Grep and replace fuzzy pattern of string

Question

I have a few files with python code and decorators like this:

@trace('api.module.function_name', info=None, custom_args=False)

The only difference between these decorators is the string 'api.module.function_name' - func name and module are different. And depending on the this param name sometimes this decorator is one-lined, some times it is two- or three-lined.

I want to replace these decorators with the other one - more simple, like "@my_new_decorator".

I thought about some regex but I have no idea if it's possible for such "fuzzy" search. I tried ^@trace([A-Za-z0-9]\, custom_args=False)$ But it doesn't work.

Is there a way to do it?

Well a regex like that wouldn't work, you'd need something more like ^\s*@trace$.+$$ because you're matching one alphanumeric character and only matching at all if this is all at column 0. — moopet
– moopet, Commented Jul 29, 2016 at 12:01

Steven Doggart · Accepted Answer · 2016-07-29 12:10:10Z

1

Something like this should work:

(\n|^)\s*@trace\(\s*'[^']*',\s*info=None,\s*custom_args=False\s*\)\s*(\r|\n|$)

See the demo

answered Jul 29, 2016 at 12:10

Steven Doggart

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user6620326 · Accepted Answer · 2016-07-29 12:13:56Z

1

Use ^@trace$'api\.(.+)\.(.+)', info=None, custom_args=False$$ with a multiline flag.

You may want to use re.sub :

>>> import re
>>> pattern = re.compile('^@trace\('api\.(.+)\.(.+)', info=None, custom_args=False\)$', re.M)
>>> re.sub(pattern, '@my_new_decorator('\1', '\2')', '@trace('api.module.function_name', info=None, custom_args=False)')
@my_new_decorator('module', 'function_name')

See this for the demo of the regex

As you can see \1 expand to the first group in the regex (.+)

answered Jul 29, 2016 at 12:13

user6620326

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