I am trying to understand how to parallelize some of my code using R. So, in the following example I want to use k-means to cluster data using 2,3,4,5,6 centers, while using 20 iterations.
Here is the code:
library(parallel)
library(BLR)
data(wheat)
parallel.function <- function(i) {
kmeans( X[1:100,100], centers=?? , nstart=i )
}
out <- mclapply( c(5, 5, 5, 5), FUN=parallel.function )
How can we parallel simultaneously the iterations and the centers?
How to track the outputs, assuming I want to keep all the outputs from k-means across all, iterations and centers, just to learn how?
This looked very simple to me at first ... and then i tried it. After a lot of monkey typing and face palming during my lunch break however, I arrived at this:
library(parallel)
library(BLR)
data(wheat)
mc = mclapply(2:6, function(x,centers)kmeans(x, centers), x=X)
It looks right though I didn't check how sensible the clustering was.
> summary(mc)
Length Class Mode
[1,] 9 kmeans list
[2,] 9 kmeans list
[3,] 9 kmeans list
[4,] 9 kmeans list
[5,] 9 kmeans list
On reflection the command syntax seems sensible - although a lot of other stuff that failed seemed reasonable too...The examples in the help documentation are maybe not that great.
Hope it helps.
EDIT
As requested here is that on two variables nstart and centers
(pars = expand.grid(i=1:3, cent=2:4))
i cent
1 1 2
2 2 2
3 3 2
4 1 3
5 2 3
6 3 3
7 1 4
8 2 4
9 3 4
L=list()
# zikes horrible
pars2=apply(pars,1,append, L)
mc = mclapply(pars2, function(x,pars)kmeans(x, centers=pars$cent,nstart=pars$i ), x=X)
> summary(mc)
Length Class Mode
[1,] 9 kmeans list
[2,] 9 kmeans list
[3,] 9 kmeans list
[4,] 9 kmeans list
[5,] 9 kmeans list
[6,] 9 kmeans list
[7,] 9 kmeans list
[8,] 9 kmeans list
[9,] 9 means list
How'd you like them apples?
There's a CRAN package called knor that is derived from a research paper that improves the performance using a memory efficient variant of Elkan's pruning algorithm. It's an order of magnitude faster than everything in these answers.
install.packages("knor")
require(knor)
iris.mat <- as.matrix(iris[,1:4])
k <- length(unique(iris[, dim(iris)[2]])) # Number of unique classes
nthread <- 4
kms <- Kmeans(iris.mat, k, nthread=nthread)
You may use parallel to try K-Means from different random starting points on multiple cores.
The code below is an example. (K=K in K-means, N= number of random starting points, C = number of cores you would like to use)
suppressMessages( library("Matrix") )
suppressMessages( library("irlba") )
suppressMessages( library("stats") )
suppressMessages( library("cluster") )
suppressMessages( library("fpc") )
suppressMessages( library("parallel") )
#Calculate KMeans results
calcKMeans <- function(matrix, K, N, C){
#Parallel running from various of random starting points (Using C cores)
results <- mclapply(rep(N %/% C, C), FUN=function(nstart) kmeans(matrix, K, iter.max=15, nstart=nstart), mc.cores=C);
#Find the solution with smallest total within sum of square error
tmp <- sapply(results, function(r){r[['tot.withinss']]})
km <- results[[which.min(tmp)]]
#return cluster, centers, totss, withinss, tot.withinss, betweenss, size
return(km)
}
runKMeans <- function(fin_uf, K, N, C,
#fout_center, fout_label, fout_size,
fin_record=NULL, fout_prediction=NULL){
uf = read.table(fin_uf)
km = calcKMeans(uf, K, N, C)
rm(uf)
#write.table(km$cluster, file=fout_label, row.names=FALSE, col.names=FALSE)
#write.table(km$center, file=fout_center, row.names=FALSE, col.names=FALSE)
#write.table(km$size, file=fout_size, row.names=FALSE, col.names=FALSE)
str(km)
return(km$center)
}
Hope it helps!
Related
With an igraph object I would like to capture some features of each node's neighbours, for example the average degree of its neighbours.
I come up with this code, which is inelegant and quite slow.
