使用解析将**运算符更改为幂函数？

我的要求是将**运算符更改为幂函数

例如

1.Input -"B**2"
Output - power(B,2)
2."B**2&&T**2*X"
Output - power(B,2)

我写了以下正则表达式来解决这个问题

   rx=r"([a-zA-Z0-9]+)\*\*([a-zA-Z0-9()]+)"
        result = regex.sub(rx, r"power(\1,\2)", expression, 0, regex.IGNORECASE | regex.MULTILINE)

但是上面的代码成功转换了类似于示例1和示例2的表达式,但是未能转换类似于(a 1)** 2或((a b)* c)** 2之类的表达式.我意识到正则表达式不是处理此类情况的最佳方法.代替该解析将是处理该问题的最佳方法.
我对python有点陌生.请指导我如何解决此问题.

解决方法:

这听起来很熟悉,我认为我在pyparsing邮件列表中处理了类似的问题,但目前找不到.但是尝试这样的事情：

from pyparsing import *

# define some basic operand expressions
number = Regex(r'\d+(\.\d*)?([Ee][+-]?\d+)?')
ident = Word(alphas+'_', alphanums+'_')

# forward declare our overall expression, since a slice could 
# contain an arithmetic expression
expr = Forward()
slice_ref = '[' + expr + ']'

# define our arithmetic operand
operand = number | Combine(ident + Optional(slice_ref))

# parse actions to convert parsed items
def convert_to_pow(tokens):
    tmp = tokens[0][:]
    ret = tmp.pop(-1)
    tmp.pop(-1)
    while tmp:
        base = tmp.pop(-1)
        # hack to handle '**' precedence ahead of '-'
        if base.startswith('-'):
            ret = '-pow(%s,%s)' % (base[1:], ret)
        else:
            ret = 'pow(%s,%s)' % (base, ret)
        if tmp:
            tmp.pop(-1)
    return ret

def unary_as_is(tokens):
    return '(%s)' % ''.join(tokens[0])

def as_is(tokens):
    return '%s' % ''.join(tokens[0])

# simplest infixNotation - may need to add a few more operators, but start with this for now
arith_expr = infixNotation( operand,
    [
    ('-', 1, opAssoc.RIGHT, as_is),
    ('**', 2, opAssoc.LEFT, convert_to_pow),
    ('-', 1, opAssoc.RIGHT, unary_as_is),
    (oneOf("* /"), 2, opAssoc.LEFT, as_is),
    (oneOf("+ -"), 2, opAssoc.LEFT, as_is),
    ])

# now assign into forward-declared expr
expr <<= arith_expr.setParseAction(lambda t: '(%s)' % ''.join(t))

assert "2**3" == expr
assert "2**-3" == expr

# test it out
tests = [
    "2**3",
    "2**-3",
    "2**3**x5",
    "2**-3**x6[-1]",
    "2**-3**x5+1",
    "(a+1)**2",
    "((a+b)*c)**2",
    "B**2",
    "-B**2",
    "(-B)**2",
    "B**-2",
    "B**(-2)",
    "B**2&&T**2*X",
    ]

x5 = 2
a,b,c = 1,2,3
B = 4
x6 = [3,2]
for test in tests:
    print test
    xform = expr.transformString(test)[1:-1]
    print xform
    print '**' not in xform and eval(xform) == eval(test)
    print

打印：

2**3
pow(2,3)
True

2**-3
pow(2,-3)
True

2**3**x5
pow(2,pow(3,x5))
True

2**-3**x6[-1]
pow(2,-pow(3,x6[((-1))]))
True

2**-3**x5+1
pow(2,-pow(3,x5))+1
True

(a+1)**2
pow((a+1),2)
True

((a+b)*c)**2
pow(((a+b)*c),2)
True

B**2
pow(B,2)
True

-B**2
(-pow(B,2))
True

(-B)**2
pow(((-B)),2)
True

B**-2
pow(B,-2)
True

B**(-2)
pow(B,((-2)))
True

B**2&&T**2*X
pow(B,2))&&(pow(T,2)*X
Traceback (most recent call last):
  File "convert_to_pow.py", line 85, in <module>
    print '**' not in xform and eval(xform) == eval(test)
  File "<string>", line 1
    pow(B,2))&&(pow(T,2)*X
            ^
SyntaxError: invalid syntax

如果要转换的代码中有更多的极端情况,则可能只需要对操作数表达式进行一些调整,或者在infixNotation表达式中添加更多的运算符(例如&&&).

(请注意,您必须像写a **(b ** c)那样转换a ** b ** c,因为链式幂运算是从右到左而不是从左到右求值的.)

编辑：

引入了可正确处理“-”和“ **”之间优先级的hack.扩展测试以实际评估字符串之前/之后.现在看起来更加牢固.

标签：pyparsing,python,python-2-7,regex,abstract-syntax-tree
来源： https://codeday.me/bug/20191011/1895995.html