how to calculate the Euclidean norm of a vector in R?

Question

I tried norm, but I think it gives the wrong result. (the norm of c(1, 2, 3) is sqrt(1*1+2*2+3*3), but it returns 6..

Answer 1

norm(c(1,1), type="2")     # 1.414214
norm(c(1, 1, 1), type="2")  # 1.732051

Answer 2

I'mma throw this out there too as an equivalent R expression

norm_vec(x) <- function(x){sqrt(crossprod(x))}

Don't confuse R's crossprod with a similarly named vector/cross product. That naming is known to cause confusion especially for those with a physics/mechanics background.

Answer 3

If you have a data.frame or a data.table 'DT', and want to compute the Euclidian norm (norm 2) across each row, the apply function can be used.

apply(X = DT, MARGIN = 1, FUN = norm, '2')

Example:

>DT 

        accx       accy       accz
 1: 9.576807 -0.1629486 -0.2587167
 2: 9.576807 -0.1722938 -0.2681506
 3: 9.576807 -0.1634264 -0.2681506
 4: 9.576807 -0.1545590 -0.2681506
 5: 9.576807 -0.1621254 -0.2681506
 6: 9.576807 -0.1723825 -0.2682434
 7: 9.576807 -0.1723825 -0.2728810
 8: 9.576807 -0.1723825 -0.2775187

> apply(X = DT, MARGIN = 1, FUN = norm, '2')
 [1] 9.581687 9.582109 9.581954 9.581807 9.581932 9.582114 9.582245 9.582378

Answer 4

Following AbdealiJK's answer,

I experimented further to gain some insight.

Here's one.

x = c(-8e+299, -6e+299, 5e+299, -8e+298, -5e+299)
sqrt(sum(x^2))
norm(x, type='2')

The first result is Inf and the second one is 1.227355e+300 which is quite correct as I show you in the code below.

library(Rmpfr)
y <- mpfr(x, 120)
sqrt(sum(y*y))    

The result is 1227354879.... I didn't count the number of trailing numbers but it looks all right. I know there another way around this OVERFLOW problem which is first applying log function to all numbers and summing up, which I do not have time to implement!

Answer 5

This is a trivial function to write yourself:

norm_vec <- function(x) sqrt(sum(x^2))

Answer 6

We can also find the norm as :

Result<-sum(abs(x)^2)^(1/2)

OR Even You can also try as:

Result<-sqrt(t(x)%*%x)

Both will give the same answer