这是一个示例解析操作,它将获取标记的平面列表并将它们嵌套,就像左递归解析一样:
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]
印刷:
(普通的)
['A', '+', 'B', '+', 'C', '+', 'D', '+', 'E']
['A', '+', 'B', '+', ['C', '*', 'D'], '+', 'E']
[['12', 'A', 'X'], '+', ['34', 'B', 'Y'], '+', ['C', '*', ['5', 'D', 'Z']], '+', 'E']
(类似LR)
[[[['A', '+', 'B'], '+', 'C'], '+', 'D'], '+', 'E']
[[['A', '+', 'B'], '+', ['C', '*', 'D']], '+', 'E']
[[[[['12', 'A'], 'X'], '+', [['34', 'B'], 'Y']], '+', ['C', '*', [['5', 'D'], 'Z']]], '+', 'E']