Optimizing K-means clustering using Genetic Algorithm - r

I have the following dataset (obtained here):
----------item survivalpoints weight
1 pocketknife 10 1
2 beans 20 5
3 potatoes 15 10
4 unions 2 1
5 sleeping bag 30 7
6 rope 10 5
7 compass 30 1
I can cluster this dataset into three clusters with kmeans() using a binary string as my initial choice of centers. For eg:
## 1 represents the initial centers
chromosome = c(1,1,1,0,0,0,0)
## exclude first column (kmeans only support continous data)
cl <- kmeans(dataset[, -1], dataset[chromosome == 1, -1])
## check the memberships
cl$clusters
# [1] 1 3 3 1 2 1 2
Using this fundamental concept, I tried it out with GA package to conduct the search where I am trying to optimize(minimize) Davies-Bouldin (DB) Index.
library(GA) ## for ga() function
library(clusterSim) ## for index.DB() function
## defining my fitness function (Davies-Bouldin)
DBI <- function(x) {
## converting matrix to vector to access each row
binary_rep <- split(x, row(x))
## evaluate the fitness of each chromsome
for(each in 1:nrow(x){
cl <- kmeans(dataset, dataset[binary_rep[[each]] == 1, -1])
dbi <- index.DB(dataset, cl$cluster, centrotypes = "centroids")
## minimizing db
return(-dbi)
}
}
g<- ga(type = "binary", fitness = DBI, popSize = 100, nBits = nrow(dataset))
Of course (I have no idea what's happening), I received error message of
Warning messages:
Error in row(x) : a matrix-like object is required as argument to 'row'
Here are my questions:
How can correctly use the GA package to solve my problem?
How can I make sure the randomly generated chromosomes contains the same number of 1s which corresponds to k number of clusters (eg. if k=3 then the chromosome must contain exactly three 1s)?

I can't comment on the sense of combining k-means with ga, but I can point out that you had issue in your fitness function. Also, errors are produced when all genes are on or off, so fitness is only calculated when that is not the case:
DBI <- function(x) {
if(sum(x)==nrow(dataset) | sum(x)==0){
score <- 0
} else {
cl <- kmeans(dataset[, -1], dataset[x==1, -1])
dbi <- index.DB(dataset[,-1], cl=cl$cluster, centrotypes = "centroids")
score <- dbi$DB
}
return(score)
}
g <- ga(type = "binary", fitness = DBI, popSize = 100, nBits = nrow(dataset))
plot(g)
g#solution
g#fitnessValue
Looks like several gene combinations produced the same "best" fitness value

