1
$\begingroup$

I have a matrix with thousands of rows and want the submatrix comprising the rows of the original matrix that have, e.g. a negative element in column 3. How to do that?

Improve this question
$\endgroup$
7
$\begingroup$

I think this question is reasonably a duplicate of: How to find rows that have maximum value? (or several similar questions) and I will delete this answer if it is closed as such. Nevertheless, again for reference:

SeedRandom[1];
a = RandomInteger[{-3, 5}, {20, 5}]

Pick[a, Negative @ a[[All, 3]]]

{{-2, 1, -3, 4, -3}, {-2, 0, -1, -2, 3}, {1, 0, -3, -2, 0}, {2, 0, -3, 0, -1},
 {0, 2, -2, 2, -1}, {0, -2, -3, 1, 1}, {-2, 2, -1, 4, 5}, {0, -1, -2, -2, 3},
 {2, 3, -3, 4, -2}, {0, -2, -1, 2, 5}, {1, 4, -3, 4, 4}}

Faster in versions 8+ should be to use UnitStep, due to packed array optimizations:

Pick[a, UnitStep @ a[[All, 3]], 0]

In version 7 optimal speed may be had with:

a[[SparseArray[BitXor[UnitStep @ a[[All, 3]], 1]]["AdjacencyLists"]]]

Timings compared to other methods proposed, in version 7:

a = RandomInteger[{-3, 5}, {1500000, 5}];

Pick[a, Negative @ a[[All, 3]]] // Timing // First

a[[SparseArray[BitXor[UnitStep @ a[[All, 3]], 1]]["AdjacencyLists"]]] // Timing // First

(col3 = a[[All, 3]]; rowsToGet = Flatten[Position[col3, _?((# < 0) &)]]; a[[rowsToGet]]) //
   Timing // First

Select[a, #[[3]] < 0 &] // Timing // First

0.234

0.031

0.905

1.17

Improve this answer
$\endgroup$
10
  • $\begingroup$ I'm curious, what computer do you run? For Pick picks a second almost on my i7 (+ Win7 and Mma 9). Can I tweak my system any towards the performance of yours? Another slower method, by a factor of 3.5 with my timing, is to: Scan with levelspec {-2} and Sow rows with negative third element. $\endgroup$ – BoLe May 20 '13 at 14:07
  • $\begingroup$ @BoLe I use an i5-2500K running at a maximum (single core) of 4.6 GHz. I know there have been changes to Pick since version 7 but I thought they would make things faster rather than slower. $\endgroup$ – Mr.Wizard May 20 '13 at 14:50
  • $\begingroup$ @BoLe I added two new methods optimized for speed. Please try both and let me know how they perform. I expect Pick[a, UnitStep @ a[[All, 3]], 0] to be best on v9. $\endgroup$ – Mr.Wizard May 20 '13 at 14:58
  • $\begingroup$ @Mr.Wizard I get (on v9.0.0) {1.896415, 0.146656, 3.255081, 3.815207} for your 4 tested values and 0.051483 for Pick[a, UnitStep @ a[[All, 3]]] $\endgroup$ – Jacob Akkerboom May 20 '13 at 16:57
  • $\begingroup$ @Mr.Wizard It must be clocking and a newer model I guess, mine is a bit old already (2009) and running at 2.93 GHz. My times are {0.952, 0.0780, 1.84, 2.04} which make approximately same ratios as yours, somewhere 10 : 1 : 25 : 30. P.S. Pick with UnitStep is the fastest here with 0.047. $\endgroup$ – BoLe May 20 '13 at 17:51
4
$\begingroup$ 
m  = RandomInteger[{-1, 1}, {10, 10}];
m1 = Select[m, #[[3]] < 0 &];

Show it:

ArrayPlot[#, ColorRules -> {1 -> Red, 0 -> Blue, -1 -> Yellow, _ -> Gray}] & /@ {m, m1}

enter image description here

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.