dNMF)In this vignette, we consider approximating a binary or non-negative matrix as a product of two non-negative low-rank matrices (a.k.a., factor matrices).
Test data is available from toyModel.
library("dcTensor")## Warning: no DISPLAY variable so Tk is not availableX <- dcTensor::toyModel("dNMF")You will see that there are five blocks in the data matrix as follows.
suppressMessages(library("fields"))
image.plot(X, main="Original Data", legend.mar=8)
Here, we consider the approximation of a binary data matrix \(X\) (\(N \times M\)) as a matrix product of \(U\) (\(N \times J\)) and \(V\) (\(M \times J\)):
\[ X \approx U V' \ \mathrm{s.t.}\ U,V \in \{0,1\} \]
This is known as binary matrix factorization (BMF). Zhang (2007) et al. developed BMF by adding
binary regularization term to non-negative matrix factorization (NMF
(Lee and Seung 1999; CICHOCK 2009)). For
the details of NMF, see also NMF function of nnTensor
package.
In BMF, a rank parameter \(J\)
(\(\leq \min(N, M)\)) 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 dNMF, see ?dNMF. After the calculation,
various objects are returned by dNMF.
set.seed(123456)
out_BMF <- dNMF(X, Bin_U=1, Bin_V=1, J=5)
str(out_BMF, 2)## List of 6
## $ U : num [1:100, 1:5] 0.979 0.979 0.979 0.979 0.979 ...
## $ V : num [1:300, 1:5] 0.999 0.999 0.999 0.999 0.999 ...
## $ RecError : Named num [1:101] 1.00e-09 1.34e+01 1.34e+01 1.34e+01 1.34e+01 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TrainRecError: Named num [1:101] 1.00e-09 1.34e+01 1.34e+01 1.34e+01 1.34e+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 1.19e-04 1.43e-04 1.49e-04 1.48e-04 ...
## ..- 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_BMF$RecError[-1]), type="b", main="Reconstruction Error")
plot(log10(out_BMF$RelChange[-1]), type="b", main="Relative Change")
The product of \(U\) and \(V\) shows whether the original data is
well-recovered by dNMF.
recX <- out_BMF$U %*% t(out_BMF$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 \(U\) and \(V\) show that both \(U\) and \(V\) take values close to 0 and 1.
layout(t(1:2))
hist(out_BMF$U, breaks=100)
hist(out_BMF$V, breaks=100)
Note that these \(U\) and \(V\) do not always take the values of 0 and 1 completely. This is because the binarization in BMF is based on the regularization to softly set the values as close to {0,1} as possible, and is not a hard binarization.
head(out_BMF$U)## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.979044 0 0 3.765852e-104 0
## [2,] 0.979044 0 0 4.277338e-106 0
## [3,] 0.979044 0 0 2.301292e-104 0
## [4,] 0.979044 0 0 1.337948e-104 0
## [5,] 0.979044 0 0 9.347010e-105 0
## [6,] 0.979044 0 0 8.698401e-105 0head(out_BMF$V)## [,1] [,2] [,3] [,4] [,5]
## [1,] 0.9988044 1.072698e-56 0 0.0013906118 0.000000e+00
## [2,] 0.9988248 5.888980e-54 0 0.0003300704 1.735858e-209
## [3,] 0.9988038 4.861490e-57 0 0.0014216661 0.000000e+00
## [4,] 0.9988042 5.637380e-58 0 0.0014015649 0.000000e+00
## [5,] 0.9988029 1.288196e-54 0 0.0014663748 0.000000e+00
## [6,] 0.9988025 6.724228e-58 0 0.0014901051 0.000000e+00If you want to get the {0,1} values, use the round
function as below:
head(round(out_BMF$U, 0))## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 0 0
## [2,] 1 0 0 0 0
## [3,] 1 0 0 0 0
## [4,] 1 0 0 0 0
## [5,] 1 0 0 0 0
## [6,] 1 0 0 0 0head(round(out_BMF$V, 0))## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 0 0 0
## [2,] 1 0 0 0 0
## [3,] 1 0 0 0 0
## [4,] 1 0 0 0 0
## [5,] 1 0 0 0 0
## [6,] 1 0 0 0 0Next, we consider the approximation of a non-negative data matrix \(X\) (\(N \times M\)) as the matrix product of binary matrix \(U\) (\(N \times J\)) and non-negative matrix \(V\) (\(M \times J\)):
\[ X \approx U V' \ \mathrm{s.t.}\ U \in \{0,1\}, V \geq 0 \]
Here, we define this formalization as semi-binary matrix factorization (SBMF). SBMF can capture discrete patterns from a non-negative matrix.
