Jim Slonder - 1 year ago 170

R Question

I have a n x n symmetrix toeplitz matrix

`T`

`v`

`T%*%v`

`T%*%v`

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

The function below works. Note that the `ifft()`

function requires the `pracma`

library.

```
toepmult <- function(A,v){
n <- nrow(A)
x <- as.matrix(c(A[1,],0,A[1,][n:2]))
p <- c(v,rep(0,n))
h <- as.vector(fft(p)*fft(x))
out <- Re(pracma::ifft(h)[1:n])
return( matrix(out,n) )
}
```

For a vector/matrix of size 1000, the `toepmult`

function takes about 18% of the time `A%*%v`

takes.

```
A <- toeplitz(runif(1000))
v <- runif(1000)
microb(A%*%v,toepmult(A,v),times=1000)
#Unit: microseconds
# expr min lq mean median uq max neval
# A %*% v 1515.858 1597.345 1809.3517 1693.533 1957.4350 3868.788 1000
# toepmult(A, v) 185.901 215.395 331.2928 298.435 347.7335 4611.285 1000
#[[1]]
# [,1] [,2]
#median 1693.533 298.435
#ratio 1.000 0.176
#diff 0.000 -1395.098
```

For a vector/matrix of size 10,000, the `toepmult`

function takes about 2.5% of the time `A%*%v`

takes.

```
A <- toeplitz(runif(10000))
v <- runif(10000)
microb(A%*%v,toepmult(A,v),times=1000)
#Unit: milliseconds
# expr min lq mean median uq max neval
# A %*% v 145.834304 160.395663 181.842779 170.396014 186.221449 495.2003 1000
# toepmult(A, v) 2.802058 4.077408 4.990894 4.322707 4.911103 180.4926 1000
#[[1]]
# [,1] [,2]
#median 170.396 4.323
#ratio 1.000 0.025
#diff 0.000 -166.073
```

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**