トークンのフラットリストを取得し、左再帰的に解析されたかのようにそれらをネストする解析アクションの例を次に示します。
from pyparsing import *
# parse action -maker
def makeLRlike(numterms):
if numterms is None:
# None operator can only by binary op
initlen = 2
incr = 1
else:
initlen = {0:1,1:2,2:3,3:5}[numterms]
incr = {0:1,1:1,2:2,3:4}[numterms]
# define parse action for this number of terms,
# to convert flat list of tokens into nested list
def pa(s,l,t):
t = t[0]
if len(t) > initlen:
ret = ParseResults(t[:initlen])
i = initlen
while i < len(t):
ret = ParseResults([ret] + t[i:i+incr])
i += incr
return ParseResults([ret])
return pa
# setup a simple grammar for 4-function arithmetic
varname = oneOf(list(alphas))
integer = Word(nums)
operand = integer | varname
# ordinary opPrec definition
arith1 = operatorPrecedence(operand,
[
(None, 2, opAssoc.LEFT),
(oneOf("* /"), 2, opAssoc.LEFT),
(oneOf("+ -"), 2, opAssoc.LEFT),
])
# opPrec definition with parseAction makeLRlike
arith2 = operatorPrecedence(operand,
[
(None, 2, opAssoc.LEFT, makeLRlike(None)),
(oneOf("* /"), 2, opAssoc.LEFT, makeLRlike(2)),
(oneOf("+ -"), 2, opAssoc.LEFT, makeLRlike(2)),
])
# parse a few test strings, using both parsers
for arith in (arith1, arith2):
print arith.parseString("A+B+C+D+E")[0]
print arith.parseString("A+B+C*D+E")[0]
print arith.parseString("12AX+34BY+C*5DZ+E")[0]
プリント:
(正常)
['A', '+', 'B', '+', 'C', '+', 'D', '+', 'E']
['A', '+', 'B', '+', ['C', '*', 'D'], '+', 'E']
[['12', 'A', 'X'], '+', ['34', 'B', 'Y'], '+', ['C', '*', ['5', 'D', 'Z']], '+', 'E']
(LRのような)
[[[['A', '+', 'B'], '+', 'C'], '+', 'D'], '+', 'E']
[[['A', '+', 'B'], '+', ['C', '*', 'D']], '+', 'E']
[[[[['12', 'A'], 'X'], '+', [['34', 'B'], 'Y']], '+', ['C', '*', [['5', 'D'], 'Z']]], '+', 'E']