Related

Combinatorial optimization with discrete options in R

I have a function with five variables that I want to maximize using only an specific set of parameters for each variable.
Are there any methods in R that can do this, other than by brutal force? (e.g. Particle Swarm Optimization, Genetic Algorithm, Greedy, etc.). I have read a few packages but they seem to create their own set of parameters from within a given range. I am only interested in optimizing the set of options provided.
Here is a simplified version of the problem:
#Example of 5 variable function to optimize
Fn<-function(x){
a=x[1]
b=x[2]
c=x[3]
d=x[4]
e=x[5]
SUM=a+b+c+d+e
return(SUM)
}
#Parameters for variables to optimize
Vars=list(
As=c(seq(1.5,3, by = 0.3)), #float
Bs=c(1,2), #Binary
Cs=c(seq(1,60, by=10)), #Integer
Ds=c(seq(60,-60, length.out=5)), #Negtive
Es=c(1,2,3)
)
#Full combination
FullCombn= expand.grid(Vars)
Results=data.frame(I=as.numeric(), Sum=as.numeric())
for (i in 1:nrow(FullCombn)){
ParsI=FullCombn[i,]
ResultI=Fn(ParsI)
Results=rbind(Results,c(I=i,Sum=ResultI))
}
#Best iteration (Largest result)
Best=Results[Results[, 2] == max(Results[, 2]),]
#Best parameters
FullCombn[Best$I,]
Two more possibilities. Both minimize by default, so I flip the sign in your objective function (i.e. return -SUM).
#Example of 5 variable function to optimize
Fn<-function(x, ...){
a=x[1]
b=x[2]
c=x[3]
d=x[4]
e=x[5]
SUM=a+b+c+d+e
return(-SUM)
}
#Parameters for variables to optimize
Vars=list(
As=c(seq(1.5,3, by = 0.3)), #float
Bs=c(1,2), #Binary
Cs=c(seq(1,60, by=10)), #Integer
Ds=c(seq(60,-60, length.out=5)), #Negtive
Es=c(1,2,3)
)
First, a grid search. Exactly what you did, just convenient. And the implementation allows you to distribute the evaluations of the objective function.
library("NMOF")
gridSearch(fun = Fn,
levels = Vars)[c("minfun", "minlevels")]
## 5 variables with 6, 2, 6, 5, ... levels: 1080 function evaluations required.
## $minfun
## [1] -119
##
## $minlevels
## [1] 3 2 51 60 3
An alternative: a simple Local Search. You start with a valid initial guess, and then move randomly through possible feasible solutions. The key ingredient is the neighbourhood function. It picks one element randomly and then, again randomly, sets this element to one allowed value.
nb <- function(x, levels, ...) {
i <- sample(length(levels), 1)
x[i] <- sample(levels[[i]], 1)
x
}
(There would be better algorithms for neighbourhood functions; but this one is simple and so demonstrates the idea well.)
LSopt(Fn, list(x0 = c(1.8, 2, 11, 30, 2), ## a feasible initial solution
neighbour = nb,
nI = 200 ## iterations
),
levels = Vars)$xbest
## Local Search.
## ##...
## Best solution overall: -119
## [1] 3 2 51 60 3
(Disclosure: I am the maintainer of package NMOF, which provides functions gridSearch and LSopt.)
In response to the comment, a few remarks on Local Search and the neighbourhood function above (nb). Local Search, as implemented in
LSopt, will start with an arbitrary solution, and
then change that solution slightly. This new solution,
called a neighbour, will be compared (by its
objective-function value) to the old solution. If the new solution is
better, it becomes the current solution; otherwise it
is rejected and the old solution remains the current one.
Then the algorithm repeats, for a number of iterations.
So, in short, Local Search is not random sampling, but
a guided random-walk through the search space. It's
guided because only better solutions get accepted, worse one's get rejected. In this sense, LSopt will narrow down on good parameter values.
The implementation of the neighbourhood is not ideal
for two reasons. The first is that a solution may not
be changed at all, since I sample from feasible
values. But for a small set of possible values as here,
it might often happen that the same element is selected
again. However, for larger search spaces, this
inefficiency is typically negligible, since the
probability of sampling the same value becomes
smaller. Often so small, that the additional code for
testing if the solution has changed becomes more
expensive that the occasionally-wasted iteration.
A second thing could be improved, albeit through a more
complicated function. And again, for this small problem it does not matter. In the current neighbourhood, an
element is picked and then set to any feasible value.
