Geometric median, also known as L1-median, is a solution to the following problem $$\textrm{argmin} \sum_{i=1}^n \| x_i - y \|_2 $$ for a given data \(x_1,x_2,\ldots,x_n \in R^p\).
weiszfeld(X, weights = NULL, maxiter = 496, abstol = 1e-06)Arguments
- X
an \((n\times p)\) matrix for \(p\)-dimensional signal. If vector is given, it is assumed that \(p=1\).
- weights
NULLfor equal weightrep(1/n,n); otherwise, it has to be a vector of length \(n\).- maxiter
maximum number of iterations.
- abstol
stopping criterion
Examples
## generate sin(x) data with noise for 100 replicates
set.seed(496)
t = seq(from=0,to=10,length.out=20)
X = array(0,c(100,20))
for (i in 1:100){
X[i,] = sin(t) + stats::rnorm(20, sd=0.5)
}
## compute L1-median and L2-mean
vecL2 = base::colMeans(X)
vecL1 = weiszfeld(X)
## visualize
opar <- par(no.readonly=TRUE)
par(mfrow=c(1,3), pty="s")
matplot(t(X[1:5,]), type="l", main="5 generated data", ylim=c(-2,2))
plot(t, vecL2, type="l", col="blue", main="L2-mean", ylim=c(-2,2))
plot(t, vecL1, type="l", col="red", main="L1-median", ylim=c(-2,2))
par(opar)