I am trying to understand NNMF (Non-Negative Matrix Factorization). This is not a built-in function in Mathematica, but there is a package that implements it, which is refered to in this post. The package is loaded by:


The problem that NNMF tries to solve is this: given a matrix $X$, factor it as $W.H$ where $W$ and $H$ both have all positive entries.

But when I try to apply this using the package, I cannot figure out what is happening. First, construct a matrix $x$ -- I build it random, but of low rank (rank 5):

xKer = RandomInteger[{0, 10}, {5, 5}];
xL = RandomInteger[{0, 10}, {50, 5}];
xR = RandomInteger[{0, 10}, {5, 100}];
x = xL.xKer.xR;

So you can see $x$ is 50 by 100, but is of rank only 5. Applying the NNMF command from the package:

{w, h} = GDCLS[x, 5, "MaxSteps" -> 1000];

So we can see that $w.h$ has the same dimensions as $x$. But

Norm[w.h - x]

is very large, so $w.h$ is not a good approximation to $x$.

Thus my questions: why doesn't NNMF seem to work? Am I expecting the wrong thing?

  • 4
    $\begingroup$ Maybe x simply cannot be factored this way? Moreover, it is more realistic to condsider a relative error measure. E.g., Norm[w.h - x, "Frobenius"]/Norm[x, "Frobenius"] returns 0.00326206 which is not that bad... With MaxSteps -> 10000, one can get down to 0.00075928 or so. $\endgroup$ – Henrik Schumacher Mar 6 at 20:41
  • $\begingroup$ If you create x = xL.xR then it for sure can be expressed as w.h, and there is still significant error in the Norm. But maybe you are right, the error is small compared to the size of x. $\endgroup$ – bill s Mar 6 at 20:48
  • 1
    $\begingroup$ @HenrikSchumacher beat me to it! (BTW, the automatic precision goal is 4.) $\endgroup$ – Anton Antonov Mar 6 at 20:58

Thank you for using that package!

The stopping criteria is based on relative precision. Find the lines:

 normV = Norm[V, "Frobenius"]; diffNorm = 10 normV;
 If[ pgoal === Automatic, pgoal = 4 ];      
 While[nSteps < maxSteps && TrueQ[! NumberQ[pgoal] || NumberQ[pgoal] && (normV > 0) && diffNorm/normV > 10^(-pgoal)],

in the implementation code. Note the condition diffNorm/normV > 10^(-pgoal).

Here is an example based on question’s code:

xKer = RandomInteger[{0, 10}, {5, 5}];
xL = RandomInteger[{0, 10}, {50, 5}];
xR = RandomInteger[{0, 10}, {5, 100}];
x = xL.xKer.xR;

(* {50, 100} *)

(* 5 *)


(* {"MaxSteps" -> 200, "NonNegative" -> True, 
 "Epsilon" -> 1.*10^-9, "RegularizationParameter" -> 0.01, 
 PrecisionGoal -> Automatic, "PrintProfilingInfo" -> False} *)

 {w, h} = GDCLS[x, 5, PrecisionGoal -> 3, "MaxSteps" -> 100000];
 {Dimensions[w], Dimensions[h]}

(* {19.759, {{50, 5}, {5, 100}}} *)

Norm[w.h - x]/Norm[x]

(* 0.000939317 *)
  • 1
    $\begingroup$ This algorithm seems to be a bit slow. 100000 iterations is quite a lot. I am pretty sure that one can make a semi-smooth Newton method with much higher convergence rate work for the underlying optimization problem. If you like, I can elaborate on this. Are you interested? $\endgroup$ – Henrik Schumacher Mar 6 at 22:59
  • $\begingroup$ Thanks for writing the package! And of course, thanks also for helping me understand how to use it. I am hoping to replace some SVD calculations with NNMF. $\endgroup$ – bill s Mar 7 at 0:04
  • $\begingroup$ @bills Thanks, good to hear! You might be also interested in Independent Component Analysis (ICA) discussed (together with NNMF) in MSE's question "How to do Independent Component Analysis?", and the Community posts "Independent component analysis for multidimensional signals" and "Comparison of dimension reduction algorithms over mandala images generation". $\endgroup$ – Anton Antonov Mar 7 at 0:24
  • 1
    $\begingroup$ @HenrikSchumacher 1) Yes, this NNMF implementation is slow and NNMF should be fast (enough) since it is usually run several times, since NNMF is prone to go into local minima. 2) "100000 iterations is quite a lot." -- I mostly use NNMF to NLP (topic extraction) and I rarely run NNMF more than 12-20 steps. 3) Of course, all improvement suggestions are welcome. I would say, it would be best if you write a package and post it in GitHub. $\endgroup$ – Anton Antonov Mar 7 at 0:47

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.