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Groups > comp.lang.python > #69574 > unrolled thread

Compact way to assign values by slicing list in Python

Started byMark Lawrence <breamoreboy@yahoo.co.uk>
First post2014-04-02 00:17 +0100
Last post2014-04-04 11:52 +1000
Articles 3 — 3 participants

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  Compact way to assign values by slicing list in Python Mark Lawrence <breamoreboy@yahoo.co.uk> - 2014-04-02 00:17 +0100
    Re: Compact way to assign values by slicing list in Python Marco Buttu <marco.buttu@gmail.com> - 2014-04-03 15:50 +0200
      Re: Compact way to assign values by slicing list in Python alex23 <wuwei23@gmail.com> - 2014-04-04 11:52 +1000

#69574 — Compact way to assign values by slicing list in Python

FromMark Lawrence <breamoreboy@yahoo.co.uk>
Date2014-04-02 00:17 +0100
SubjectCompact way to assign values by slicing list in Python
Message-ID<mailman.8828.1396530054.18130.python-list@python.org>
Came across this 
http://stackoverflow.com/questions/22756632/compact-way-to-assign-values-by-slicing-list-in-python?newsletter=1&nlcode=245176|202f 
- just curious what you guys and gals thought of the answers.

-- 
My fellow Pythonistas, ask not what our language can do for you, ask 
what you can do for our language.

Mark Lawrence

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#69583

FromMarco Buttu <marco.buttu@gmail.com>
Date2014-04-03 15:50 +0200
Message-ID<lhjovl$sbn$1@speranza.aioe.org>
In reply to#69574
On 04/02/2014 01:17 AM, Mark Lawrence wrote:
> Came across this
> http://stackoverflow.com/questions/22756632/compact-way-to-assign-values-by-slicing-list-in-python?newsletter=1&nlcode=245176|202f
> - just curious what you guys and gals thought of the answers.

I prefere this one:

bar = ['a','b','c','x','y','z']
v1, _, _, v2, v3, _ = bar

I also like the solution with itemgetter:

v1, v2, v3 = itemgetter(0, 3, 4)(bar)

but I think it is less readable than the previous one

-- 
Marco Buttu

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#69628

Fromalex23 <wuwei23@gmail.com>
Date2014-04-04 11:52 +1000
Message-ID<lhl38q$ung$1@dont-email.me>
In reply to#69583
On 3/04/2014 11:50 PM, Marco Buttu wrote:
> I prefere this one:
>
> bar = ['a','b','c','x','y','z']
> v1, _, _, v2, v3, _ = bar
>
> I also like the solution with itemgetter:
>
> v1, v2, v3 = itemgetter(0, 3, 4)(bar)
>
> but I think it is less readable than the previous one

What if you wanted the 2nd, 37th, and 1007th items from a list?

Personally, I find the 2nd form far more readable, once I got past my 
initial surprise that it would access a list of items.

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