I think I remember seeing a shorthand for solve(t(X) %*% X) in R, but I can't remember what it was. Is there something like that? Just a way to do that in fewer keystrokes?
Maybe you're thinking of crossprod()? It's not fewer keystrokes, but is a bit more elegant and, according to its help file, it can be slightly faster than the naive construction.
x <- matrix(rnorm(9), ncol=3)
solve(crossprod(x))
# [,1] [,2] [,3]
# [1,] 1.34638151 -0.02957435 0.8010735
# [2,] -0.02957435 0.32780020 -0.1786295
# [3,] 0.80107345 -0.17862950 1.4533671
solve(t(x) %*% x)
# [,1] [,2] [,3]
# [1,] 1.34638151 -0.02957435 0.8010735
# [2,] -0.02957435 0.32780020 -0.1786295
# [3,] 0.80107345 -0.17862950 1.4533671
Related
I´m trying to get different elements from multiple diagonal saved as lists. My data looks something like this:
res <- list()
res[[1]] <- matrix(c(0.04770856,0.02854005,0.02854005,0.03260190), nrow=2, ncol=2)
res[[2]] <- matrix(c(0.05436957,0.04887182,0.04887182, 0.10484454), nrow=2, ncol=2)
> res
[[1]]
[,1] [,2]
[1,] 0.04770856 0.02854005
[2,] 0.02854005 0.03260190
[[2]]
[,1] [,2]
[1,] 0.05436957 0.04887182
[2,] 0.04887182 0.10484454
> diag(res[[1]])
[1] 0.04770856 0.03260190
> diag(res[[2]])
[1] 0.05436957 0.10484454
I would like to save the first and second elements of each diagonal of a given list into a vector similar to this:
d.1st.el <- c(0.04770856, 0.05436957)
d.2nd.el <- c(0.03260190, 0.10484454)
My issue is to write the function that runs for all given lists and get the diagonals. For some reason, when I use unlist() to extract the values of each matrix for a given level, it doesn't get me the number but the full matrix.
Does anyone have a simple solution?
sapply(res, diag)
[,1] [,2]
[1,] 0.04770856 0.05436957
[2,] 0.03260190 0.10484454
# or
lapply(res, diag)
[[1]]
[1] 0.04770856 0.03260190
[[2]]
[1] 0.05436957 0.10484454
If you want the vectors for some reason in your global environment:
alld <- lapply(res, diag)
names(alld) <- sprintf("d.%d.el", 1:length(alld))
list2env(alld, globalenv())
In two steps you can do:
# Step 1 - Get the diagonals
all_diags <- sapply(res, function(x) diag(t(x)))
print(all_diags)
[,1] [,2]
[1,] 0.04770856 0.05436957
[2,] 0.03260190 0.10484454
# Step 2 - Append to vectors
d.1st.el <- all_diags[1,]
d.2nd.el <- all_diags[2,]
As the title suggests, I am looking for a way to get the standard deviation per element from two separate matrices. However, I am quite the beginner at R and I can't seem to figue out how to do this. Below is an example of what I am trying to accomplish with a small sample of my data (first three rows)
I have two matrices with coordinates (df143 and df143_2, or matrices A and B as you will)
A:
[1,] 21.729504 -55.66055 -37.26477
[2,] 39.445610 -67.67449 -32.19464
[3,] 57.604027 -54.16734 -28.48679
B:
[1,] 21.706865 -55.50722 -37.57840
[2,] 39.553314 -67.68414 -31.95995
[3,] 57.286247 -54.13008 -28.44446
I am looking for an matrix output that shows the standard deviation per element of the two combined matrices.
Or you can do base R:
matrix(mapply(function(x,y) sd(c(x,y)),A, B), ncol=ncol(A))
# [,1] [,2] [,3]
#[1,] 0.01600819 0.10842068 0.22176990
#[2,] 0.07615823 0.00682358 0.16595089
#[3,] 0.22470439 0.02634680 0.02993183
I believe this is what you're looking to do:
library(abind)
a <- c(21.729504, -55.66055, -37.26477, 39.445610, -67.67449, -32.19464, 57.604027, -54.16734, -28.48679)
a <- matrix(a, ncol=3, byrow=TRUE)
b <- c(21.706865, -55.50722, -37.57840, 39.553314, -67.68414, -31.95995, 57.286247, -54.13008, -28.44446)
b <- matrix(b, ncol=3, byrow=TRUE)
m <- abind(a, b, along=3)
apply(m, 1:2, sd)
## [,1] [,2] [,3]
## [1,] 0.01600819 0.10842068 0.22176990
## [2,] 0.07615823 0.00682358 0.16595089
## [3,] 0.22470439 0.02634680 0.02993183
I want to generate a random matrix which should be symmetric.
I have tried this:
matrix(sample(0:1, 25, TRUE), 5, 5)
but it is not necessarily symmetric.
How can I do that?
Another quite interesting opportunity is based on the following mathematical fact: if A is some matrix, then A multiplied by its transpose is always symmetric.