But that means that changes from one solution to the
next might be large. Instead of picking any feasible values of the As,
in realistic problems it will often be better to pick a
value close to the current value. For example, when you are at 2.1, either move to 1.8 or 2.4, but not to 3.0. (This reasoning is only relevant, of course, if the variable in question is on a numeric or at least ordinal scale.)
Ultimately, what implementation works well can be
tested only empirically. Many more details are in this tutorial.
Here is one alternative implementation. A solution is now a vector of positions for the original values, e.g. if x[1] is 2, it "points" to 1.8, if x[2] is 2, it points to 1, and so on.
## precompute lengths of vectors in Vars
lens <- lengths(Vars)
nb2 <- function(x, lens, ...) {
i <- sample(length(lens), 1)
if (x[i] == 1L) {
x[i] <- 2
} else if (x[i] == lens[i]) {
x[i] <- lens[i] - 1
} else
x[i] <- x[i] + sample(c(1, -1), 1)
x
}
## the objective function now needs to map the
## indices in x back to the levels in Vars
Fn2 <- function(x, levels, ...){
y <- mapply(`[`, levels, x)
## => same as
## y <- numeric(length(x))
## y[1] <- Vars[[1]][x[1]]
## y[2] <- Vars[[2]][x[2]]
## ....
SUM <- sum(y)
return(-SUM)
}
xbest <- LSopt(Fn2,
list(x0 = c(1, 1, 1, 1, 1), ## an initial solution
neighbour = nb2,
nI = 200 ## iterations
),
levels = Vars,
lens = lens)$xbest
## Local Search.
## ....
## Best solution overall: -119
## map the solution back to the values
mapply(`[`, Vars, xbest)
## As Bs Cs Ds Es
## 3 2 51 60 3
Here is a genetic algorithm solution with package GA.
The key is to write a function decode enforcing the constraints, see the package vignette.
library(GA)
#> Loading required package: foreach
#> Loading required package: iterators
#> Package 'GA' version 3.2.2
#> Type 'citation("GA")' for citing this R package in publications.
#>
#> Attaching package: 'GA'
#> The following object is masked from 'package:utils':
#>
#> de
decode <- function(x) {
As <- Vars$As
Bs <- Vars$Bs
Cs <- Vars$Cs
Ds <- rev(Vars$Ds)
# fix real variable As
i <- findInterval(x[1], As)
if(x[1L] - As[i] < As[i + 1L] - x[1L])
x[1L] <- As[i]
else x[1L] <- As[i + 1L]
# fix binary variable Bs
if(x[2L] - Bs[1L] < Bs[2L] - x[2L])
x[2L] <- Bs[1L]
else x[2L] <- Bs[2L]
# fix integer variable Cs
i <- findInterval(x[3L], Cs)
if(x[3L] - Cs[i] < Cs[i + 1L] - x[3L])
x[3L] <- Cs[i]
else x[3L] <- Cs[i + 1L]
# fix integer variable Ds
i <- findInterval(x[4L], Ds)
if(x[4L] - Ds[i] < Ds[i + 1L] - x[4L])
x[4L] <- Ds[i]
else x[4L] <- Ds[i + 1L]
# fix the other, integer variable
x[5L] <- round(x[5L])
setNames(x , c("As", "Bs", "Cs", "Ds", "Es"))
}
Fn <- function(x){
x <- decode(x)
# a <- x[1]
# b <- x[2]
# c <- x[3]
# d <- x[4]
# e <- x[5]
# SUM <- a + b + c + d + e
SUM <- sum(x, na.rm = TRUE)
return(SUM)
}
#Parameters for variables to optimize
Vars <- list(
As = seq(1.5, 3, by = 0.3), # Float
Bs = c(1, 2), # Binary
Cs = seq(1, 60, by = 10), # Integer
Ds = seq(60, -60, length.out = 5), # Negative
Es = c(1, 2, 3)
)
res <- ga(type = "real-valued",
fitness = Fn,
lower = c(1.5, 1, 1, -60, 1),
upper = c(3, 2, 51, 60, 3),
popSize = 1000,
seed = 123)
summary(res)
#> ── Genetic Algorithm ───────────────────
#>
#> GA settings:
#> Type = real-valued
#> Population size = 1000
#> Number of generations = 100
#> Elitism = 50
#> Crossover probability = 0.8
#> Mutation probability = 0.1
#> Search domain =
#> x1 x2 x3 x4 x5
#> lower 1.5 1 1 -60 1
#> upper 3.0 2 51 60 3
#>
#> GA results:
#> Iterations = 100
#> Fitness function value = 119
#> Solutions =
#> x1 x2 x3 x4 x5
#> [1,] 2.854089 1.556080 46.11389 49.31045 2.532682
#> [2,] 2.869408 1.638266 46.12966 48.71106 2.559620
#> [3,] 2.865254 1.665405 46.21684 49.04667 2.528606
#> [4,] 2.866494 1.630416 46.12736 48.78017 2.530454
#> [5,] 2.860940 1.650015 46.31773 48.92642 2.521276
#> [6,] 2.851644 1.660358 46.09504 48.81425 2.525504
#> [7,] 2.855078 1.611837 46.13855 48.62022 2.575492
#> [8,] 2.857066 1.588893 46.15918 48.60505 2.588992
#> [9,] 2.862644 1.637806 46.20663 48.92781 2.579260
#> [10,] 2.861573 1.630762 46.23494 48.90927 2.555612
#> ...
#> [59,] 2.853788 1.640810 46.35649 48.87381 2.536682
#> [60,] 2.859090 1.658127 46.15508 48.85404 2.590679
apply(res#solution, 1, decode) |> t() |> unique()
#> As Bs Cs Ds Es
#> [1,] 3 2 51 60 3
Created on 2022-10-24 with reprex v2.0.2

