Replace variable in expression

Question

I'm using regexp to find and replace a variable in python expression in string format. I don't want the 'var' replaced to be the part of another function or variable name. (I banned the solution using if 'var' in expr and expr.replace("var", etc.)).

So, I check the previous characters (allowed) and the following characters(allowed) with the following regexp:

pattern = re.compile(r'(^var)(?=\+|\-|\*|\/| |$)|(?<=\+|\=|\[|\-|\*|\/|,| |\()var(?=\+|\-|\*|\/|$| |,|\))')

ô_O, it seems to be complicated but it works on the following test, replacing 'var' by '###'

expr     = ' var + variable + avar + var[x] + fun(var,variable) + fun2(variable, var, var1) + fun3(variable,var)+ var  -var/var+var*var*(var)'
expected = ' ### + variable + avar + var[x] + fun(###,variable) + fun2(variable, ###, var1) + fun3(variable,###)+ ###  -###/###+###*###*(###)'

I use regexp:

to ckeck if 'var' is in expression
to replace 'var' in expression
to check if 'var' is not in expression after being replaced.

if pattern.search(expr):
  new_expr = re.sub(pattern, '###', expr)  
  assert not pattern.search(new_expr), 'Replace failed'

I use the code a lot of time and I'm wondering if something simpler/faster exists ?

regex is not your tool of choice for parsing programming languages. Look into the ast module. — georg
– georg, Commented Oct 21, 2013 at 13:26
It seems odd that you are not replacing var in var[x] - is that intentional, or a typo? — PaulMcG
– PaulMcG, Commented Oct 21, 2013 at 13:31

Inbar Rose · Accepted Answer · 2013-10-21 13:27:45Z

1

Well, the pattern you need is: r'\bvar\b', the \b is "border" which lets us define the full "string" we want to replace without replacing things like "variable"

However, upon testing your "expected" string, I found it had a mistake in it:

expected = ' ### + variable + avar + var[x] ' # <- this last 'var' should be ###

Anyway. Solution:

>>> import re
>>> re.sub(r'\bvar\b', '###', expr)
' ### + variable + avar + ###[x] + fun(###,variable) + fun2(variable, ###, var1) + fun3(variable,###)+ ###  -###/###+###*###*(###)'

answered Oct 21, 2013 at 13:27

Inbar Rose

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5 Comments

user2652620 Over a year ago

it's not a mistake, I don't want the var to be a dict. It's looks like to be a border with exception.

Inbar Rose Over a year ago

I am not sure what you are saying, but valid code will not handle two variables with the same name in a namespace. var is var is var.

user2652620 Over a year ago

do I wrote a complex regexp to match a dump expression ? T_T. I'll check the code, I'll give you the answer.

user2652620 Over a year ago

Well, you're right, it's not a real situation. A better way to tests regexp is to use lists of expressions and lists of expected ones. The solution works with '\b'.

user2652620 Over a year ago

I wrote this sample to do not match the following situation : avar[x] => a###[x], even var[x] but var[x] will never exist in my code. '\b' is enough. thx.

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