[Tutor] In-place expansion of list members... possible?

[Ricardo Aráoz]

Kent Johnson wrote:
> Marc Tompkins wrote:
>> I'm working with delimited files (ANSI X12 EDI nonsense, to be 
>> precise.)  First I load the records to a list:
>>         tmpSegs = inString.split(self.SegTerm)
>>
>> Now, I want to replace each string in that list with a string:
>>
>>         for seg in tmpSegs:       # 'seg' is short for 'segment'
>>             seg = seg.split(self.ElemSep)
>>
>> It doesn't work.  If I check the contents of tmpSegs, it's still a list 
>> of strings - not a list of lists.  So I do this instead:
>>         tmpSegs2 = []
>>         for seg in tmpSegs:
>>             tmpSegs2.append(seg.split(self.sElemSep))
>>         del tmpSegs
>>
>> This works.  And, for the size of files that I'm likely to run into, 
>> creating the extra list and burning the old one is probably not going to 
>> swamp the machine - but it feels clumsy to me, like if I just knew the 
>> secret I could be much more elegant about it.
> 
> Creating a new list is fine, actually. You can do it more elegantly with 
> a list comprehension, and it's fine to assign to the same name, which 
> will implicitly delete the old list:
> tmpSegs = [ seg.split(self.ElemSep) for seg in tmpSegs ]
> 
> If you really want to replace elements in the same list you can use 
> enumerate to get indices:
> 
> for i, seg in enumerate(tmpSegs):
>    tmpSegs[i] = seg.split(self.ElemSep)
> 
> or use slice assignment:
> tmpSegs[:] = [ seg.split(self.ElemSep) for seg in tmpSegs ]

Won't the slice assignment create a new list and then assign it to the
old list (which the OP is trying to avoid)?