Amit Khemka <[EMAIL PROTECTED]> wrote:
> On 3 Apr 2007 11:20:33 -0700, bahoo <[EMAIL PROTECTED]> wrote:
> > Hi,
> >
> > I have a list like ['0024', 'haha', '0024']
> > and as output I want ['haha']
> >
> > If I
> > myList.remove('0024')
> >
> > then only the first instance of '0024' is removed.
>
> To remove all items with multiple occurances in myList:
>
> list(set(myList) - set([x for x in myList if myList.count(x)>1]))
This approach is unfortunately quite inefficient, because each call to
myList.count is O(N), and there are O(N) such calls, so the whole thing
is O(N squared). [also, the inner square brackets are not needed, and
cause more memory to be consumed than omitting them would].
A "we-don't-need-no-stinkin'-one-liners" more relaxed approach:
import collections
d = collections.defaultdict(int)
for x in myList: d[x] += 1
list(x for x in myList if d[x]==1)
yields O(N) performance (give that dict-indexing is about O(1)...).
Collapsing this kind of approach back into a "one-liner" while keeping
the O(N) performance is not easy -- whether this is a fortunate or
unfortunate ocurrence is of course debatable. If we had a "turn
sequence into bag" function somewhere (and it might be worth having it
for other reasons):
def bagit(seq):
import collections
d = collections.defaultdict(int)
for x in seq: d[x] += 1
return d
then one-linerness might be achieved in the "gimme nonduplicated items"
task via a dirty, rotten trick...:
list(x for bag in [bagit(myList)] for x in myList if bag[x] == 1)
...which I would of course never mention in polite company...
Alex
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