list unpack trick?

aurora aurora00 at
Sat Jan 22 19:03:27 CET 2005

Thanks. I'm just trying to see if there is some concise syntax available  
without getting into obscurity. As for my purpose Siegmund's suggestion  
works quite well.

The few forms you have suggested works. But as they refer to list multiple  
times, it need a separate assignment statement like

   list = s.split('=',1)

I am think more in the line of string.ljust(). So if we have a  
list.ljust(length, filler), we can do something like

   name, value = s.split('=',1).ljust(2,'')

I can always break it down into multiple lines. The good thing about list  
unpacking is its a really compact and obvious syntax.

On Sat, 22 Jan 2005 08:34:27 +0100, Fredrik Lundh <fredrik at>  
>> So more generally, is there an easy way to pad a list into length of n   
>> with filler items appended
>> at the end?
> some variants (with varying semantics):
>     list = (list + n*[item])[:n]
> or
>     list += (n - len(list)) * [item]
> or (readable):
>     if len(list) < n:
>         list.extend((n - len(list)) * [item])
> etc.
> </F>

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