Error in x[[jj]][iseq] <- vjj : replacement has length zero in R (KlaR package)

I have a dataset with 188 columns and 100 rows (plus a header row). I'm trying to apply the kmodes clustering method (from klaR package) in R to this matrix.
There are two types of data in the array data structure: string and binary. Both have null values.
For example:
Q27_history Q28
1 <NA>
<NA> yes, sometimes
function to compute total within-cluster sum of square:
set.seed (96743)
# function to compute total within-cluster sum of square
wss <- function(k) {
sum((kmodes( whois_data, k)$withindiff))
}
# Compute and plot wss for k = 1 to k = 15
k.values <- 2:15
# extract wss for 2-15 clusters
wss_values <- map_dbl(k.values, wss)
print(wss_values)
The text of error:
Error in x[[jj]][iseq] <- vjj : replacement has length zero
Afret that:
Error in print(wss_values) :object 'wss_values' is not found
I've tried to put kmodes(na.fill(data, fill=""), k) in:
wss <- function(k) {
sum((kmodes( whois_data, k)$withindiff))
kmodes(na.fill(data, fill=""), k)
}
But after that library(purrr) stop working and does not find variable map_dbl
How should I inline rows with empty data?
I don't think you can have NAs when using kmodes, it should throw an error:
set.seed(111)
whois_data = data.frame(Q1 = rbinom(100,1,0.5),
Q2 = sample(c("Y","N"),100,replace=TRUE),
Q3 = sample(c(NA,1:3),100,replace=TRUE))
kmodes(whois_data,3)
Error in old.cluster != cluster :
comparison of these types is not implemented
Makes more sense to do kmodes without the NAs :
wss <- function(k,df) {
sum((kmodes(df, k)$withindiff))
}
library(purrr)
map_dbl(2:5, wss,df = whois_data[complete.cases(whois_data),])
[1] 91 58 70 42

Elbow/knee in a curve in R

I've got this data processing:
library(text2vec)
##Using perplexity for hold out set
t1 <- Sys.time()
perplex <- c()
for (i in 3:25){
set.seed(17)
lda_model2 <- LDA$new(n_topics = i)
doc_topic_distr2 <- lda_model2$fit_transform(x = dtm, progressbar = F)
set.seed(17)
sample.dtm2 <- itoken(rawsample$Abstract,
preprocessor = prep_fun,
tokenizer = tok_fun,
ids = rawsample$id,
progressbar = F) %>%
create_dtm(vectorizer,vtype = "dgTMatrix", progressbar = FALSE)
set.seed(17)
new_doc_topic_distr2 <- lda_model2$transform(sample.dtm2, n_iter = 1000,
convergence_tol = 0.001, n_check_convergence = 25,
progressbar = FALSE)
perplex[i] <- text2vec::perplexity(sample.dtm2, topic_word_distribution =
lda_model2$topic_word_distribution,
doc_topic_distribution = new_doc_topic_distr2)
}
print(difftime(Sys.time(), t1, units = 'sec'))
I know there are a lot of questions like this, but I haven't been able to exactly find the answer to my situation. Above you see perplexity calculation from 3 to 25 topic number for a Latent Dirichlet Allocation model. I want to get the most sufficient value among those, meaning that I want to find the elbow or knee, for those values that might only be considered as a simple numeric vector which outcome looks like this:
1 NA
2 NA
3 222.6229
4 210.3442
5 200.1335
6 190.3143
7 180.4195
8 174.2634
9 166.2670
10 159.7535
11 153.7785
12 148.1623
13 144.1554
14 141.8250
15 138.8301
16 134.4956
17 131.0745
18 128.8941
19 125.8468
20 123.8477
21 120.5155
22 118.4426
23 116.4619
24 113.2401
25 114.1233
plot(perplex)
This is how plot looks like
I would say that the elbow would be 13 or 16, but I'm not completely sure and I want the exact number as an outcome. I saw in this paper that f''(x) / (1+f'(x)^2)^1.5 is the knee formula, which I tried like this and says it's 18:
> d1 <- diff(perplex) # first derivative
> d2 <- diff(d1) / diff(perplex[-1]) # second derivative
> knee <- (d2)/((1+(d1)^2)^1.5)
Warning message:
In (d2)/((1 + (d1)^2)^1.5) :
longer object length is not a multiple of shorter object length
> which.min(knee)
[1] 18
I can't fully figure this thing out. Would someone like to share how I could get the exact ideal topics number according to perplexity as an outcome?
Found this: "The LDA model with the optimal coherence score, obtained with an elbow method (the point with maximum absolute second derivative) (...)" in this paper, so this coding does the work: d1 <- diff(perplex); k <- which.max(abs(diff(d1) / diff(perplex[-1])))

