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From: Garapata on 22 May 2010 00:42
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: Bill Rowe on 23 May 2010 03:17
On 5/22/10 at 12:42 AM, warsaw95826(a)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



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