New submission from wang xuancong <xuancon...@gmail.com>:
A very common use of defaultdict is that if the key exist, use the corresponding mapped target, if the key does not exist, use the key itself. However, current Python 2/3 defaultdict does not support parametric lambda function: >>> from collections import * >>> aa=defaultdict(lambda t:t) >>> aa defaultdict(<function <lambda> at 0x10a55c950>, {}) >>> aa[0] Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: <lambda>() missing 1 required positional argument: 't' >>> I would like to suggest that use the dict's query key as the first argument in the default lambda function. And use the dict itself as the 2nd argument in the default lambda function (e.g., if the key exist, use the mapped target, otherwise, use the size of the current defaultdict). I think that will make Python much more powerful than any other programming language. Anyone can think of any additional information for the default lambda function? Thanks! ---------- messages: 354255 nosy: xuancong84 priority: normal severity: normal status: open title: defaultdict does not support parametric lambda type: enhancement versions: Python 2.7, Python 3.5, Python 3.6, Python 3.7, Python 3.8, Python 3.9 _______________________________________ Python tracker <rep...@bugs.python.org> <https://bugs.python.org/issue38420> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com