CODEWITHSUNDEEP
rfor-loopmatrix
quintinm
quintinm

Reputation: 129

How to use a for loop on a matrix

Say I have a matrix:

A <- matrix(c(2,4,3,1,5,7), nrow=3, ncol=2)
colnames(A) <- c ("x", "y")

A
     x y
[1,] 2 1
[2,] 4 5
[3,] 3 7

Is there a way to access each row of the matrix using a for loop?

What I'm trying to do is total the euclidean distance between each successive point (x,y). So in this example, I would find the total distance between:

(2,1) and (4,5)
(4,5) and (3,7)

So first I would find the distance between each of the two points, ie:

(2,1) and (4,5)    =>   (|4-2|,|5-1|)    =>      (2,4)
(4,5) and (3,7)    =>   (|3-4|,|7-5|)    =>      (1,2)

Then I would turn it into euclidean distance:

(2,4)   =>    sqrt(2^2 + 4^2)    => 4.47
(1,2)   =>    sqrt(1^2 + 2^2)    => 2.24

And total the distance

4.47 + 2.24 = 6.71

I'm quite confident that if I can access each row of the matrix as a vector, I can easily code this. However, I would love to hear any better ways of doing this.

I was also looking into turning the matrix into a list of lists (ie a list of (x,y) points, where each point is a list of the x and y value), or a list of points (x,y).

I'm not very experienced in programming and I've just started using R, so sorry if I'm not making sense.

Upvotes: 2

Views: 516

Answers (2)

Krycke
Krycke

Reputation: 3186

You can try the following

for (i in 1:nrow(A))
{
    row = A[i,]
    % Do something with the row
}

Upvotes: 2

thelatemail
thelatemail

Reputation: 93938

However, I would love to hear any better ways of doing this.

R has built in functions for distance calculations such as dist, e.g.:

out <- as.matrix(dist(A))
#         1        2        3
#1 0.000000 4.472136 6.082763
#2 4.472136 0.000000 2.236068
#3 6.082763 2.236068 0.000000

You can extract the off-diagonal, which is the values you want, using:

row(out) - col(out)==1
#      [,1]  [,2]  [,3]
#[1,] FALSE FALSE FALSE
#[2,]  TRUE FALSE FALSE
#[3,] FALSE  TRUE FALSE

Thus:

out[row(out) - col(out)==1]
#[1] 4.472136 2.236068
sum(out[row(out) - col(out)==1])
#[1] 6.708204

Upvotes: 1

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