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I have huge matrices in the form of

mtx1 = {{24+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8]},{24+24 FF[5,10] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[7,8] GG[7,8]},{24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]+24 FF[7,8] GG[7,8],24+24 FF[5,10] GG[5,10]+24 FF[6,9] GG[5,10]+24 FF[7,8] GG[5,10]+24 FF[5,10] GG[6,9]+24 FF[6,9] GG[6,9]+24 FF[7,8] GG[6,9]+24 FF[5,10] GG[7,8]+24 FF[6,9] GG[7,8]}};

but the matrices I use are much bigger. Now I want to get rid of each term that contains FF[___] or GG[___]. Both always come together, therefore I used

mtx2 = mtx1 /. FF[___] -> 0; 
(* mtx2={{24, 24, 24}, {24, 24, 24}, {24, 24, 24}} *)

and got the desired result in mtx2. Unfortunatly it turns out that this zeroing is extremly time-consuming. For my huge matrices, it takes on the order of 100 seconds.

Question:

Is there a more time-efficient way to zero all FF[___]-terms in mtx1?

Comparison:

I compare several approaches, for a big 3 big test-matrix. The approaches also include the construction of the matrix.

my original approach

  • {174.8417751, 65.4913582, 25.3878123} seconds

Coefficient-Creation of Matrix

  • {134.4920621, 51.4260521sec, 19.6079772} seconds

belisarius' Block-evaluation methode

  • {82.3675688, 31.5639078, 12.3822025} seconds

eldo's Join/Partition

  • {77.8615328, 29.0973367, 11.2742769} seconds

kguler's Block-evaluation

  • {75.8906436, 29.1315892, 11.6544345} seconds

Mr.Wizard's mtx1[[All, All, 1]]

  • {75.5589726, 29.0378220, 11.9491954} seconds

Edit

The full problem, including the matrix-creation is posted here: Time-efficient creation of matrix

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  • $\begingroup$ Why not a scope-aware redefinition FF[x__]:=0? $\endgroup$
    – Dr. belisarius
    Aug 12, 2014 at 16:18
  • $\begingroup$ Thought about that aswell. I am using FF[__] as undefined parameters, which with I continue calculating later. it should stay undefined. But maybe an intermediate definition and evaluation? $\endgroup$
    – Mario Krenn
    Aug 12, 2014 at 16:20
  • 1
    $\begingroup$ Block[{FF}, FF[__] = 0; a = mtx1] $\endgroup$
    – Dr. belisarius
    Aug 12, 2014 at 16:40
  • $\begingroup$ Wow, even the matrix-construction is slower, your Block-evaluation outperforms my approach alot, see above. To have a fair comparison, i should also post the matrix construction method - for this i need some time. $\endgroup$
    – Mario Krenn
    Aug 12, 2014 at 16:57
  • $\begingroup$ ... or a = Block[{F = 0 &}, mtx1] $\endgroup$
    – kglr
    Aug 13, 2014 at 10:56

2 Answers 2

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For the example you gave FF[___] and GG[___] are the only non-number terms, therefore by polynomial sort order you could use simply:

mtx1[[All, All, 1]]

{{24, 24, 24}, {24, 24, 24}, {24, 24, 24}}

I shall now look at your newer question where I anticipate a more representative example.

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3
  • $\begingroup$ LOL, you don't mind that I leave my answer for comparison purposes? $\endgroup$
    – eldo
    Aug 12, 2014 at 20:16
  • $\begingroup$ @eldo I didn't notice you were doing the same thing but verbosely. I found the question odd so I did no more than glance at your answer before posting. For some reason I thought you were doing something with Position but I misread. $\endgroup$
    – Mr.Wizard
    Aug 12, 2014 at 20:20
  • $\begingroup$ See above, your approach seems to be the fastest one. Nice solution, thanks! $\endgroup$
    – Mario Krenn
    Aug 13, 2014 at 11:47
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Given the structure of your SQUARE example matrix this should be fast:

dim = First @ Dimensions @ mtx1

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tup = Join [#, {1}] & /@ Tuples[Range@dim, 2]

{{1, 1, 1}, {1, 2, 1}, {1, 3, 1}, {2, 1, 1}, {2, 2, 1}, {2, 3, 1}, {3,1, 1}, {3, 2, 1}, {3, 3, 1}}

Partition[mtx1[[#1, #2, #3]] & @@@ tup, dim]

{{24, 24, 24}, {24, 24, 24}, {24, 24, 24}}

Oneliner:

With[{dim = First@Dimensions@mtx1}, 
 Partition[mtx1[[#1, #2, #3]] & @@@ #, dim] &[
  Join [#, {1}] & /@ Tuples[Range@dim, 2]]]
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1
  • $\begingroup$ This works very fast, thanks. I will write the compareing results down later. I posted the full problem (construction of a matrix, see above), for which this is a subproblem. Maybe you also have a solution for that? Thanks!! $\endgroup$
    – Mario Krenn
    Aug 12, 2014 at 17:35

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