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SubjectAuthor
* A missing iterator on itertools module?ast
+- Re: A missing iterator on itertools module?ast
+* Re: A missing iterator on itertools module?Stefan Ram
|`* Re: A missing iterator on itertools module?ast
| `* Re: A missing iterator on itertools module?Stefan Ram
|  +- Re: A missing iterator on itertools module? (Posting On Python-List Prohibited)Lawrence D'Oliveiro
|  `- Re: A missing iterator on itertools module?Mark Bourne
+- Re: A missing iterator on itertools module?Antoon Pardon
`- RE: A missing iterator on itertools module?<avi.e.gross

1
A missing iterator on itertools module?

<66059eb6$0$7522$426a34cc@news.free.fr>

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 by: ast - Thu, 28 Mar 2024 16:45 UTC

Hello

Suppose I have these 3 strings:

s1 = "AZERTY"
s2 = "QSDFGH"
s3 = "WXCVBN"

and I need an itertor who delivers

A Q W Z S C E D C ...

I didn't found anything in itertools to do the job.

So I came up with this solution:

list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN")))

['A', 'Q', 'W', 'Z', 'S', 'X', 'E', 'D', 'C', 'R', 'F', 'V', 'T', 'G',
'B', 'Y', 'H', 'N']

Do you havbe a neat solution ?

Re: A missing iterator on itertools module?

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 by: ast - Thu, 28 Mar 2024 16:47 UTC

Le 28/03/2024 à 17:45, ast a écrit :

> A Q W Z S C E D C ...

sorry
A Q W Z S X E D C

Re: A missing iterator on itertools module?

<solution-20240328180624@ram.dialup.fu-berlin.de>

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From: ram@zedat.fu-berlin.de (Stefan Ram)
Newsgroups: comp.lang.python
Subject: Re: A missing iterator on itertools module?
Date: 28 Mar 2024 17:07:25 GMT
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 by: Stefan Ram - Thu, 28 Mar 2024 17:07 UTC

ast <none@none.fr> wrote or quoted:
>s1 = "AZERTY"
>s2 = "QSDFGH"
>s3 = "WXCVBN"
>and I need an itertor who delivers
>A Q W Z S C E D C ...
>I didn't found anything in itertools to do the job.
>So I came up with this solution:
>list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN")))

Maybe you meant "zip(s1,s2,s3)" as the definition of s1, s2,
and s3 otherwise would not be required. Also the "list" is not
necessary because "chain.from_iterable" already is an iterable.
You could also use "*" instead of "list" to print it. So,

import itertools as _itertools
s =[ "AZERTY", "QSDFGH", "WXCVBN" ]
print( *_itertools.chain.from_iterable( zip( *s )))

. But these are only minor nitpicks; you have found a nice solution!

Re: A missing iterator on itertools module?

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 by: ast - Thu, 28 Mar 2024 17:12 UTC

Le 28/03/2024 à 18:07, Stefan Ram a écrit :
> ast <none@none.fr> wrote or quoted:
>> s1 = "AZERTY"
>> s2 = "QSDFGH"
>> s3 = "WXCVBN"
>> and I need an itertor who delivers
>> A Q W Z S C E D C ...
>> I didn't found anything in itertools to do the job.
>> So I came up with this solution:
>> list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN")))
>
> Maybe you meant "zip(s1,s2,s3)" as the definition of s1, s2,
> and s3 otherwise would not be required. Also the "list" is not
> necessary because "chain.from_iterable" already is an iterable.
> You could also use "*" instead of "list" to print it. So,
>
> import itertools as _itertools
> s =[ "AZERTY", "QSDFGH", "WXCVBN" ]
> print( *_itertools.chain.from_iterable( zip( *s )))
>
> . But these are only minor nitpicks; you have found a nice solution!

Why did you renamed itertools as _itertools ?

Re: A missing iterator on itertools module?

<underscore-20240328182256@ram.dialup.fu-berlin.de>

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From: ram@zedat.fu-berlin.de (Stefan Ram)
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Subject: Re: A missing iterator on itertools module?
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 by: Stefan Ram - Thu, 28 Mar 2024 17:23 UTC

ast <none@none.fr> wrote or quoted:
>Why did you renamed itertools as _itertools ?

Assume I have a module A.py:

import math
def f(): pass

. Assume I have an additional module B.py:

import A

. Now, when I'm editing "B.py" in IDLE and type "A.", IIRC
IDLE will offer me two possible completions: "A.math" and
"A.f". The "A.math" makes no sense to me. I want it to go
away. Therefore, I rewrite A.py as:

import math as _math
def f(): pass

. Now, Idle will only offer the completion "A.f".

So, I sometimes use this "import math as _math" style. But then,
it is simpler for me to /always/ use this style; after all: you
can't know whether someone eventually will import your module!

