Issue 28939: Groupby Is Roughly Equivalent To ... (original) (raw)

Issue28939

Created on 2016-12-11 20:16 by greg.solomon, last changed 2022-04-11 14:58 by admin. This issue is now closed.

Messages (3)
msg282943 - (view)	Author: Greg Solomon (greg.solomon)	Date: 2016-12-11 20:16
https://docs.python.org/3/library/itertools.html#itertools.groupby I found the "equivalent" code a bit hard to read. Is there any merit in changing it to something a bit like the following ? Kind Rgds, Greg class groupby: def __init__( self , iterable , key_function = ( lambda x: x ) ): self.iterable = iter( iterable ) self.key_function = key_function self.FINISHED = object() try: self.next_value = next( self.iterable ) except StopIteration: self.next_value = self.FINISHED def __iter__( self ): return self def __next__( self ): if self.next_value == self.FINISHED: raise StopIteration self.group_key_value = self.key_function( self.next_value ) return ( self.group_key_value , self._group() ) def _group( self ): while self.next_value != self.FINISHED \ and self.group_key_value == self.key_function( self.next_value ): yield self.next_value try: self.next_value = next( self.iterable ) except StopIteration: self.next_value = self.FINISHED return
msg282947 - (view)	Author: Julien Palard (mdk) *	Date: 2016-12-11 20:54
I renamed your function groupby2 to compare it with itertools.groupby and tested but: >>> print(list(groupby2(['A', 'B']))) does not returns, looks like your implementation have a bug, so I tried: >>> for k in groupby2(['A', 'B']): ... print(k) and I'm getting loads of: ('A', <generator object groupby2._group at 0x7f0476809f10>) ('A', <generator object groupby2._group at 0x7f0476851f68>) ('A', <generator object groupby2._group at 0x7f0476809f10>) ('A', <generator object groupby2._group at 0x7f0476851f68>) ('A', <generator object groupby2._group at 0x7f0476809f10>) ('A', <generator object groupby2._group at 0x7f0476851f68>) ('A', <generator object groupby2._group at 0x7f0476809f10>) ('A', <generator object groupby2._group at 0x7f0476851f68>) ('A', <generator object groupby2._group at 0x7f0476809f10>) ('A', <generator object groupby2._group at 0x7f0476851f68>) You may also want to test your implementation against https://github.com/python/cpython/blob/master/Lib/test/test_itertools.py#L699
msg282968 - (view)	Author: Greg Solomon (greg.solomon)	Date: 2016-12-12 07:52
Oh, I get it. There needs to be a next(self.it) loop in __next__ as well as in _grouper in case the user doesn't call _grouper. My test was for ( k , g ) in groupby( L ): print ( k , len( list( g ) ) ) so I was executing _grouper on every row. Thanks !!!