The question is confusing, but it appears to be a longish and confused description of simple formatting or template substitution. It's hard to be sure, but it sounds like one of Those Things™ (TT).
Most of Those Things (TT) are standard problems with standard solutions. Until you've seen a lot TT's, it seems like your problem is unique and special. It's hard to see TT's for what they are.
In this case, the problem appears to be solved by Python's string.Template class with minor modifications. The documentation for customizing string.Template isn't clear, so here's an example.
from string import Template
class MyTemplate( Template ):
delimiter= '@'
pattern= r"@(?P<escaped>@)|@(?P<named>[_a-z][_a-z0-9]*)@|@(?P<braced>[_a-z][_a-z0-9]*)@|@(?P<invalid>)"
That appears to be the standard solution to the standard problem. Define a new delimiter ('@') and some slightly different delimiter parsing rules and away you go.
This can be used as follows to replace any '@x@' variables in any template file. What's important is that very little actual code is needed, since it's one of Those Things that's already been solved.
with open( 'a.txt', 'r' ) as source:
t = MyTemplate(source.read())
result= t.substitute( x=15 )
print result
I think this is a rather over-engineered solution. As the accepted answer on the question shows, the replace method is vastly more obvious and less verbose than that frightening regular expression.
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