I'm trying to translate a code from MatLab into R, but I'm stuck on the following line:
SqO=U.* sqrt(D)*V'
I feel like I'm close:
SqO<-Conj(t(U%*%sqrt(D)*V))
...but the output still isn't matching up. All the variables (Sq0, U, D, and V) are 20x20 matrices if that helps.
Hmmm, I'm no expert in R, but I do know a bit of Matlab. In Matlab the sub-expression
U.* sqrt(D)
does an element-by-element multiplication of U and the square root of D. That is, element (i,j) in U is multiplied by element (i,j) in sqrt(D); so this is not the usual matrix multiplication. Is that what your U%*%sqrt(D) does ? sqrt(D) also operates on the individual elements, that is sqrt(D)~=D^(1/2)*D^(1/2).
Then the Matlab code multiplies the result of the previous operation by the transpose of V (if V is a real array); again my R is too weak to know whether you've done this or an equivalent operation.
From what HighPerformanceMark wrote the translation should be:
SqO=U.* sqrt(D)*V' # Matlab
SqO <- U* sqrt(D) %*% t(V) # R
Related
I am new to R (also not too good at math) and I am trying to calculate this equation in R with some difficulties:
X is some integer data I have, with 550 samples.
Any help is appreciated since I am unsure how to do this. I think I have to use a for loop and the sum() function but other than that I don;t know.
R supports vectorisation, which means you very rarely need to implement for loops.
For example, you can solve your equation like so:
## I'm just making up a long numerical vector for x - obviously you can use anything
x <- 1:1000
solution <- sum(20/x)^0.5
Unless the brackets denote the integral, rather than the sum? In which case:
solution <- sum( (20/x)^0.5 )
The following code causes a memory error:
diag(1:100000)
Is there any alternative for diag which allows producing a huge diagonal matrix?
Longer answer: I suggest not creating a diagonal matrix, because in most situations you can do without it. To make that clear, consider the most typical matrix operations:
Multiply the diagonal matrix D by a vector v to produce Dv. Instead of maintaining a matrix, keep your "matrix" as a vector d of the diagonal elements, and then multiply d elementwise by v. Same result.
Invert the matrix. Again, easy: invert each element (of course, only for diagonal matrices is this generally the correct inverse).
Various decompositions/eigenvalues/determinants/trace. Again, these can all be done on the vector d.
In short, though it requires a bit of attention in your code, you can always represent a diagonal matrix as a vector, and that should solve your memory issues.
Shorter answer: Now, having said all that, of course people have already implemented the above steps implicitly using sparse matrices, which does the above steps under the hood. In R, the Matrix package is nice for sparse matrices: https://cran.r-project.org/web/packages/Matrix/Matrix.pdf
Say I have a function func that takes two scalar numeric inputs and delivers a scalar numeric result, and I have the following code to calculate a result vector u, based on input numeric vector v and initial value u0 for the result vector:
u<-rep(u0,1+length(v))
for (k in 2:length(u)){
u[k]<-func(u[k-1],v[k-1])
}
Note how a component of the result vector depends not only on the corresponding element of the input vector but also on the immediately prior element of the result vector. I can see no obvious way to vectorise this.
It is common to do this sort of thing in financial simulations, for instance when projecting forward company accounts, rolling them up with interest or inflation and adding in operational cash flows each year.
For some specific instances, it is possible to find a case-specific, non-iterative coding, but I would like to know if there's a general solution.
The problem can also be coded by recursion, as follows:
calc.u<-function(v,u0){
if (length(v)<2){
func(u0,v[1]) }
else {
u.prior<-func(u0,v[-length(v),drop=FALSE])
c(u.prior,func(u.prior[length(u.prior)],v[length(v)]) )
}
u<-calc.u(v,u0)
Is there an R tactic for doing this without using either iteration or recursion, ie for vectorising it?
Answered: Thank you #MrFlick for introducing me to the Reduce function, which does exactly what I was wanting. I see that
Reduce('+',v,0,accumulate=T)[-1]
gives me
cumsum(v)
and
Reduce('*',v,0,accumulate=T)[-1]
gives me
cumprod(v)
as expected, where the [-1] is to discard the initial value.
Very nice indeed! Thanks again.
If you have this example
u0 <- 5
v <- (1:5)*2
func <- function(u,v) {u/2+v}
u <- rep(u0,1+length(v))
for (k in 2:length(u)){
u[k]<-func(u[k-1],v[k-1])
}
this is equivalent to
w <- Reduce(func, v, u0, accumulate=TRUE)
And we can check that
all(u==w)
# [1] TRUE
I have been learning R for the past few days, and want to find out whether the problem below can be solved in a better manner (compacter code perhaps) than my solution.
Problem: A vector V of N (~ 1000) numeric elements, needs to be transformed in the following way.
Choose M (~ 100) elements at random.
Replace each such element x with f(x).
My Solution: for (i in sample(1:N, M)) V[i] = f(V[i])
Edit: The function f takes as input a single numeric value, and also outputs a single numeric value. Something like: f <- function (x) x^3 + 2
Edit: Thanks for everyone's contributions! I now understand the power of vectorized functions. :)
How about this
i <- sample(1:N, M)
V[i] <- f(V[i])
No need for loop since [<- is a vectorized function. See ?"[<-" to get further details on that.
It depends on the type of your function. If f is vectorised then
V <- f(V) # V is a vector with random numbers
will do the job. If f takes and returns a single value then:
V <- sapply(V, f)
Thankfully, in R most of the function are vectorised, so the first approach would work quite often.
I've got a big square matrix, which I've taken the first row for testing purposes....
so the initial matrix is 1x63000, which is pretty big. Every time i try to multiply it by itself, using
a %*% b
Every time I do this, I get...
Error in fooB %*% fooB : non-conformable arguments
However, this works with smaller matrices. Are there any packages for handling mathematical functions of large matrices? or is there a trick I'm missing somewhere?
cheers
You are looking for the crossproduct, i.e. a %*% t(a) and there is a base R function for this. Try:
crossprod(a)