New submission from Madison May: I often find myself trying to access a file relative to the module I'm working on. When this occurs, I'll often use something like the following:
``` os.path.abspath(os.path.join(os.path.dirname(__file__), "data/resource.pkl")) ``` I have good reason to believe that I'm not the only one, as searching for other examples of this code on github returns ~20k exact matches: https://github.com/search?utf8=%E2%9C%93&q=%22os.path.abspath%28os.path.join%28os.path.dirname%28__file__%29%22+&type=Code&ref=searchresults Low priority, but a more concise way of achieving the same result would be much appreciated. ---------- components: Library (Lib) messages: 244920 nosy: madison.may priority: normal severity: normal status: open title: Provide convenience function for paths relative to the current module type: enhancement versions: Python 3.5, Python 3.6 _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue24396> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