How should I rethink it for large and complex networks?
library(igraph)
# Toy example
set.seed(123)
g <- erdos.renyi.game(10,0.2)
# Loop to calculate average degree of each node's neighbourhood
s <- character(0)
for(i in 1:gorder(g)){
n <- ego_size(g, nodes = i, order = 1, mindist = 1)
node_of_interest <- unique(unlist(ego(g, nodes = i, order = 1, mindist = 1)))
m <- mean(degree(g, v = node_of_interest, loops = TRUE, normalized = FALSE)-1)
s <- rbind(s,data.frame(node = i, neighbours = n, mean = m))
}
Expanding the data structure with rbind in a loop can get quite slow in R, because at every step it needs to allocate the space for the new object, and then copy it (see section 24.6 here). Also, you might be computing the degree of a node many times, if it s the neighbor of multiple nodes.
A possibly better alternative could be:
# add vertex id (not really necessary)
V(g)$name <- V(g)
# add degree to the graph
V(g)$degree <- degree(g, loops = TRUE, normalized = FALSE)
# get a list of neighbours, for each node
g_ngh <- neighborhood(g, mindist = 1)
# write a function that gets the means
get.mean <- function(x){
mean(V(g)$degree[x]-1)
}
# apply the function, add result to the graph
V(g)$av_degr_nei <- sapply(g_ngh, get.mean)
# get data into dataframe, if necessary
d_vert_attr <- as_data_frame(g, what = "vertices")
d_vert_attr
name degree av_degr_nei
1 1 0 NaN
2 2 1 2.0000000
3 3 2 1.0000000
4 4 1 1.0000000
5 5 2 1.0000000
6 6 1 1.0000000
7 7 3 0.6666667
8 8 1 0.0000000
9 9 1 0.0000000
10 10 0 NaN
i am working with r language and on umbalanced dataset and i need to know how can get the k nearest neighbors of a dataset becaue i need them to create new synthetic examples .
set.seed(123)
test <- 1:100
train.gc <- gc.subset[-test,]
test.gc <- gc.subset[test,]
train.def <- gc$Default[-test]
test.def <- gc$Default[test]
library(class)
knn.5 <- knn(train.gc, test.gc, train.def, k=5)
#how can i get the five nearest neighbours????????
Although it doesn't seem to be documented, the help for knn hints that the attributes may store something:
train <- rbind(iris3[1:25,,1], iris3[1:25,,2], iris3[1:25,,3])
test <- rbind(iris3[26:50,,1], iris3[26:50,,2], iris3[26:50,,3])
cl <- factor(c(rep("s",25), rep("c",25), rep("v",25)))
k = knn(train, test, cl, k = 3, prob=TRUE)
names(attributes(k))
# [1] "levels" "class" "prob" "nn.index" "nn.dist"
and I'd hazard a guess that nn.index is the index of the neighbours:
> head(attr(k,"nn.index"))
[,1] [,2] [,3]
[1,] 10 2 13
[2,] 24 8 18
[3,] 1 18 8
[4,] 1 18 8
I'd guess those are the 3 nearest neighbours of the first four data points.
I have to calculate cosine similarity (patient similarity metric) in R between 48k patients data with some predictive variables. Here is the equation: PSM(P1,P2) = P1.P2/ ||P1|| ||P2||
where P1 and P2 are the predictor vectors corresponding to two different patients, where for example P1 index patient and P2 will be compared with index (P1) and finally pairwise patient similarity metric PSM(P1,P2) will be calculated.
This process will go on for all 48k patients.
I have added sample data-set for 300 patients in a .csv file. Please find the sample data-set here.https://1drv.ms/u/s!AhoddsPPvdj3hVTSbosv2KcPIx5a
First things first: You can find more rigorous treatments of cosine similarity at either of these posts:
Find cosine similarity between two arrays
Creating co-occurrence matrix
Now, you clearly have a mixture of data types in your input, at least
decimal
integer
categorical
I suspect that some of the integer values are Booleans or additional categoricals. Generally, it will be up to you to transform these into continuous numerical vectors if you want to use them as input into the similarity calculation. For example, what's the distance between admission types ELECTIVE and EMERGENCY? Is it a nominal or ordinal variable? I will only be modelling the columns that I trust to be numerical dependent variables.
