For a better understanding of MAKL
library, we
build a simple example in this document. We first create a synthetic
dataset that consists of 1000 rows and 6 features, using standard
Gaussian distribution.
library(MAKL)
set.seed(64327) #midas
df <- matrix(rnorm(6000, 0, 1), nrow = 1000)
colnames(df) <- c("F1", "F2", "F3", "F4", "F5", "F6")
As to membership
argument of makl_train()
,
we prepare a list consisting of two groups such that the first one
contains the features F1, F5 and F6; the second one contains the rest.
Note that the column names of the input dataset should be a superset of
the union of all feature names in the groups
list.
# check colnames(df) for them to be matching with group members
groups <- list()
groups[[1]] <- c("F1", "F5", "F6")
groups[[2]] <- c("F2", "F3", "F4")
We then create the response vector y
such that it will
be dependent on the second, the third and the fourth features, namely
F2, F3 and F4: If, for a data instance, the sum of entries in the
second, the third and the fourth columns is positive, the corresponding
response is assigned +1, else, it is assigned -1.
y <- c()
for(i in 1:nrow(df)) {
if((df[i, 2] + df[i, 3] + df[i, 4]) > 0) {
y[i] <- +1
} else {
y[i] <- -1
}
}
We use the synthetic dataset df
and response vector
y
as our train dataset and train response vector in
makl_train()
, we choose the number of random features
D
equal to 2 which makes sense knowing that our train
dataset is 6 dimensional. We choose the number of rows to be used for
distance matrix calculation, sigma_N
equal to 1000, and
lambda_set
consisting of 0.9, 0.8, 0.7, 0.6 for sparse
solutions. As membership list, we use the groups
list that
we created above.
makl_model <- makl_train(X = df, y = y, D = 2, sigma_N = 1000, CV = 1, membership = groups, lambda_set = c(0.9, 0.8, 0.7, 0.6))
#> Lambda: 155.0901 nr.var: 5
#> Lambda: 137.8579 nr.var: 5
#> Lambda: 120.6257 nr.var: 5
#> Lambda: 103.3934 nr.var: 5
When we check the coefficients of our model, we see that the chosen
kernel for prediction by makl_train()
was the kernel of the
second group. This was an expected result since we created the response
vector y
to be dependent on the second group members of the
groups
list.
makl_model$model$coefficients
#> 155.090126229481 137.857889981761 120.625653734041 103.39341748632
#> [1,] 0.00000000 0.0000000 0.0000000 0.0000000
#> [2,] 0.00000000 0.0000000 0.0000000 0.0000000
#> [3,] 0.00000000 0.0000000 0.0000000 0.0000000
#> [4,] 0.00000000 0.0000000 0.0000000 0.0000000
#> [5,] -0.29314353 -0.5938544 -0.9106226 -1.2539243
#> [6,] 0.06703617 0.1352210 0.2057486 0.2799665
#> [7,] 0.24539658 0.4973664 0.7630398 1.0509792
#> [8,] -0.36108294 -0.7320709 -1.1246002 -1.5535840
#> [9,] 0.12450233 0.1542956 0.1858601 0.2195980
Now, let us create a synthetic dataset df_test
and a
synthetic test response vector y_test
to use in
makl_test()
to check the results.
df_test <- matrix(rnorm(600, 0, 1), nrow = 100)
colnames(df_test) <- c("F1", "F2", "F3", "F4", "F5", "F6")
y_test <- c()
for(i in 1:nrow(df_test)) {
if((df_test[i, 2] + df_test[i, 3] + df_test[i, 4]) > 0) {
y_test[i] <- +1
} else {
y_test[i] <- -1
}
}
result <-makl_test(X = df_test, y = y_test, makl_model = makl_model)
The list result
contains two elements: 1) The
predictions for the test response vector y_test
and 2) The
area under the ROC curve (AUROC) versus the number of selected kernels
values for each element in the lambda_set
if
CV
is not applied; the area under the ROC curve versus the
number of selected kernels value for the best lambda
in the
lambda_set
if CV
is applied.