El domingo, 23 de febrero de 2014 18:09:36 UTC-6, Jeff Bezanson escribió:
>
> This was just changed on master so that you'd do Set([i*2 for i in 1:5]).
>
OK, that seems reasonable, thanks.
> On Sun, Feb 23, 2014 at 12:45 AM, David P. Sanders
> >
> wrote:
> >
> >
> > El viernes, 21 de feb
This was just changed on master so that you'd do Set([i*2 for i in 1:5]).
On Sun, Feb 23, 2014 at 12:45 AM, David P. Sanders wrote:
>
>
> El viernes, 21 de febrero de 2014 16:03:19 UTC-6, Steven G. Johnson
> escribió:
>>
>>
>>
>> On Friday, February 21, 2014 9:02:48 AM UTC-5, David P. Sanders wro
El viernes, 21 de febrero de 2014 16:03:19 UTC-6, Steven G. Johnson
escribió:
>
>
>
> On Friday, February 21, 2014 9:02:48 AM UTC-5, David P. Sanders wrote:
>>
>> OK, I think I have answered my own question: a Set is the good structure.
>> And to create a set from an array I can do something li
On Friday, February 21, 2014 9:02:48 AM UTC-5, David P. Sanders wrote:
>
> OK, I think I have answered my own question: a Set is the good structure.
> And to create a set from an array I can do something like
>
> s = Set([3, 4, 5]...)
>
> or
>
> s = Set([i*2 for i in 1:5]...)
>
> which would be
El viernes, 21 de febrero de 2014 07:49:27 UTC-6, David P. Sanders escribió:
>
> Hi,
>
> Suppose I have a collection of elements that I will create once at the
> beginining, visit in an unknown order and delete one by one when they are
> visited.
> What is the most efficient data structure for