New submission from Dmitriy <miron...@icloud.com>:
There is an error in documentation about string concatenation: https://docs.python.org/3/library/stdtypes.html --- Common Sequence Operations [...] 6. Concatenating immutable sequences always results in a new object. This means that building up a sequence by repeated concatenation will have a quadratic runtime cost in the total sequence length. To get a linear runtime cost, you must switch to one of the alternatives below: - if concatenating str objects, you can build a list and use str.join() at the end or else write to an io.StringIO instance and retrieve its value when complete --- It is not true for str objects anymore. Example using timeit module shows that time grows linearly: >>> timeit('a+="a"', setup='a=""', number=10000) 0.0005754000012530014 >>> timeit('a+="a"', setup='a=""', number=100000) 0.005819800004246645 But for bytes it is still right: >>> timeit('a+=b"a"', setup='a=b""', number=10000) 0.0017669000080786645 >>> timeit('a+=b"a"', setup='a=b""', number=100000) 0.20758410000416916 ---------- assignee: docs@python components: Documentation messages: 347390 nosy: dmitriym, docs@python priority: normal severity: normal status: open title: Error in the documentation about string concatenation type: enhancement versions: Python 3.7 _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue37512> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
