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devel / comp.lang.python / Re: A problem with itertools.groupby

SubjectAuthor
* A problem with itertools.groupbyast
+- Re: A problem with itertools.groupbyAntoon Pardon
+- Re: A problem with itertools.groupbyChris Angelico
`- Re: A problem with itertools.groupbyPeter Pearson

1
A problem with itertools.groupby

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 by: ast - Fri, 17 Dec 2021 08:25 UTC

Python 3.9.9

Hello

I have some troubles with groupby from itertools

from itertools import groupby

for k, grp in groupby("aahfffddssssnnb"):
print(k, list(grp))
print(k, list(grp))

a ['a', 'a']
a []
h ['h']
h []
f ['f', 'f', 'f']
f []
d ['d', 'd']
d []
s ['s', 's', 's', 's']
s []
n ['n', 'n']
n []
b ['b']
b []

It works as expected.
itertools._grouper objects are probably iterators
so they provide their datas only once. OK

but:

li = [grp for k, grp in groupby("aahfffddssssnnb")]
list(li[0])

[]

list(li[1])

[]

It seems empty ... I don't understand why, this is
the first read of an iterator, it should provide its
data.

This one works:

["".join(grp) for k, grp in groupby("aahfffddssssnnb")]

['aa', 'h', 'fff', 'dd', 'ssss', 'nn', 'b']

regards

Re: A problem with itertools.groupby

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 by: Antoon Pardon - Fri, 17 Dec 2021 15:41 UTC

> but:
>
> li = [grp for k, grp in groupby("aahfffddssssnnb")]
> list(li[0])
>
> []
>
> list(li[1])
>
> []
>
> It seems empty ... I don't understand why, this is
> the first read of an iterator, it should provide its
> data.

The group-iterators are connected. Each group-iterator is a wrapper
around the original iterator
with an extra termination condition. So in order to start the next
group-iterator the previous
group-iterator is exhausted, because the original iterator has to be
ready to produce values
for the next group-iterator.

--
Antoon Pardon.

Re: A problem with itertools.groupby

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 by: Chris Angelico - Fri, 17 Dec 2021 15:44 UTC

On Sat, Dec 18, 2021 at 2:32 AM ast <ast@invalid> wrote:
>
> Python 3.9.9
>
> Hello
>
> I have some troubles with groupby from itertools
>
> from itertools import groupby
>
> li = [grp for k, grp in groupby("aahfffddssssnnb")]
> list(li[0])
>
> []
>
> list(li[1])
>
> []
>
> It seems empty ... I don't understand why, this is
> the first read of an iterator, it should provide its
> data.
>

https://docs.python.org/3/library/itertools.html#itertools.groupby

Check the explanatory third paragraph :)

ChrisA

Re: A problem with itertools.groupby

<j23qriFosbrU1@mid.individual.net>

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https://www.novabbs.com/devel/article-flat.php?id=16434&group=comp.lang.python#16434

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From: pkpear...@nowhere.invalid (Peter Pearson)
Newsgroups: comp.lang.python
Subject: Re: A problem with itertools.groupby
Date: 17 Dec 2021 16:06:10 GMT
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 by: Peter Pearson - Fri, 17 Dec 2021 16:06 UTC

On Fri, 17 Dec 2021 09:25:03 +0100, ast <ast@invalid> wrote:
[snip]
>
> but:
>
> li = [grp for k, grp in groupby("aahfffddssssnnb")]
> list(li[0])
>
> []
>
> list(li[1])
>
> []
>
> It seems empty ... I don't understand why, this is
> the first read of an iterator, it should provide its
> data.

Baffling. Here's a shorter and less readable illustration:

>>> list(groupby("aabbb"))
[('a', <itertools._grouper object at 0xb7018c0c>),
('b', <itertools._grouper object at 0xb7018a0c>)]
>>> list(groupby("aabbb"))[0]
('a', <itertools._grouper object at 0xb71d3e4c>)
>>> list(list(groupby("aabbb"))[0][1])
[]

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