Also, what have you done to ensure that some of your columns don't correlate with others? Using just a little awareness of data science and biomedical terminology, it seems likely that the following are all correlated:
diasbp_max, diasbp_min, meanbp_max, meanbp_min, sysbp_max and sysbp_min
I suggest going to a print shop and ordering a poster-size printout of psm_pairs.pdf. :-) Your eyes are better at detecting meaningful (but non-linear) dependencies between variable. Including multiple measurements of the same fundamental phenomenon may over-weight that phenomenon in your similarity calculation. Don't forget that you can derive variables like
diasbp_rage <- diasbp_max - diasbp_min
Now, I'm not especially good at linear algebra, so I'm importing a cosine similarity function form the lsa text analysis package. I'd love to see you write out the formula in your question as an R function. I would write it to compare one row to another, and use two nested apply loops to get all comparisons. Hopefully we'll get the same results!
After calculating the similarity, I try to find two different patients with the most dissimilar encounters.
Since you're working with a number of rows that's relatively large, you'll want to compare various algorithmic methodologies for efficiency. In addition, you could use SparkR/some other Hadoop solution on a cluster, or the parallel package on a single computer with multiple cores and lots of RAM. I have no idea whether the solution I provided is thread-safe.
Come to think of it, the transposition alone (as I implemented it) is likely to be computationally costly for a set of 1 million patient-encounters. Overall, (If I remember my computational complexity correctly) as the number of rows in your input increases, the performance could degrade exponentially.
library(lsa)
library(reshape2)
psm_sample <- read.csv("psm_sample.csv")
row.names(psm_sample) <-
make.names(paste0("patid.", as.character(psm_sample$subject_id)), unique = TRUE)
temp <- sapply(psm_sample, class)
temp <- cbind.data.frame(names(temp), as.character(temp))
names(temp) <- c("variable", "possible.type")
numeric.cols <- (temp$possible.type %in% c("factor", "integer") &
(!(grepl(
pattern = "_id$", x = temp$variable
))) &
(!(
grepl(pattern = "_code$", x = temp$variable)
)) &
(!(
grepl(pattern = "_type$", x = temp$variable)
))) | temp$possible.type == "numeric"
psm_numerics <- psm_sample[, numeric.cols]
row.names(psm_numerics) <- row.names(psm_sample)
psm_numerics$gender <- as.integer(psm_numerics$gender)
psm_scaled <- scale(psm_numerics)
pair.these.up <- psm_scaled
# checking for independence of variables
# if the following PDF pair plot is too big for your computer to open,
# try pair-plotting some random subset of columns
# keep.frac <- 0.5
# keep.flag <- runif(ncol(psm_scaled)) < keep.frac
# pair.these.up <- psm_scaled[, keep.flag]
# pdf device sizes are in inches
dev <-
pdf(
file = "psm_pairs.pdf",
width = 50,
height = 50,
paper = "special"
)
pairs(pair.these.up)
dev.off()
#transpose the dataframe to get the
#similarity between patients
cs <- lsa::cosine(t(psm_scaled))
# this is super inefficnet, because cs contains
# two identical triangular matrices
cs.melt <- melt(cs)
cs.melt <- as.data.frame(cs.melt)
names(cs.melt) <- c("enc.A", "enc.B", "similarity")
extract.pat <- function(enc.col) {
my.patients <-
sapply(enc.col, function(one.pat) {
temp <- (strsplit(as.character(one.pat), ".", fixed = TRUE))
return(temp[[1]][[2]])
})
return(my.patients)
}
cs.melt$pat.A <- extract.pat(cs.melt$enc.A)
cs.melt$pat.B <- extract.pat(cs.melt$enc.B)
same.pat <- cs.melt[cs.melt$pat.A == cs.melt$pat.B ,]
different.pat <- cs.melt[cs.melt$pat.A != cs.melt$pat.B ,]
most.dissimilar <-
different.pat[which.min(different.pat$similarity),]
dissimilar.pat.frame <- rbind(psm_numerics[rownames(psm_numerics) ==
