[英]pyparsing - How to parse string with comparison operators?
因此,我有一個NumericStringParser
類(從此處提取),定義如下:
from __future__ import division
from pyparsing import Literal, CaselessLiteral, Word, Combine, Group, Optional, ZeroOrMore, Forward, nums, alphas, oneOf, ParseException
import math
import operator
class NumericStringParser(object):
def __push_first__(self, strg, loc, toks):
self.exprStack.append(toks[0])
def __push_minus__(self, strg, loc, toks):
if toks and toks[0] == "-":
self.exprStack.append("unary -")
def __init__(self):
point = Literal(".")
e = CaselessLiteral("E")
fnumber = Combine(Word("+-" + nums, nums) +
Optional(point + Optional(Word(nums))) +
Optional(e + Word("+-" + nums, nums)))
ident = Word(alphas, alphas + nums + "_$")
plus = Literal("+")
minus = Literal("-")
mult = Literal("*")
floordiv = Literal("//")
div = Literal("/")
mod = Literal("%")
lpar = Literal("(").suppress()
rpar = Literal(")").suppress()
addop = plus | minus
multop = mult | floordiv | div | mod
expop = Literal("^")
pi = CaselessLiteral("PI")
tau = CaselessLiteral("TAU")
expr = Forward()
atom = ((Optional(oneOf("- +")) +
(ident + lpar + expr + rpar | pi | e | tau | fnumber).setParseAction(self.__push_first__))
| Optional(oneOf("- +")) + Group(lpar + expr + rpar)
).setParseAction(self.__push_minus__)
factor = Forward()
factor << atom + \
ZeroOrMore((expop + factor).setParseAction(self.__push_first__))
term = factor + \
ZeroOrMore((multop + factor).setParseAction(self.__push_first__))
expr << term + \
ZeroOrMore((addop + term).setParseAction(self.__push_first__))
self.bnf = expr
self.opn = {
"+": operator.add,
"-": operator.sub,
"*": operator.mul,
"/": operator.truediv,
"//": operator.floordiv,
"%": operator.mod,
"^": operator.pow,
"=": operator.eq,
"!=": operator.ne,
"<=": operator.le,
">=": operator.ge,
"<": operator.lt,
">": operator.gt
}
self.fn = {
"sin": math.sin,
"cos": math.cos,
"tan": math.tan,
"asin": math.asin,
"acos": math.acos,
"atan": math.atan,
"exp": math.exp,
"abs": abs,
"sqrt": math.sqrt,
"floor": math.floor,
"ceil": math.ceil,
"trunc": math.trunc,
"round": round,
"fact": factorial,
"gamma": math.gamma
}
def __evaluate_stack__(self, s):
op = s.pop()
if op == "unary -":
return -self.__evaluate_stack__(s)
if op in ("+", "-", "*", "//", "/", "^", "%", "!=", "<=", ">=", "<", ">", "="):
op2 = self.__evaluate_stack__(s)
op1 = self.__evaluate_stack__(s)
return self.opn[op](op1, op2)
if op == "PI":
return math.pi
if op == "E":
return math.e
if op == "PHI":
return phi
if op == "TAU":
return math.tau
if op in self.fn:
return self.fn[op](self.__evaluate_stack__(s))
if op[0].isalpha():
raise NameError(f"{op} is not defined.")
return float(op)
我有一個evaluate()
函數,定義如下:
def evaluate(expression, parse_all=True):
nsp = NumericStringParser()
nsp.exprStack = []
try:
nsp.bnf.parseString(expression, parse_all)
except ParseException as error:
raise SyntaxError(error)
return nsp.__evaluate_stack__(nsp.exprStack[:])
evaluate()
是一個函數,它將解析字符串以計算數學運算,例如:
>>> evaluate("5+5")
10
>>> evaluate("5^2+1")
26
問題是它無法計算比較運算符( =
, !=
, <
, >
, <=
, >=
),並且當我嘗試: evaluate("5=5")
,它將引發SyntaxError: Expected end of text (at char 1), (line:1, col:2)
而不是返回True
。 函數如何計算這六個比較運算符?
正如@rici指出的那樣,您已添加了評估部分,但未添加解析部分。
解析器在以下幾行中定義:
factor = atom + \
ZeroOrMore((expop + factor).setParseAction(self.__push_first__))
term = factor + \
ZeroOrMore((multop + factor).setParseAction(self.__push_first__))
expr <<= term + \
ZeroOrMore((addop + term).setParseAction(self.__push_first__))
這些語句的順序很重要,因為它們使解析器識別您在中學數學中學到的運算的優先級。 也就是說,乘冪最高,然后是乘法和除法,然后是加法和減法。
您需要按照相同的模式將關系運算符插入此解析器定義。 添加之后,來自C語言運算符優先級的約定(我找到此參考-https : //www.tutorialspoint.com/cprogramming/c_operators_precedence.htm )為:
relational operations - <=, >=, >, <
equality operations - ==, !=
在您的情況下,您選擇使用'='而不是'==',在這種情況下應該可以。 我建議您使用pyparsing的oneOf
助手來定義這些運算符組,因為它將處理短字符串可能掩蓋較長字符串的情況(例如,在上一篇文章中'/'掩蓋了'//')。
請注意,通過將所有這些操作混合到一個表達式解析器中,您將獲得5 + 2 > 3
。 由於'>'的優先級較低,因此將首先計算5 + 2並給出7,然后將評估7> 3,並且operator.__gt__
將返回1或0。
將此示例擴展到其他運算符的困難是導致我在infixNotation
編寫infixNotation
helper方法的原因。 您可能想看看。
編輯:
您詢問有關使用Literal('<=') | Literal('>=) | etc.
Literal('<=') | Literal('>=) | etc.
Literal('<=') | Literal('>=) | etc.
,正如您所寫的那樣,它將正常工作。 您只需要注意要在較短的運算符之前尋找較長的運算符。 如果您寫Literal('>') | Literal('>=') | ...
Literal('>') | Literal('>=') | ...
Literal('>') | Literal('>=') | ...
然后匹配'> ='將失敗,因為第一個匹配將匹配'>',然后您將剩下'='。 使用oneOf
可以幫助您解決此問題。
要添加其他解析器步驟,只需要對最后一級執行expr <<= ...
步驟。 再次查看語句模式。 更改expr <<= term + etc.
,以arith_expr = term + etc.
,遵循它來添加水平relational_expr
和equality_expr
,然后完成expr <<= equality_expr
。
此模式基於:
factor := atom (^ atom)...
term := factor (mult_op factor)...
arith_expr := term (add_op term)...
relation_expr := arith_expr (relation_op arith_expr)...
equality_expr := relation_expr (equality_op relation_expr)...
嘗試自行轉換為Python / pyparsing。
factor << atom + \
ZeroOrMore((expop + factor).setParseAction(self.__push_first__))
term = factor + \
ZeroOrMore((multop + factor).setParseAction(self.__push_first__))
arith_expr = term + \
ZeroOrMore((addop + term).setParseAction(self.__push_first__))
relational = arith_expr + \
ZeroOrMore((diffop + arith_expr).setParseAction(self.__push_first__))
expr <<= relational + \
ZeroOrMore((compop + relational).setParseAction(self.__push_first__))
所以我測試了,它有效! 非常感謝PaulMcG! :)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.