Re: Intersection of sublists on date and making a 2D list from a 3D one
- To: mathgroup at smc.vnet.net
- Subject: [mg109927] Re: Intersection of sublists on date and making a 2D list from a 3D one
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Sun, 23 May 2010 03:17:20 -0400 (EDT)
On 5/22/10 at 12:42 AM, warsaw95826 at mypacks.net (Garapata) wrote:
>I have a list with the following dimensionality:
> Dimensions[list]
> Dimensions[list[[1]]]
> Dimensions[list[[2]]]
> Dimensions[list[[3]]]
You can get exactly the same result using Map (shorthand /@) here. That is
Dimensions[list[
Dimensions/@list
yields the same result as you posted
>I think of it as 3 groups, each with a different number of rows, but
>each of these rows has 3 columns. Of the 3 columns, the third has
>dates in a DateList format.
>So, I can simulate the list with the following:
>a = Table[{RandomInteger[10], RandomInteger[10], DatePlus[{2010, 1,
>1}, RandomInteger[{1, 150}]]}, {1}, {191}]; list = Join[a[[All, 4 ;;
>133, All]], a[[All, 1 ;; 126, All]], a];
>Ultimately I'd like to find all the rows that intersect on the same
>dates and put the whole thing into a 2 dimensional structure that
>would have the following columns:
>commonDates, group1Data1, group1Data2, group2Data1, group2Data2,
>group3Data1, group3Data2,
>I can use Intersect[] to find the common dates:
>Intersection[list[[1, All,3]], list[[2, All, 3]], list[[3, All,
>3]]];
>but this seems a bit cumbersome given I may have a greater or lesser
>number of groups. Seems like I need to start this in a better way,
>but since Intersection[] doesn't take a list of lists I don't know
>where to take this.
much less cumbersome is using Map and Apply, that is
commonDates=Intersection@@(#[[All,3]]&/@list);
Once you have the common dates, you can select the desired data using Cases as follows
Cases[SortBy[#,Last],{__,_?(MemberQ[commonDates,#]&)]]&/@list