> A <- matrix(runif(25), 5, 5)
> A %*% t(A)
[,1] [,2] [,3] [,4] [,5]
[1,] 1.727769 1.0337816 1.2195505 1.4661507 1.1041355
[2,] 1.033782 1.0037048 0.7368944 0.9073632 0.7643080
[3,] 1.219551 0.7368944 1.8383986 1.3309980 0.9867812
[4,] 1.466151 0.9073632 1.3309980 1.3845322 1.0034140
[5,] 1.104135 0.7643080 0.9867812 1.0034140 0.9376534
Try this from the Matrix package
library(Matrix)
x<-Matrix(rnorm(9),3)
x
3 x 3 Matrix of class "dgeMatrix"
[,1] [,2] [,3]
[1,] -0.9873338 0.8965887 -0.6041742
[2,] -0.3729662 -0.5882091 -0.2383262
[3,] 2.1263985 -0.3550972 0.1067264
X<-forceSymmetric(x)
X
3 x 3 Matrix of class "dsyMatrix"
[,1] [,2] [,3]
[1,] -0.9873338 0.8965887 -0.6041742
[2,] 0.8965887 -0.5882091 -0.2383262
[3,] -0.6041742 -0.2383262 0.1067264
If you don't want to use a package:
n=3
x <- matrix(rnorm(n*n), n)
ind <- lower.tri(x)
x[ind] <- t(x)[ind]
x
I like this one:
n <- 3
aux <- matrix(NA, nrow = n, ncol = n)
for(i in c(1:n)){
for(j in c(i:n)){
aux[i,j] <- sample(c(1:n), 1)
aux[j,i] <- aux[i,j]
}
}
How do I get actual matrix using Singular value decomposition(SVD)
efficiently in R ,
cause A=svd$u %*% svd$d %*% t(svd$v) This is not an efficient way to get matrix A
Try svd(A)$u%*%diag(svd(A)$d)%*%t(svd(A)$v).
set.seed(12345)
A <- matrix(data=runif(n=9, min=1, max=9), nrow=3)
A
[,1] [,2] [,3]
[1,] 6.767231 8.088997 3.600763
[2,] 8.006186 4.651848 5.073795
[3,] 7.087859 2.330974 6.821642
s <- svd(A)
D <- diag(s$d)
s$u %*% D %*% t(s$v)
[,1] [,2] [,3]
[1,] 6.767231 8.088997 3.600763
[2,] 8.006186 4.651848 5.073795
[3,] 7.087859 2.330974 6.821642
Improving upon the answer by #MYaseen208
(s$u) %*% (t(s$v)*s$d)
This has one less matrix multiplication (which is an O(n^3) operation).
I have a matrix, named "mat", and a smaller matrix, named "center".
temp = c(1.8421,5.6586,6.3526,2.904,3.232,4.6076,4.8,3.2909,4.6122,4.9399)
mat = matrix(temp, ncol=2)
[,1] [,2]
[1,] 1.8421 4.6076
[2,] 5.6586 4.8000
[3,] 6.3526 3.2909
[4,] 2.9040 4.6122
[5,] 3.2320 4.9399
center = matrix(c(3, 6, 3, 2), ncol=2)
[,1] [,2]
[1,] 3 3
[2,] 6 2
I need to compute the distance between each row of mat with every row of center. For example, the distance of mat[1,] and center[1,] can be computed as
diff = mat[1,]-center[1,]
t(diff)%*%diff
[,1]
[1,] 3.92511
Similarly, I can find the distance of mat[1,] and center[2,]
diff = mat[1,]-center[2,]
t(diff)%*%diff
[,1]
[1,] 24.08771
Repeat this process for each row of mat, I will end up with
[,1] [,2]
[1,] 3.925110 24.087710
[2,] 10.308154 7.956554
[3,] 11.324550 1.790750
[4,] 2.608405 16.408805
[5,] 3.817036 16.304836
I know how to implement it with for-loops. I was really hoping someone could tell me how to do it with some kind of an apply() function, maybe mapply() I guess.
Thanks
apply(center, 1, function(x) colSums((x - t(mat)) ^ 2))
# [,1] [,2]
# [1,] 3.925110 24.087710
# [2,] 10.308154 7.956554
# [3,] 11.324550 1.790750
# [4,] 2.608405 16.408805
# [5,] 3.817036 16.304836
If you want the apply for expressiveness of code that's one thing but it's still looping, just different syntax. This can be done without any loops, or with a very small one across center instead of mat. I'd just transpose first because it's wise to get into the habit of getting as much as possible out of the apply statement. (The BrodieG answer is pretty much identical in function.) These are working because R will automatically recycle the smaller vector along the matrix and do it much faster than apply or for.
tm <- t(mat)
apply(center, 1, function(m){
colSums((tm - m)^2) })
Use dist and then extract the relevant submatrix:
ix <- 1:nrow(mat)
as.matrix( dist( rbind(mat, center) )^2 )[ix, -ix]
6 7
# 1 3.925110 24.087710
# 2 10.308154 7.956554
# 3 11.324550 1.790750
# 4 2.608405 16.408805
# 5 3.817036 16.304836
REVISION: simplified slightly.
You could use outer as well
d <- function(i, j) sum((mat[i, ] - center[j, ])^2)
outer(1:nrow(mat), 1:nrow(center), Vectorize(d))
This will solve it
t(apply(mat,1,function(row){
d1<-sum((row-center[1,])^2)
d2<-sum((row-center[2,])^2)
return(c(d1,d2))
}))
Result:
[,1] [,2]
[1,] 3.925110 24.087710
[2,] 10.308154 7.956554
[3,] 11.324550 1.790750
[4,] 2.608405 16.408805
[5,] 3.817036 16.304836