How to reduce dimension of gene expression matrix by calculating correlation coefficients?

I am in interested in finding Pearson correlation coefficients between a list of genes. Basically, I have Affymetrix gene level expression matrix (genes in the rows and sample ID on the columns), and I have annotation data of microarray experiment observation where sample ID in the rows and description identification on the columns.
data
> expr_mat[1:8, 1:3]
Tarca_001_P1A01 Tarca_003_P1A03 Tarca_004_P1A04
1_at 6.062215 6.125023 5.875502
10_at 3.796484 3.805305 3.450245
100_at 5.849338 6.191562 6.550525
1000_at 3.567779 3.452524 3.316134
10000_at 6.166815 5.678373 6.185059
100009613_at 4.443027 4.773199 4.393488
100009676_at 5.836522 6.143398 5.898364
10001_at 6.330018 5.601745 6.137984
> anodat[1:8, 1:3]
V1 V2 V3
1 SampleID GA Batch
2 Tarca_001_P1A01 11 1
3 Tarca_013_P1B01 15.3 1
4 Tarca_025_P1C01 21.7 1
5 Tarca_037_P1D01 26.7 1
6 Tarca_049_P1E01 31.3 1
7 Tarca_061_P1F01 32.1 1
8 Tarca_051_P1E03 19.7 1
goal:
I intend to see how the genes in each sample are correlated with GA value of corresponding samples in the annotation data, then generate sub expression matrix of keeping high correlated genes with target observation data anodat$GA.
my attempt:
gene_corrs <- function(expr_mat, anno_mat){
stopifnot(ncol(expr_mat)==nrow(anno_mat))
res <- list()
lapply(colnames(expr_mat), function(x){
lapply(x, rownames(y){
if(colnames(x) %in% rownames(anno_mat)){
cor_mat <- stats::cor(y, anno_mat$GA, method = "pearson")
ncor <- ncol(cor_mat)
cmatt <- col(cor_mat)
ord <- order(-cmat, cor_mat, decreasing = TRUE)- (ncor*cmatt - ncor)
colnames(ord) <- colnames(cor_mat)
res <- cbind(ID=c(cold(ord), ID2=c(ord)))
res <- as.data.frame(cbind(out, cor=cor_mat[res]))
res <- cbind(res, cor=cor_mat[out])
res <- as.dara.frame(res)
}
})
})
return(res)
}
however, my above implementation didn't return what I expected, I need to filter out the genes by finding genes which has a strong correlation with anodat$GA.
Another attempt:
I read few post about similar issue and some people discussed about using limma package. Here is my attempt by using limma. Here I used anodat$GA as a covariate to fit limma linear model:
library(limma)
fit <- limma::lmFit(expr_mat, design = model.matrix( ~ 0 + anodat$GA)
fit <- eBayes(fit)
topTable(fit, coef=2)
then I am expecting to get a correlation matrix from the above code, and would like to do following in order to get filtered sub expression matrix:
idx <- which( (abs(cor) > 0.8) & (upper.tri(cor)), arr.ind=TRUE)
idx <- unique(c(idx[, 1],idx[, 2])
correlated.genes <- matrix[idx, ]
but I still didn't get the right answer. I am confident about using limma approach but I couldn't figure out what went wrong above code again. Can anyone point me out how to make this work? Is there any efficient way to make this happen?
Don't have your data so hard to double check, but in the abstract I would try this:
library(matrixTests)
cors <- row_cor_pearson(expr_mat, anodat$GA)
which(cors$cor > 0.9) # to get the indeces of genes with correlation > 0.9

