Change one list item in place

[Steve Holden]

On 11/30/2010 8:28 PM, MRAB wrote:
> On 01/12/2010 01:08, Gnarlodious wrote:
>> This works for me:
>>
>> def sendList():
>>      return ["item0", "item1"]
>>
>> def query():
>>      l=sendList()
>>      return ["Formatting only {0} into a string".format(l[0]), l[1]]
>>
>> query()
>>
>>
>> However, is there a way to bypass the
>>
>> l=sendList()
>>
>> and change one list item in-place? Possibly a list comprehension
>> operating on a numbered item?
>>
> There's this:
> 
>     return ["Formatting only {0} into a string".format(x) if i == 0 else
> x for i, x in enumerate(sendList())]
> 
> but that's too clever for its own good. Keep it simple. :-)

I quite agree. That solution is so clever it would be asking for a fight
walking into a bar in Glasgow.

However, an unpacking assignment can make everything much more
comprehensible [pun intended] by removing the index operations. The
canonical solution would be something like:

def query():
    x, y = sendList()
    return ["Formatting only {0} into a string".format(x), y]

regards
 Steve
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