as.character(most.dissimilar$enc.A) ,],
psm_numerics[rownames(psm_numerics) ==
as.character(most.dissimilar$enc.B) ,])
print(t(dissimilar.pat.frame))
which gives
patid.68.49 patid.9
gender 1.00000 2.00000
age 41.85000 41.79000
sysbp_min 72.00000 106.00000
sysbp_max 95.00000 217.00000
diasbp_min 42.00000 53.00000
diasbp_max 61.00000 107.00000
meanbp_min 52.00000 67.00000
meanbp_max 72.00000 132.00000
resprate_min 20.00000 14.00000
resprate_max 35.00000 19.00000
tempc_min 36.00000 35.50000
tempc_max 37.55555 37.88889
spo2_min 90.00000 95.00000
spo2_max 100.00000 100.00000
bicarbonate_min 22.00000 26.00000
bicarbonate_max 22.00000 30.00000
creatinine_min 2.50000 1.20000
creatinine_max 2.50000 1.40000
glucose_min 82.00000 129.00000
glucose_max 82.00000 178.00000
hematocrit_min 28.10000 37.40000
hematocrit_max 28.10000 45.20000
potassium_min 5.50000 2.80000
potassium_max 5.50000 3.00000
sodium_min 138.00000 136.00000
sodium_max 138.00000 140.00000
bun_min 28.00000 16.00000
bun_max 28.00000 17.00000
wbc_min 2.50000 7.50000
wbc_max 2.50000 13.70000
mingcs 15.00000 15.00000
gcsmotor 6.00000 5.00000
gcsverbal 5.00000 0.00000
gcseyes 4.00000 1.00000
endotrachflag 0.00000 1.00000
urineoutput 1674.00000 887.00000
vasopressor 0.00000 0.00000
vent 0.00000 1.00000
los_hospital 19.09310 4.88130
los_icu 3.53680 5.32310
sofa 3.00000 5.00000
saps 17.00000 18.00000
posthospmort30day 1.00000 0.00000
Usually I wouldn't add a second answer, but that might be the best solution here. Don't worry about voting on it.
Here's the same algorithm as in my first answer, applied to the iris data set. Each row contains four spatial measurements of the flowers form three different varieties of iris plants.
Below that you will find the iris analysis, written out as nested loops so you can see the equivalence. But that's not recommended for production with large data sets.
Please familiarize yourself with starting data and all of the intermediate dataframes:
The input iris data
psm_scaled (the spatial measurements, scaled to mean=0, SD=1)
cs (the matrix of pairwise similarities)
cs.melt (the pairwise similarities in long format)
At the end I have aggregated the mean similarities for all comparisons between one variety and another. You will see that comparisons between individuals of the same variety have mean similarities approaching 1, and comparisons between individuals of the same variety have mean similarities approaching negative 1.
library(lsa)
library(reshape2)
temp <- iris[, 1:4]
iris.names <- paste0(iris$Species, '.', rownames(iris))
psm_scaled <- scale(temp)
rownames(psm_scaled) <- iris.names
cs <- lsa::cosine(t(psm_scaled))
# this is super inefficient, because cs contains
# two identical triangular matrices
cs.melt <- melt(cs)
cs.melt <- as.data.frame(cs.melt)
names(cs.melt) <- c("enc.A", "enc.B", "similarity")
names(cs.melt) <- c("flower.A", "flower.B", "similarity")
class.A <-
strsplit(as.character(cs.melt$flower.A), '.', fixed = TRUE)
cs.melt$class.A <- sapply(class.A, function(one.split) {
return(one.split[1])
})
class.B <-
strsplit(as.character(cs.melt$flower.B), '.', fixed = TRUE)
cs.melt$class.B <- sapply(class.B, function(one.split) {
return(one.split[1])
})
cs.melt$comparison <-
paste0(cs.melt$class.A , '_vs_', cs.melt$class.B)
cs.agg <-
aggregate(cs.melt$similarity, by = list(cs.melt$comparison), mean)
print(cs.agg[order(cs.agg$x),])
which gives
# Group.1 x
# 3 setosa_vs_virginica -0.7945321
# 7 virginica_vs_setosa -0.7945321
# 2 setosa_vs_versicolor -0.4868352
# 4 versicolor_vs_setosa -0.4868352
# 6 versicolor_vs_virginica 0.3774612
# 8 virginica_vs_versicolor 0.3774612
# 5 versicolor_vs_versicolor 0.4134413
# 9 virginica_vs_virginica 0.7622797
# 1 setosa_vs_setosa 0.8698189
If you’re still not comfortable with performing lsa::cosine() on a scaled, numerical dataframe, we can certainly do explicit pairwise calculations.
