Ronak, If you are looking for the smallest 3 unique values, and those are "d", "e", and "f", do you want all of the locations where your piddle is "d" or "e" or "f" ? Or if "d" is the smallest, just 3 "d" locations? If the latter, then what Chris has will work as-is. If the former, then you need a little bit extra at the beginning:
$matrix = pdl([0,1,2,3,4],[1,2,3,4,5],[2,3,4,5,6],[3,4,5,6,7],[4,5,6,7,8]); $k = 3; ($quant,$uniq_vals)=$matrix->flat->qsort->rle; #I am a big fan of ->qsort->rle. ->hist() might work too. $num_in_matrix = $quant(0:$k-1)->sum; #then Chris's stuff will work $smallest_k = zeroes(indx, $num_in_matrix); $matrix->flat->minimum_n_ind($smallest_k); #answer gets stuffed into $smallest_k $coords2D = pdl($matrix->one2nd($smallest_k)); $smallest_vals = $matrix->range($coords2D->transpose)->flat; cheers, Derek On Oct 17, 2014, at 10:29 AM, Chris Marshall <[email protected]> wrote: > Hi- > > First off, you can use the PDL on-line documentation to > find relevant commands to try: > > pdldoc -a maximum > > or > > pdl> apropos maximum > > in the pdl2 or perldl shells includes the following items: > > maximum Project via maximum to N-1 dimensions > maximum_ind Like maximum but returns the index rather than the value > maximum_n_ind Returns the index of `m' maximum elements > > where maximum_n_ind is what you are looking for. Here is a > sample pdl2 session showing the calculation. You can use > the PDL shells and the online documentation to understand > fully how it works: > > > pdl> floor(random(10,10)*10) > pdl> p $m > [ > [4 7 5 0 6 7 8 4 6 7] > [0 7 0 2 5 3 0 4 6 6] > [5 4 3 6 9 5 8 4 4 1] > [9 6 7 7 4 1 4 2 0 8] > [5 1 3 3 7 7 1 2 8 5] > [2 0 8 7 3 1 5 7 7 6] > [5 9 6 2 7 2 6 7 0 2] > [6 3 1 1 8 2 1 9 9 9] > [6 7 9 3 5 2 2 5 9 3] > [4 5 0 4 7 9 3 3 1 6] > ] > > pdl> $top5 = zeros indx, 5; # size of $top5 determines n > pdl> $m->flat->maximum_n_ind($top5) # get top 5 element indexes > pdl> p pdl($m->one2nd($top5)) # convert linear to ND index > [ > [4 0 1 7 8] > [2 3 6 7 7] > ] > > > > Hope this helps, > Chris > > > On Fri, Oct 17, 2014 at 11:54 AM, Ronak Agrawal <[email protected]> wrote: > First I would Thank You for your constant help...it has helped me a lot in > improving my skills > ------- > > I have generated a Hankel Matrix by the following operation > > // $a is a svd and therby this operation will always form Hankel Matrix > $matrix = $a x transpose($a); > > > > I need to find the top K minimum elements ( 2D) with their 2 Dimesional > indices .. > Can you suggest me some good approach for this > > I though to do following but it is not optimized > Converting the Lower triangular Matrix as Bad Values or 0 > Then Finding the mimimum row_wise and column_wise using minimum function > > Thanks > > > > > > _______________________________________________ > Perldl mailing list > [email protected] > http://mailman.jach.hawaii.edu/mailman/listinfo/perldl > > > _______________________________________________ > Perldl mailing list > [email protected] > http://mailman.jach.hawaii.edu/mailman/listinfo/perldl
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