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`

.

library("dcTensor") X <- 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 $U$ ($N \times J1$), $S$ ($J1 \times J2$), 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 [@nmtf1; @nmtf2; @nmtf3]) and Binary Matrix Factorization (BMF [@bmf]). 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`

.

set.seed(123456) out_BMTF <- dNMTF(X, Bin_U=10, Bin_V=10, rank=c(5,5)) str(out_BMTF, 2)

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.

out_BMTF$U[1:3,1:3] out_BMTF$S out_BMTF$V[1:3,1:3]

If you want to get the {0,1} values, use the `round`

function as below:

round(out_BMTF$U[1:3,1:3], 0) round(out_BMTF$S, 0) round(out_BMTF$V[1:3,1:3], 0)

Next, 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.

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 BMTF to SBMTF is quite easy; SBMTF is achieved by specifying the binary regularization parameter as a large value like below:

set.seed(123456) out_SBMTF <- dNMTF(X2, Bin_U=1E+6, rank=c(5,5)) str(out_SBMTF, 2)

`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:

set.seed(123456) out_STMTF <- dNMTF(X2, Ter_U=1E+5, rank=c(5,5)) str(out_STMTF, 2)

`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)

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