The formula you gave for PSM, or cosine similarity of patients, is expressed in two formats at Wikipedia
Remembering that vectors A and B represent the ordered list of attributes for PatientA and PatientB, the PSM is the dot product of A and B, divided by (the scalar product of [the magnitude of A] and [the magnitude of B])
The terse way of saying that in R is
cosine.sim <- function(A, B) { A %*% B / sqrt(A %*% A * B %*% B) }
But we can rewrite that to look more similar to your post as
cosine.sim <- function(A, B) { A %*% B / (sqrt(A %*% A) * sqrt(B %*% B)) }
I guess you could even re-write that (the calculations of similarity between a single pair of individuals) as a bunch of nested loops, but in the case of a manageable amount of data, please don’t. R is highly optimized for operations on vectors and matrices. If you’re new to R, don’t second guess it. By the way, what happened to your millions of rows? This will certainly be less stressful now that your down to tens of thousands.
Anyway, let’s say that each individual only has two elements.
individual.1 <- c(1, 0)
individual.2 <- c(1, 1)
So you can think of individual.1 as a line that passes between the origin (0,0) and (0, 1) and individual.2 as a line that passes between the origin and (1, 1).
some.data <- rbind.data.frame(individual.1, individual.2)
names(some.data) <- c('element.i', 'element.j')
rownames(some.data) <- c('individual.1', 'individual.2')
plot(some.data, xlim = c(-0.5, 2), ylim = c(-0.5, 2))
text(
some.data,
rownames(some.data),
xlim = c(-0.5, 2),
ylim = c(-0.5, 2),
adj = c(0, 0)
)
segments(0, 0, x1 = some.data[1, 1], y1 = some.data[1, 2])
segments(0, 0, x1 = some.data[2, 1], y1 = some.data[2, 2])
So what’s the angle between vector individual.1 and vector individual.2? You guessed it, 0.785 radians, or 45 degrees.
cosine.sim <- function(A, B) { A %*% B / (sqrt(A %*% A) * sqrt(B %*% B)) }
cos.sim.result <- cosine.sim(individual.1, individual.2)
angle.radians <- acos(cos.sim.result)
angle.degrees <- angle.radians * 180 / pi
print(angle.degrees)
# [,1]
# [1,] 45
Now we can use the cosine.sim function I previously defined, in two nested loops, to explicitly calculate the pairwise similarities between each of the iris flowers. Remember, psm_scaled has already been defined as the scaled numerical values from the iris dataset.
cs.melt <- lapply(rownames(psm_scaled), function(name.A) {
inner.loop.result <-
lapply(rownames(psm_scaled), function(name.B) {
individual.A <- psm_scaled[rownames(psm_scaled) == name.A, ]
individual.B <- psm_scaled[rownames(psm_scaled) == name.B, ]
similarity <- cosine.sim(individual.A, individual.B)
return(list(name.A, name.B, similarity))
})
inner.loop.result <-
do.call(rbind.data.frame, inner.loop.result)
names(inner.loop.result) <-
c('flower.A', 'flower.B', 'similarity')
return(inner.loop.result)
})
cs.melt <- do.call(rbind.data.frame, cs.melt)
Now we repeat the calculation of cs.melt$class.A, cs.melt$class.B, and cs.melt$comparison as above, and calculate cs.agg.from.loops as the mean similarity between the various types of comparisons:
cs.agg.from.loops <-
aggregate(cs.agg.from.loops$similarity, by = list(cs.agg.from.loops $comparison), mean)
print(cs.agg.from.loops[order(cs.agg.from.loops$x),])
# Group.1 x
# 3 setosa_vs_virginica -0.7945321
# 7 virginica_vs_setosa -0.7945321
# 2 setosa_vs_versicolor -0.4868352
# 4 versicolor_vs_setosa -0.4868352
# 6 versicolor_vs_virginica 0.3774612
# 8 virginica_vs_versicolor 0.3774612
# 5 versicolor_vs_versicolor 0.4134413
# 9 virginica_vs_virginica 0.7622797
# 1 setosa_vs_setosa 0.8698189
Which, I believe is identical to the result we got with lsa::cosine.
So what I'm trying to say is... why wouldn't you use lsa::cosine?
Maybe you should be more concerned with
selection of variables, including removal of highly correlated variables
scaling/normalizing/standardizing the data
performance with a large input data set
identifying known similars and dissimilars for quality control
as previously addressed
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
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