dNMTF)In this vignette, we consider approximating a binary or non-negative matrix as a product of three non-negative low-rank matrices (a.k.a., factor matrices).
Test data is available from toyModel.
You will see that there are five blocks in the data matrix as follows.
Here, we consider the approximation of a binary data matrix \(X\) (\(N \times M\)) as a matrix product of \(U\) (\(N \times J1\)), \(S\) (\(J1 \times J2\)), and \(V\) (\(M \times J2\)):
\[ X \approx U S V' \ \mathrm{s.t.}\ U,V \in \{0,1\}, S \geq 0 \]
Here, we call this Binary Matrix Tri-Factorization (BMTF). BMTF is
based on Non-negative Matrix Tri-Factorization (NMTF (Copar e2019; Long 2005; Ding 2006)) and Binary
Matrix Factorization (BMF (Zhang 2007)).
For the details of NMTF, see also NMTF function of nnTensor
package.
In BMTF, two rank parameters \(J1\)
(\(\leq N\)) and \(J2\) (\(\leq
M)\)) is needed to be set in advance. Other settings such as the
number of iterations (num.iter) or factorization algorithm
(algorithm) are also available. For the details of
arguments of dNMTF, see ?dNMTF. After the calculation,
various objects are returned by dNMTF.
## List of 8
## $ U : num [1:100, 1:5] 0.124 0.124 0.124 0.124 0.124 ...
## $ S : num [1:5, 1:5] 9.54e-15 7.71e-02 7.65e-01 4.55 3.01 ...
## $ V : num [1:300, 1:5] 9.80e-08 2.12e-13 1.36e-56 1.07e-07 1.34e-11 ...
## $ rank : num [1:2] 5 5
## $ RecError : Named num [1:101] 1.00e-09 6.99e+01 6.59e+01 6.35e+01 6.19e+01 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TrainRecError: Named num [1:101] 1.00e-09 6.99e+01 6.59e+01 6.35e+01 6.19e+01 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TestRecError : Named num [1:101] 1e-09 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ RelChange : Named num [1:101] 1.00e-09 7.22 6.06e-02 3.89e-02 2.46e-02 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...The reconstruction error (RecError) and relative error
(RelChange, the amount of change from the reconstruction
error in the previous step) can be used to diagnose whether the
calculation is converged or not.
layout(t(1:2))
plot(log10(out_BMTF$RecError[-1]), type="b", main="Reconstruction Error")
plot(log10(out_BMTF$RelChange[-1]), type="b", main="Relative Change")
The product of \(U\), \(S\), and \(V\) shows whether the original data is
well-recovered by dNMTF.
recX <- out_BMTF$U %*% out_BMTF$S %*% t(out_BMTF$V)
layout(t(1:2))
image.plot(X, main="Original Data", legend.mar=8)
image.plot(recX, main="Reconstructed Data (BMF)", legend.mar=8)
The histograms of \(U\), \(S\), and \(V\) show that these take values close to 0 and 1.
layout(t(1:3))
hist(out_BMTF$U, breaks=100)
hist(out_BMTF$S, breaks=100)
hist(out_BMTF$V, breaks=100)
Note that these factor matrices do not always take the values of 0 and 1 completely. This is because the binarization in BMTF is based on the regularization to softly set the values as close to {0,1} as possible, and is not a hard binarization.
## [,1] [,2] [,3]
## [1,] 0.1238387 2.015980e-45 8.933262e-20
## [2,] 0.1238387 3.694257e-44 5.997028e-20
## [3,] 0.1238387 3.067534e-44 7.937174e-19## [,1] [,2] [,3] [,4] [,5]
## [1,] 9.535194e-15 7.902857e+00 1.213441e-40 0.006942176 8.308241e-06
## [2,] 7.707013e-02 4.897875e-28 7.171875e+00 0.001055494 1.150494e-01
## [3,] 7.649903e-01 8.425523e-34 2.629017e-01 0.021129672 3.997469e-01
## [4,] 4.546750e+00 1.047616e-15 8.178169e-02 0.173047727 1.173014e-01
## [5,] 3.014648e+00 3.017382e-31 2.199562e-01 0.100951552 1.204483e-01## [,1] [,2] [,3]
## [1,] 9.798525e-08 1.002167 7.419010e-10
## [2,] 2.121454e-13 1.002167 6.227816e-12
## [3,] 1.362156e-56 1.001178 4.707794e-72If you want to get the {0,1} values, use the round
function as below:
## [,1] [,2] [,3]
## [1,] 0 0 0
## [2,] 0 0 0
## [3,] 0 0 0## [,1] [,2] [,3] [,4] [,5]
## [1,] 0 8 0 0 0
## [2,] 0 0 7 0 0
## [3,] 1 0 0 0 0
## [4,] 5 0 0 0 0
## [5,] 3 0 0 0 0## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 1 0
## [3,] 0 1 0Next, we consider the approximation of a non-negative data matrix \(X\) (\(N \times M\)) as the matrix product of binary matrix \(U\) (\(N \times J1\)) and non-negative matrices, \(S\) (\(J1 \times J2\)) and \(V\) (\(M \times J2\)):
\[ X \approx U S V' \ \mathrm{s.t.}\ U \in \{0,1\}, S, V \geq 0 \]
Here, we define this formalization as Semi-Binary Matrix Tri-Factorization (SBMTF). SBMTF can capture discrete patterns from a non-negative matrix.
