Compare Matrices in R efficiently - r

Question

I have an array a with some matrices in it. Now i need to efficiently check how many different matrices I have and what indices (in ascending order) they have in the array. My approach is the following: Paste the columns of the matrixes as character vectors and have a look at the frequency table like this:
n <- 10 #observations
a <- array(round(rnorm(2*2*n),1),
c(2,2,n))
paste_a <- apply(a, c(3), paste, collapse=" ") #paste by column
names(paste_a) <- 1:n
freq <- as.numeric( table(paste_a) ) # frequencies of different matrices (in ascending order)
indizes <- as.numeric(names(sort(paste_a[!duplicated(paste_a)])))
nr <- length(freq) #number of different matrices
However, as you increase n to large numbers, this gets very inefficient (it's mainly paste() that's getting slower and slower). Does anyone have a better solution?
Here is a "real" dataset with 100 observations where some matrices are actual duplicates (as opposed to my example above): https://pastebin.com/aLKaSQyF
Thank you very much.

Answer 1

Since your actual data is made up of the integers 0,1,2,3, why not take advantage of base 4? Integers are much faster to compare than entire matrix objects. (All occurrences of a below are of the data found in the real data set from the link.)
Base4Approach <- function() {
toBase4 <- sapply(1:dim(a)[3], function(x) {
v <- as.vector(a[,,x])
pows <- which(v > 0)
coefs <- v[pows]
sum(coefs*(4^pows))
})
myDupes <- which(duplicated(toBase4))
a[,,-(myDupes)]
}
And since the question is about efficiency, let's benchmark:
MartinApproach <- function() {
### commented this out for comparison reasons
# dimnames(a) <- list(1:dim(a)[1], 1:dim(a)[2], 1:dim(a)[3])
a <- a[,,!duplicated(a, MARGIN = 3)]
nr <- dim(a)[3]
a
}
identical(MartinApproach(), Base4Approach())
[1] TRUE
microbenchmark(Base4Approach(), MartinApproach())
Unit: microseconds
expr min lq mean median uq max neval
Base4Approach() 291.658 303.525 339.2712 325.4475 352.981 636.361 100
MartinApproach() 983.855 1000.958 1160.4955 1071.9545 1187.321 3545.495 100
The approach by #d.b. doesn't really do the same thing as the previous two approaches (it simply identifies and doesn't remove duplicates).
DBApproach <- function() {
a[, , 9] = a[, , 1]
#Convert to list
mylist = lapply(1:dim(a)[3], function(i) a[1:dim(a)[1], 1:dim(a)[2], i])
temp = sapply(mylist, function(x) sapply(mylist, function(y) identical(x, y)))
temp2 = unique(apply(temp, 1, function(x) sort(which(x))))
#The indices in 'a' where the matrices are same
temp2[lengths(temp2) > 1]
}
However, Base4Approach still dominates:
microbenchmark(Base4Approach(), MartinApproach(), DBApproach())
Unit: microseconds
expr min lq mean median uq max neval
Base4Approach() 298.764 324.0555 348.8534 338.899 356.0985 476.475 100
MartinApproach() 1012.601 1087.9450 1204.1150 1110.662 1162.9985 3224.299 100
DBApproach() 9312.902 10339.4075 11616.1644 11438.967 12413.8915 17065.494 100
Update courtesy of #alexis_laz
As mentioned in the comments by #alexis_laz, we can do much better.
AlexisBase4Approach <- function() {
toBase4 <- colSums(a * (4 ^ (0:(prod(dim(a)[1:2]) - 1))), dims = 2)
myDupes <- which(duplicated(toBase4))
a[,,-(myDupes)]
}
microbenchmark(Base4Approach(), MartinApproach(), DBApproach(), AlexisBase4Approach(), unit = "relative")
Unit: relative
expr min lq mean median uq max neval
Base4Approach() 11.67992 10.55563 8.177654 8.537209 7.128652 5.288112 100
MartinApproach() 39.60408 34.60546 27.930725 27.870019 23.836163 22.488989 100
DBApproach() 378.91510 342.85570 262.396843 279.190793 231.647905 108.841199 100
AlexisBase4Approach() 1.00000 1.00000 1.000000 1.000000 1.000000 1.000000 100
## Still gives accurate results
identical(MartinApproach(), AlexisBase4Approach())
[1] TRUE

Answer 2

My first attempt was actually really slow. So here is slightly changed version of yours:
dimnames(a) <- list(1:dim(a)[1], 1:dim(a)[2], 1:dim(a)[3])
a <- a[,,!duplicated(a, MARGIN = 3)]
nr <- dim(a)[3] #number of different matrices
idx <- dimnames(a)[[3]] # indices of left over matrices

Answer 3

I don't know if this is exactly what you want but here is a way you can extract indices where the matrices are same. More processing may be necessary to get what you want
#DATA
n <- 10
a <- array(round(rnorm(2*2*n),1), c(2,2,n))
a[, , 9] = a[, , 1]
temp = unique(apply(X = sapply(1:dim(a)[3], function(i)
sapply(1:dim(a)[3], function(j) identical(a[, , i], a[, , j]))),
MARGIN = 1,
FUN = function(x) sort(which(x))))
temp[lengths(temp) > 1]
#[[1]]
#[1] 1 9