Depth of a node in partykit

I am building a tree using the partykit R package, and I am wondering if there is a simple, efficient way to determine the depth number at each internal node. For example, the root node would have depth 0, the first two kid nodes have depth 1, the next kid nodes have depth 2, and so forth. This will eventually be used to calculate the minimal depth of a variable. Below is a very basic example (taken from vignette("constparty", package="partykit")):
library("partykit")
library("rpart")
data("Titanic", package = "datasets")
ttnc<-as.data.frame(Titanic)
ttnc <- ttnc[rep(1:nrow(ttnc), ttnc$Freq), 1:4]
names(ttnc)[2] <- "Gender"
rp <- rpart(Survived ~ ., data = ttnc)
ttncTree<-as.party(rp)
plot(ttncTree)
#This is one of my many attempts which does NOT work
internalNodes<-nodeids(ttncTree)[-nodeids(ttncTree, terminal = TRUE)]
depth(ttncTree)-unlist(nodeapply(ttncTree, ids=internalNodes, FUN=function(n){depth(n)}))
In this example, I want to output something similar to:
nodeid = 1 2 4 7
depth = 0 1 2 1
I apologize if my question is too specific.
Here's a possible solution which should be efficient enough as usually the trees have no more than several dozens of nodes.
I'm ignoring node #1, as it is always 0 an hence no point neither calculating it or showing it (IMO)
Inters <- nodeids(ttncTree)[-nodeids(ttncTree, terminal = TRUE)][-1]
table(unlist(sapply(Inters, function(x) intersect(Inters, nodeids(ttncTree, from = x)))))
# 2 4 7
# 1 2 1
I had to revisit this problem recently. Below is a function to determine the depth of each node. I count the depth based on the number of times a vertical line | appears running the print.party() function.
library(stringr)
idDepth <- function(tree) {
outTree <- capture.output(tree)
idCount <- 1
depthValues <- rep(NA, length(tree))
names(depthValues) <- 1:length(tree)
for (index in seq_along(outTree)){
if (grepl("\\[[0-9]+\\]", outTree[index])) {
depthValues[idCount] <- str_count(outTree[index], "\\|")
idCount = idCount + 1
}
}
return(depthValues)
}
> idDepth(ttncTree)
1 2 3 4 5 6 7 8 9
0 1 2 2 3 3 1 2 2
There definitely seems to be a simpler, faster solution, but this is faster than using the intersect() function. Below is an example of the computation time for a large tree (around 1,500 nodes)
# Compare computation time for large tree #
library(mlbench)
set.seed(470174)
dat <- data.frame(mlbench.friedman1(5000))
rp <- rpart(as.formula(paste0("y ~ ", paste(paste0("x.", 1:10), collapse=" + "))),
data=dat, control = rpart.control(cp = -1, minsplit=3, maxdepth = 10))
partyTree <- as.party(rp)
> length(partyTree) #Number of splits
[1] 1503
>
> # Intersect() computation time
> Inters <- nodeids(partyTree)[-nodeids(partyTree, terminal = TRUE)][-1]
> system.time(table(unlist(sapply(Inters, function(x) intersect(Inters, nodeids(partyTree, from = x))))))
user system elapsed
22.38 0.00 22.44
>
> # Proposed computation time
> system.time(idDepth(partyTree))
user system elapsed
2.38 0.00 2.38

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