[Tutor] In-place expansion of list members... possible?
Eric Brunson
brunson at brunson.com
Wed Nov 7 21:28:02 CET 2007
Marc Tompkins wrote:
> This question has probably been asked and answered many times, but I
> can't figure out how to word my question to get relevant results from
> Google. So I thought I'd try some human beings, eh?
>
> 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)
A list is an array of pointers to objects, what you've done here is
create a new name referencing an item in the list, then making that new
name point to something different.
Try this (untested code):
for index in xrange(0, len(tmpSegs)):
tmpSegs[index] = tmpSegs[index].split(self.ElemSep)
This will iterate through the list and set each reference in the list to
point to your new object.
Hope that helps,
e.
>
> 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. I don't want the
> next guy who reads my code to scratch his head and say 'What was he
> thinking?' - especially if it's me in a year's time!
>
> What puzzles me most is that I can replace 'seg' with just about
> anything else -
> for seg in tmpSegs:
> seg = seg.strip()
> or
> for seg in tmpSegs:
> seg = 'bananas'
> or
> for seg in tmpSegs:
> seg = seg + ' bananas'
> and it works. So why can't I replace it with its own split?
>
> It's not a burning question - development continues - but I'm really
> curious.
>
> Thanks!
> --
> www.fsrtechnologies.com <http://www.fsrtechnologies.com>
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