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From: Ray Koopman on 22 May 2010 23:34
Intersection@@(#[[All,3]]&/@list)

will handle a variable number of groups.

On May 21, 9:42 pm, Garapata <warsaw95...(a)mypacks.net> wrote:
> I have a list with the following dimensionality:
>
> Dimensions[list]
> Dimensions[list[[1]]]
> Dimensions[list[[2]]]
> Dimensions[list[[3]]]
>
> {3}
> {130, 3}
> {126, 3}
> {191, 3}
>
> 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.
>
> Probably easier if I had a more regular structure but I don't.
>
> Any help in pointing me in the right direction much appreciated.
>
> Thx to all.

From: Garapata on 23 May 2010 03:17
A friend working serendipitously on a very similar problem got a step
of the solution from Wolfram Support (credit where credit's due, they
seem to have made some real improvements over there).

Anyway, since I have unique dates within each of the groups described
in my original post I can use Part to pull out all of the dates,
something like this:

groupDates = list[[All,All, 3]]

Then I can use this elegant solution via Wolfram support:

datesIntersection = Apply[Intersection, groupDates]

Doesn't get me all the way to what I need, but looks like a good
start.

Remember I still need to get to this (from my original post):
....

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
....

So any help still very much appreciated.

Thx


From: Ray Koopman on 23 May 2010 03:17
Further to my earlier post, the whole thing would then be

Function[i,Flatten@{i,Cases[#,{x_,y_,i}:>{x,y},{1},1]&/@list}] /@
Intersection@@(#[[All,3]]&/@list)

On May 21, 9:42 pm, Garapata <warsaw95...(a)mypacks.net> wrote:
> I have a list with the following dimensionality:
>
> Dimensions[list]
> Dimensions[list[[1]]]
> Dimensions[list[[2]]]
> Dimensions[list[[3]]]
>
> {3}
> {130, 3}
> {126, 3}
> {191, 3}
>
> 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.
>
> Probably easier if I had a more regular structure but I don't.
>
> Any help in pointing me in the right direction much appreciated.
>
> Thx to all.

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