You can get almost the same result using pattern matching. For example, your
"foo:bar;baz".partition(":", ";")
can be done by a well-known matching idiom:
re.match(r'([^:]*):([^;]*);(.*)', 'foo:bar;baz').groups()
_______________________________________________
Python-ideas mailing list -- python-ideas@python.org
To unsubscribe send an email to python-ideas-le...@python.org
https://mail.python.org/mailman3/lists/python-ideas.python.org/
Message archived at
https://mail.python.org/archives/list/python-ideas@python.org/message/TVGHFAVNQU4BTLCAPK34DBU5ESN464Z3/
Code of Conduct: http://python.org/psf/codeofconduct/