Re: A missing iterator on itertools module? (Posting On Python-List Prohibited)

<uu4qb1$3r1c8$2@dont-email.me>

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 by: Lawrence D'Oliv - Thu, 28 Mar 2024 22:18 UTC

On 28 Mar 2024 17:23:14 GMT, Stefan Ram wrote:

> . Now, when I'm editing "B.py" in IDLE and type "A.", IIRC
> IDLE will offer me two possible completions: "A.math" and
> "A.f". The "A.math" makes no sense to me.

It works, though.

Re: A missing iterator on itertools module?

<uu68ip$8tb8$1@dont-email.me>

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Subject: Re: A missing iterator on itertools module?
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 by: Mark Bourne - Fri, 29 Mar 2024 11:27 UTC

Stefan Ram wrote:
> ast <none@none.fr> wrote or quoted:
>> Why did you renamed itertools as _itertools ?
>
> Assume I have a module A.py:
>
> import math
> def f(): pass
>
> . Assume I have an additional module B.py:
>
> import A
>
> . Now, when I'm editing "B.py" in IDLE and type "A.", IIRC
> IDLE will offer me two possible completions: "A.math" and
> "A.f". The "A.math" makes no sense to me.

`import math` imports the `math` module and binds it to `math` in the
global namespace of the `A` module. Since it doesn't have a leading
underscore, by default it's considered to be a public attribute of the
`A` module, and IDLE is offering all the public attributes of the `A`
module for completion.

> I want it to go
> away. Therefore, I rewrite A.py as:
>
> import math as _math
> def f(): pass
>
> . Now, Idle will only offer the completion "A.f".
>
> So, I sometimes use this "import math as _math" style. But then,
> it is simpler for me to /always/ use this style; after all: you
> can't know whether someone eventually will import your module!

You can explicitly declare the public interface of a module by defining
`__all__`, listing the names which should be considered part of the
module's public interface; see:
- https://docs.python.org/3/reference/simple_stmts.html#the-import-statement
- https://peps.python.org/pep-0008/#public-and-internal-interfaces

Although `from A import *` is generally discouraged, if `A` defines
`__all__` then only the names listed in `__all__` are bound in the
importing module's namespace. Otherwise, all names from `A` which don't
have a leading underscore are considered to be public and bound in the
importing module.

I don't use IDLE, but it may be that it also uses `__all__` to determine
a module's public API. In that case, setting `__all__ = ["f"]` in `A`
should prevent it from offering `math` as a completion (nor any other
name that's not in the `__all__` list).

--
Mark.

Re: A missing iterator on itertools module?

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 by: Antoon Pardon - Wed, 3 Apr 2024 09:11 UTC

Op 28/03/2024 om 17:45 schreef ast via Python-list:
> Hello
>
> Suppose I have these 3 strings:
>
> s1 = "AZERTY"
> s2 = "QSDFGH"
> s3 = "WXCVBN"
>
> and I need an itertor who delivers
>
> A Q W Z S C E D C ...
>
> I didn't found anything in itertools to do the job.

The documentation mentions a roundrobin recipe.
>
> So I came up with this solution:
>
>
> list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN")))
>
> ['A', 'Q', 'W', 'Z', 'S', 'X', 'E', 'D', 'C', 'R', 'F', 'V', 'T', 'G',
> 'B', 'Y', 'H', 'N']

But if your strings are not equal, this will only produce a partial result.

RE: A missing iterator on itertools module?

<mailman.61.1712146500.3468.python-list@python.org>

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 by: <avi.e.gross@gmail.com> - Wed, 3 Apr 2024 12:14 UTC

Antoon,

Even if the suggested solution offers a partial result, you would need
specific requirements to determine what should be done if one or more of the
parts being cycled is shorter than the others. Stopping at that point is one
option. Another is to continue but only interleave ones still producing and
in the same order.

There is a function in itertools called zip_longest() that might be
considered as it keeps going but substitutes a customizable value for
"missing" parts. You could then, perhaps, make a change so that sentinel is
not passed along.

-----Original Message-----
From: Python-list <python-list-bounces+avi.e.gross=gmail.com@python.org> On
Behalf Of Antoon Pardon via Python-list
Sent: Wednesday, April 3, 2024 5:11 AM
To: python-list@python.org
Subject: Re: A missing iterator on itertools module?

Op 28/03/2024 om 17:45 schreef ast via Python-list:
> Hello
>
> Suppose I have these 3 strings:
>
> s1 = "AZERTY"
> s2 = "QSDFGH"
> s3 = "WXCVBN"
>
> and I need an itertor who delivers
>
> A Q W Z S C E D C ...
>
> I didn't found anything in itertools to do the job.

The documentation mentions a roundrobin recipe.
>
> So I came up with this solution:
>
>
> list(chain.from_iterable(zip("AZERTY", "QSDFGH", "WXCVBN")))
>
> ['A', 'Q', 'W', 'Z', 'S', 'X', 'E', 'D', 'C', 'R', 'F', 'V', 'T', 'G',
> 'B', 'Y', 'H', 'N']

But if your strings are not equal, this will only produce a partial result.
--
https://mail.python.org/mailman/listinfo/python-list

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