Mrthfx

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Using Non-Negative Matrix Factorization (NNMF)

Noise filtering in images with PCAHow to do Independent Component Analysis?Solving optimal control problem when input is constrainedMore efficient matrix-vector productWhy is arithmetic faster for inexact arithmetic?Finite precision - how to get rid of “near zeros”What if do NOT want Mathematica to normalize eigenvectors with Eigenvectors[N[matrix]]?LU factorizationGetting mathematica to factor out constants and apply euler's identity to a tableSpeed up construction of a dense matrixNonlinear Markov chain (numerical simulation)Solving a linear system with a badly conditioned matrixNearest non-collinear/non-coplanar points

3

$begingroup$

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:

Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/NonNegativeMatrixFactorization.m"]

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;

Dimensions[x]

MatrixRank[x]

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];

Dimensions[w]

Dimensions[h]

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?

matrix linear-algebra

share|improve this question

edited 1 hour ago

bill s

asked 5 hours ago

bill sbill s

53.7k376153

$endgroup$

• 
  
  
  

  
  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
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @HenrikSchumacher beat me to it! (BTW, the automatic precision goal is 4.)
  $endgroup$
  – Anton Antonov
  4 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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:

Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/NonNegativeMatrixFactorization.m"]

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;

Dimensions[x]

MatrixRank[x]

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];

Dimensions[w]

Dimensions[h]

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?

matrix linear-algebra

share|improve this question

edited 1 hour ago

bill s

asked 5 hours ago

bill sbill s

53.7k376153

$endgroup$

• 
  
  
  

  
  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
  5 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @HenrikSchumacher beat me to it! (BTW, the automatic precision goal is 4.)
  $endgroup$
  – Anton Antonov
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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:

Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/NonNegativeMatrixFactorization.m"]

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;

Dimensions[x]

MatrixRank[x]

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];

Dimensions[w]

Dimensions[h]

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?

matrix linear-algebra

share|improve this question

edited 1 hour ago

bill s

asked 5 hours ago

bill sbill s

53.7k376153

$endgroup$

Import["https://raw.githubusercontent.com/antononcube/MathematicaForPrediction/master/NonNegativeMatrixFactorization.m"]

xKer = RandomInteger[{0, 10}, {5, 5}];

xL = RandomInteger[{0, 10}, {50, 5}];

xR = RandomInteger[{0, 10}, {5, 100}];

x = xL.xKer.xR;

Dimensions[x]

MatrixRank[x]

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

Dimensions[w]

Dimensions[h]

Norm[w.h - x]

share|improve this question

edited 1 hour ago

bill s

asked 5 hours ago

bill sbill s

53.7k376153

share|improve this question

edited 1 hour ago

bill s

asked 5 hours ago

bill sbill s

53.7k376153

asked 5 hours ago

bill sbill s

53.7k376153

asked 5 hours ago

bill sbill s

53.7k376153

1 Answer
1

active

oldest

votes

5

$begingroup$

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)],

   nSteps++;

   ...

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

Here is an example based on questions code:

SeedRandom[2343]

xKer = RandomInteger[{0, 10}, {5, 5}];

xL = RandomInteger[{0, 10}, {50, 5}];

xR = RandomInteger[{0, 10}, {5, 100}];

x = xL.xKer.xR;

Dimensions[x]

MatrixRank[x]



(* {50, 100} *)



(* 5 *)



Options[GDCLS]



(* {"MaxSteps" -> 200, "NonNegative" -> True, 

 "Epsilon" -> 1.*10^-9, "RegularizationParameter" -> 0.01, 

 PrecisionGoal -> Automatic, "PrintProfilingInfo" -> False} *)



AbsoluteTiming[

 {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 *)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Anton AntonovAnton Antonov

23.9k167114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  58 mins ago
  
  
  

  
  
  
  

add a comment | 

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5

$begingroup$

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)],

   nSteps++;

   ...

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

Here is an example based on questions code:

SeedRandom[2343]

xKer = RandomInteger[{0, 10}, {5, 5}];

xL = RandomInteger[{0, 10}, {50, 5}];

xR = RandomInteger[{0, 10}, {5, 100}];

x = xL.xKer.xR;

Dimensions[x]

MatrixRank[x]



(* {50, 100} *)



(* 5 *)



Options[GDCLS]



(* {"MaxSteps" -> 200, "NonNegative" -> True, 

 "Epsilon" -> 1.*10^-9, "RegularizationParameter" -> 0.01, 

 PrecisionGoal -> Automatic, "PrintProfilingInfo" -> False} *)



AbsoluteTiming[

 {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 *)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Anton AntonovAnton Antonov

23.9k167114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  58 mins ago
  
  
  

  
  
  
  

add a comment | 

5

$begingroup$

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)],

   nSteps++;

   ...

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

Here is an example based on questions code:

SeedRandom[2343]

xKer = RandomInteger[{0, 10}, {5, 5}];

xL = RandomInteger[{0, 10}, {50, 5}];

xR = RandomInteger[{0, 10}, {5, 100}];

x = xL.xKer.xR;

Dimensions[x]

MatrixRank[x]



(* {50, 100} *)



(* 5 *)



Options[GDCLS]



(* {"MaxSteps" -> 200, "NonNegative" -> True, 

 "Epsilon" -> 1.*10^-9, "RegularizationParameter" -> 0.01, 

 PrecisionGoal -> Automatic, "PrintProfilingInfo" -> False} *)



AbsoluteTiming[

 {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 *)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Anton AntonovAnton Antonov

23.9k167114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  58 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

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)],

   nSteps++;

   ...

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

Here is an example based on questions code:

SeedRandom[2343]

xKer = RandomInteger[{0, 10}, {5, 5}];

xL = RandomInteger[{0, 10}, {50, 5}];

xR = RandomInteger[{0, 10}, {5, 100}];

x = xL.xKer.xR;

Dimensions[x]

MatrixRank[x]



(* {50, 100} *)



(* 5 *)



Options[GDCLS]



(* {"MaxSteps" -> 200, "NonNegative" -> True, 

 "Epsilon" -> 1.*10^-9, "RegularizationParameter" -> 0.01, 

 PrecisionGoal -> Automatic, "PrintProfilingInfo" -> False} *)



AbsoluteTiming[

 {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 *)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Anton AntonovAnton Antonov

23.9k167114

$endgroup$

 ....

 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)],

   nSteps++;

   ...

SeedRandom[2343]

xKer = RandomInteger[{0, 10}, {5, 5}];

xL = RandomInteger[{0, 10}, {50, 5}];

xR = RandomInteger[{0, 10}, {5, 100}];

x = xL.xKer.xR;

Dimensions[x]

MatrixRank[x]



(* {50, 100} *)



(* 5 *)



Options[GDCLS]



(* {"MaxSteps" -> 200, "NonNegative" -> True, 

 "Epsilon" -> 1.*10^-9, "RegularizationParameter" -> 0.01, 

 PrecisionGoal -> Automatic, "PrintProfilingInfo" -> False} *)



AbsoluteTiming[

 {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 *)

share|improve this answer

edited 4 hours ago

answered 4 hours ago

Anton AntonovAnton Antonov

23.9k167114

answered 4 hours ago

Anton AntonovAnton Antonov

23.9k167114

answered 4 hours ago

Anton AntonovAnton Antonov

23.9k167114