• Laeeth Isharc (12/12) Sep 09 2015 I have a DateTime[][] arg (actually my own date type, but that
• Sebastiaan Koppe (3/3) Sep 09 2015 Couldn't you use setIntersection together with reduce?
• Artem Tarasov (4/12) Sep 09 2015 I'd use something like this, it works in O(Σ|arg[i]| *
• deed (16/18) Sep 10 2015 A suggestion:
• Laeeth Isharc (4/22) Sep 11 2015 Thank you for this - exactly what I was looking for - and for the

"Laeeth Isharc" <laeethnospam nospamlaeeth.com> writes:

I have a DateTime[][] arg (actually my own date type, but that 
shouldnt matter).  Ordered by time series ticker (eg a stock) and 
then the date.

eg arg[0] might be the dates relating to IBM and arg[0][0] would 
be the date of the first point in IBM time series.

I would like to find the intersection of the dates.

so setIntersection(arg[0],arg[1],arg[2] .. arg[$-1])
except that I don't know how many series are in arg at compile 
time.

what's the most efficient way to use Phobos to find these?  (I 
could write a loop, but I am trying to learn how to use 
std.algorithm better).

"Sebastiaan Koppe" <mail skoppe.eu> writes:

Couldn't you use setIntersection together with reduce?

Doesn't seem like the most efficient solution, but its less 
typing and most likely will have no bugs.

"Artem Tarasov" <lomereiter gmail.com> writes:

On Wednesday, 9 September 2015 at 20:28:35 UTC, Laeeth Isharc 
wrote:
 so setIntersection(arg[0],arg[1],arg[2] .. arg[$-1])
 except that I don't know how many series are in arg at compile 
 time.

 what's the most efficient way to use Phobos to find these?  (I 
 could write a loop, but I am trying to learn how to use 
 std.algorithm better).

I'd use something like this, it works in O(Σ|arg[i]| * 
log|arg.length|)
 arg.nWayUnion.group.filter!(g => g[1] == arg.length).map!(g => 
 g[0]).array

"deed" <none none.none> writes:

On Wednesday, 9 September 2015 at 20:28:35 UTC, Laeeth Isharc 
wrote:
 I have a DateTime[][] arg ...
 I would like to find the intersection of the dates.

A suggestion:

auto minLength      = arg.map!(a => a.length).reduce!min;
auto minIdx         = arg.map!(a => 
a.length).countUntil(minLength);
auto intersection   = arg[minIdx].filter!(e => 
chain(arg[0..minIdx], arg[minIdx..$]).all!(a => 
a.assumeSorted.contains(e)));

reduce with setIntersection seems the most straightforward, but 
needs array AFAIK, i.e.:

auto intersection =
//reduce!((r, x) => setIntersection(r, x))(arg);      // doesn't 
work
   reduce!((r, x) => setIntersection(r, x).array)(arg);// does, 
but with array

"Laeeth Isharc" <spamnolaeeth nospamlaeeth.com> writes:

On Thursday, 10 September 2015 at 11:58:10 UTC, deed wrote:
 On Wednesday, 9 September 2015 at 20:28:35 UTC, Laeeth Isharc 
 wrote:
 I have a DateTime[][] arg ...
 I would like to find the intersection of the dates.

 A suggestion:

 auto minLength      = arg.map!(a => a.length).reduce!min;
 auto minIdx         = arg.map!(a => 
 a.length).countUntil(minLength);
 auto intersection   = arg[minIdx].filter!(e => 
 chain(arg[0..minIdx], arg[minIdx..$]).all!(a => 
 a.assumeSorted.contains(e)));

 reduce with setIntersection seems the most straightforward, but 
 needs array AFAIK, i.e.:

 auto intersection =
 //reduce!((r, x) => setIntersection(r, x))(arg);      // 
 doesn't work
   reduce!((r, x) => setIntersection(r, x).array)(arg);// does, 
 but with array

Thank you for this - exactly what I was looking for - and for the 
other answer.


Laeeth.