To demonstrate SBMTF, next we use a non-negative matrix from the
nnTensor package.
You will see that there are five blocks in the data matrix as follows.
Switching from BMTF to SBMTF is quite easy; SBMTF is achieved by specifying the binary regularization parameter as a large value like below:
## List of 8
## $ U : num [1:100, 1:5] 0.977 0.983 0.985 0.987 0.975 ...
## $ S : num [1:5, 1:5] 0.000028 0.988919 0.200886 0.101142 0.119332 ...
## $ V : num [1:300, 1:5] 1.60e-10 2.01e-10 7.10e-10 4.99e-09 2.69e-11 ...
## $ rank : num [1:2] 5 5
## $ RecError : Named num [1:101] 1.00e-09 1.02e+04 9.64e+03 9.10e+03 8.67e+03 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TrainRecError: Named num [1:101] 1.00e-09 1.02e+04 9.64e+03 9.10e+03 8.67e+03 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TestRecError : Named num [1:101] 1e-09 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ RelChange : Named num [1:101] 1.00e-09 2.35e-01 6.25e-02 5.94e-02 4.95e-02 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...RecError and RelChange can be used to
diagnose whether the calculation is converged or not.
layout(t(1:2))
plot(log10(out_SBMTF$RecError[-1]), type="b", main="Reconstruction Error")
plot(log10(out_SBMTF$RelChange[-1]), type="b", main="Relative Change")
The product of \(U\), \(S\), and \(V\) shows whether the original data is
well-recovered by dNMTF.
recX2 <- out_SBMTF$U %*% out_SBMTF$S %*% t(out_SBMTF$V)
layout(t(1:2))
image.plot(X2, main="Original Data", legend.mar=8)
image.plot(recX2, main="Reconstructed Data (SBMF)", legend.mar=8)
The histograms of \(U\), \(S\), and \(V\) show that \(U\) looks binary but \(S\) and \(V\) do not.
layout(t(1:3))
hist(out_SBMTF$U, breaks=100)
hist(out_SBMTF$S, breaks=100)
hist(out_SBMTF$V, breaks=100)
Finally, we expand the binary regularization to ternary regularization to take {0,1,2} values as below:
\[ X \approx U S V' \ \mathrm{s.t.}\ U \in \{0,1,2\}, S, V \geq 0, \] where \(X\) (\(N \times M\)) is a non-negative data matrix, \(U\) (\(N \times J1\)) is a ternary matrix, and \(S\) (\(J1 \times J2\)) and \(V\) (\(M \times J2\)) are non-negative matrices.
STMTF is achieved by specifying the ternary regularization parameter as a large value like the below:
## List of 8
## $ U : num [1:100, 1:5] 0.963 0.972 0.976 0.98 0.959 ...
## $ S : num [1:5, 1:5] 0.000277 0.86879 0.376385 0.194511 0.153944 ...
## $ V : num [1:300, 1:5] 5.27e-16 6.35e-16 7.18e-16 3.77e-15 1.08e-16 ...
## $ rank : num [1:2] 5 5
## $ RecError : Named num [1:101] 1.00e-09 1.02e+04 9.71e+03 9.26e+03 8.86e+03 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TrainRecError: Named num [1:101] 1.00e-09 1.02e+04 9.71e+03 9.26e+03 8.86e+03 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TestRecError : Named num [1:101] 1e-09 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 0e+00 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ RelChange : Named num [1:101] 1.00e-09 2.40e-01 5.06e-02 4.91e-02 4.51e-02 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...RecError and RelChange can be used to
diagnose whether the calculation is converging or not.
layout(t(1:2))
plot(log10(out_STMTF$RecError[-1]), type="b", main="Reconstruction Error")
plot(log10(out_STMTF$RelChange[-1]), type="b", main="Relative Change")
The product of \(U\), \(S\), and \(V\) shows that the original data is
well-recovered by dNMTF.
recX <- out_STMTF$U %*% out_STMTF$S %*% t(out_STMTF$V)
layout(t(1:2))
image.plot(X2, main="Original Data", legend.mar=8)
image.plot(recX, main="Reconstructed Data (STMF)", legend.mar=8)
The histograms of \(U\), \(S\), and \(V\) show that \(U\) looks ternary but \(S\) and \(V\) do not.
layout(t(1:3))
hist(out_STMTF$U, breaks=100)
hist(out_STMTF$S, breaks=100)
hist(out_STMTF$V, breaks=100)
## R version 4.3.1 (2023-06-16)
## Platform: x86_64-pc-linux-gnu (64-bit)
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## attached base packages:
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## [5] dcTensor_1.3.0
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## loaded via a namespace (and not attached):
## [1] gtable_0.3.4 jsonlite_1.8.7 highr_0.10 compiler_4.3.1
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