To demonstrate SBMF, next we use a non-negative matrix from the
nnTensor package.
suppressMessages(library("nnTensor"))
X2 <- nnTensor::toyModel("NMF")You will see that there are five blocks in the data matrix as follows.
image.plot(X2, main="Original Data", legend.mar=8)
Switching from BMF to SBMF is quite easy; SBMF is achieved by specifying the binary regularization parameter as a large value like below:
set.seed(123456)
out_SBMF <- dNMF(X2, Bin_U=1E+6, J=5)
str(out_SBMF, 2)## List of 6
## $ U : num [1:100, 1:5] 0.978 0.984 0.985 0.988 0.977 ...
## $ V : num [1:300, 1:5] 98.5 100.7 100 101.5 100.6 ...
## $ RecError : Named num [1:101] 1.00e-09 2.99e+03 2.92e+03 2.83e+03 2.77e+03 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TrainRecError: Named num [1:101] 1.00e-09 2.99e+03 2.92e+03 2.83e+03 2.77e+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 2.60e-01 2.44e-02 3.18e-02 2.27e-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_SBMF$RecError[-1]), type="b", main="Reconstruction Error")
plot(log10(out_SBMF$RelChange[-1]), type="b", main="Relative Change")
The product of \(U\) and \(V\) shows whether the original data is
well-recovered by dNMF.
recX2 <- out_SBMF$U %*% t(out_SBMF$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 \(U\) and \(V\) show that \(U\) looks binary but \(V\) does not.
layout(t(1:2))
hist(out_SBMF$U, breaks=100)
hist(out_SBMF$V, breaks=100)
Finally, we expand the binary regularization to ternary regularization to take {0,1,2} values as below:
\[ X \approx U V' \ \mathrm{s.t.}\ U \in \{0,1,2\}, V \geq 0, \] where \(X\) (\(N \times M\)) is a non-negative data matrix, \(U\) (\(N \times J\)) is a ternary matrix, and \(V\) (\(M \times J\)) is a non-negative matrix.
STMF is achieved by specifying the ternary regularization parameter as a large value like the below:
set.seed(123456)
out_STMF <- dNMF(X2, Ter_U=1E+6, J=5)
str(out_STMF, 2)## List of 6
## $ U : num [1:100, 1:5] 2.02 2.02 2.02 2.02 2.02 ...
## $ V : num [1:300, 1:5] 48.7 49.8 49.5 50.2 49.8 ...
## $ RecError : Named num [1:101] 1.00e-09 2.52e+03 2.58e+03 2.59e+03 2.58e+03 ...
## ..- attr(*, "names")= chr [1:101] "offset" "1" "2" "3" ...
## $ TrainRecError: Named num [1:101] 1.00e-09 2.52e+03 2.58e+03 2.59e+03 2.58e+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 1.22e-01 2.12e-02 3.09e-03 2.03e-03 ...
## ..- 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_STMF$RecError[-1]), type="b", main="Reconstruction Error")
plot(log10(out_STMF$RelChange[-1]), type="b", main="Relative Change")
The product of \(U\) and \(V\) shows that the original data is
well-recovered by dNMF.
recX <- out_STMF$U %*% t(out_STMF$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 \(U\) and \(V\) show that \(U\) looks ternary but \(V\) does not.
layout(t(1:2))
hist(out_STMF$U, breaks=100)
hist(out_STMF$V, breaks=100)
## R version 4.3.0 (2023-04-21)
## Platform: x86_64-pc-linux-gnu (64-bit)
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## [5] spam_2.9-1 dcTensor_1.2.0
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