Tom - 2 years ago 107

Python Question

My requirement is to change ** operator to power function

For example

`1.Input -"B**2"`

Output - power(B,2)

2."B**2&&T**2*X"

Output - power(B,2)

I have wrote following regular expression to address that problem

`rx=r"([a-zA-Z0-9]+)\*\*([a-zA-Z0-9()]+)"`

result = regex.sub(rx, r"power(\1,\2)", expression, 0, regex.IGNORECASE | regex.MULTILINE)

But above code successfully converting expression similar to the example 1 and example 2, but failed to convert expression like

`(a+1)**2 or ((a+b)*c)**2`

I bit new to python .Please guide me how to approach to solve this problem.

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Answer Source

This sounds very familiar, I think I dealt with a similar problem on the pyparsing mailing list, but I can't find it at the moment. But try something like this:

```
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
```

prints:

```
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
```

If you have more corner cases in the code that you are converting, it will probably just need a bit more tweaking of the `operand`

expression, or adding more operators (like `&&`

) to the `infixNotation`

expression.

(Note that you have to convert `a**b**c`

as if written `a**(b**c)`

, as chained exponentiation is evaluated right-to-left, not left-to-right.)

EDIT:

Introduced hack to properly handle precedence between '-' and '**'. Expanded tests to actually evaluate before/after strings. This looks more solid now.

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