Thanks Chris and Derek for the optimized method
For unique I was going to use squaretotri() and then sort it
The rle had went out of my mind...
I have a doubt, is it efficient to use the qsort method in large matrix (
say 5000 x 5000)
On Sat, Oct 18, 2014 at 11:21 PM, Chris Marshall devel.chm...@gmail.com
wrote:
Nice, Derek! Good catch (it is pretty cool to
see how powerful a few lines of PDL can be).
--Chris
On Fri, Oct 17, 2014 at 3:19 PM, Derek Lamb de...@boulder.swri.edu
wrote:
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 devel.chm...@gmail.com
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 ronagra...@gmail.com
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);
[image: [a b c d e; b c d e f; c d e f g; d e f g h; e f g h i].]
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
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