[Tutor] consider a problem of finding if two lists have any item in common

[Joel Goldstick]

On Sun, Mar 21, 2021 at 1:41 PM Peter Otten <__peter__ at web.de> wrote:
>
> On 21/03/2021 16:01, Mats Wichmann wrote:
> > On 3/21/21 8:53 AM, Peter Otten wrote:
> >> On 21/03/2021 09:52, Alan Gauld via Tutor wrote:
> >>> On 21/03/2021 07:15, Manprit Singh wrote:
> >>>
> >>>> lst1 = [2, 4, 6, 9]
> >>>> lst2 = [2, 5, 9, 0]
> >>>> this can be done using isdisjoint(other), it will return True if the
> >>>> lists
> >>>> have no items in common .The official documentation says nothing about
> >>>> other , just need to know if other can be an iterable or it should
> >>>> strictly
> >>>> be a set or frozenset?
> >>>> what should be the correct way ?
> >>>
> >>> Don't ask us, ask the interpreter, it knows best.
> >>> And it gives faster answers too.
> >>>
> >>> If its answers aren't what you'd expect *then* ask us why?
> >
> > I'd perhaps also note that while the answer to the question asked may
> > indeed be interesting, a lot of the time you actually want to know
> > _what_ they have in common, and that would be a different operation.
> > Which you can always truth-test to find out if they indeed had any
> > common members.
>
> The big difference between a.isdisjoint(b) and (not a & b) being that
> the first can stop after the first common element. Though I wouldn't go
> so far as to claim that the use cases are disjoint ;)

+1
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-- 
Joel Goldstick
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