']


countedArray(expr) matches Word(nums) + expr + expr + ... ("n" times, where
n is the leading integer).  Look at the pyparsing source code to see how
countedArray uses a Forward expression that gets its content defined within
a parse action attached to the leading integer.

-- Paul
 

-----Original Message-----
From: pyp...@li...
[mailto:pyp...@li...] On Behalf Of Andreas
Matthias
Sent: Wednesday, September 03, 2008 4:16 PM
To: pyp...@li...
Subject: [Pyparsing] Reading n words

Hello, 

I'm trying to parse a string like '3 foo bar baz', where the number in the
string specifies the number of the following words.
Is there a more elegant way to achieve this than with the following code? 

Ciao
Andreas



from pyparsing import *

def foo(st):
    p1 = Word(nums)
    n = p1.parseString(st)[0]
    p2 = Literal(n) + Word(alphas)*int(n)
    print p2.parseString(st)

foo('1 foo')
foo('2 foo baz')
foo('3 foo bar baz')


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