« Progress bars in R (part II) – a wrapper for apply functions
Playing with the ‘playwith’ package »

R vs. Matlab – a small example

15Feb10

At the institute I’m working quite a lot of people prefer using Matlab and only a few of them know about R. Today one of my colleagues — who is also an eager user of Matlab — ran into the following problem:

  • He had a vector v in hand which consisted of \frac{n(n+1)}{2} elements.
  • He wanted to reshape this data into an n×n matrix M, where the element M_{ij} is equal to v_{k+j}I(j<=n-i+1) with k=\frac{(2n-i+2)(i-1)}{2} and I(j <= n-i+1)=1 if the condition j <= n-i+1 is satisfied and 0 otherwise. In other words, the first (n-i+1)th element of the ith row of M is equal to the vector (v_{k+1},v_{k+2},\ldots,v_{k+n-i+1}) and the remaining elements are zero.

He struggled for long minutes of how he should design a loop for doing this task. Of course writing such a loop is not a highly difficult task, but why would we waste our time, if we can get the same result in a single line of R code?

For the sake of illustration, I’ve generated an input vector for the case of n=99 (the value of n was 99 in my colleague’s problem as well):

v <- rep(99:1,times=99:1)

and used the one-liner

M <- t(matrix(unlist(tapply(v,rep(1:99,times=99:1),function(x) c(x,rep(0,99-length(x))))),nrow=99))

This is the kind of compactness I like pretty much in R. At the end I would like to emphasize that this post is not against Matlab, it just points out how the different logic of the R language can simplify problem solving in many situations. As a bonus let me share the visualization of the resulted matrix using the color2D.matplot function of the plotrix package:

library(plotrix)
color2D.matplot(M,c(0,1),c(1,0),c(0,0))


Filed under: R / R-Code   |  13 Comments
Tags: , ,

13 Responses to “R vs. Matlab – a small example”

Feed for this Entry Trackback Address
  1. 1 nmwe on February 25, 2010 said:

    I’ve always wondered why Matlab despite having so much syntactic sugar for working with matrices, often forces you to resort to loops for cases such as the above. I might have missed out some stuff and have searched the manual more than once if there were any easier ways to do things but have yet to find it.

  2. 2 Rick Wicklin on September 3, 2010 said:

    Nice result. MATLAB and the SAS/IML language are similar languages. If your collegue hasn’t solved the problem yet, I wrote a SAS/IML solution at
    http://blogs.sas.com/iml/index.php?/archives/3-Filling-an-Upper-Triangular-Matrix-from-a-Vector.html

  3. 3 bub on October 11, 2010 said:

    I have two questions:

    – how much longer are ‘long’ minutes than normal minutes? :o) just kidding…but I would probably have struggled days or weeks..

    – should I use matlab or R, are they mutually exclusive in any way?

    Thanks.

  4. 4 Atul on November 15, 2010 said:

    Check out, looks much simpler than the loop suggested by you:
    sparse(i,j,s,m,n)

  5. 5 Atul on November 15, 2010 said:

    To be more specific:

    [i,j]=find(triu(ones(n)))
    M=full(sparse(i,j,v,n,n))

    and you are done. Way more compact.

  6. 6 Saurabh Chandorkar on January 10, 2011 said:

    Very occasions in MATLAB require loops. I can always write a single or two line code to do what you want without resorting to loops. In short, even moderately advanced MATLAB user will know never to touch a loop! I am no expert in R, but I can do what you need in a couple of MATLAB code lines. No problem at all.

    But, your post at least suggests that R can do what I can do in MATLAB in just a couple of lines of compact code. Thanks.

  7. 7 miguel on March 26, 2011 said:

    In Matlab! ;)
    For example, your vector for some n:
    n = 5;
    Your vector:
    v = 1:n*(n+1)/2;
    An your matrix
    M = fliplr(triu(ones(n)));
    M(logical(M)) = v;
    Ciao

  8. 8 Xiaolong on June 10, 2011 said:

    m=zeros(n);
    m(logical(triu(ones(n))))=v;

  9. 9 Michael on August 8, 2011 said:

    Is there any point in replying to such an old thread?

    I’d do:

    j=matrix(1:n,n,n); i=t(j)
    k=(2*n-i+2)*(i-1)/2
    M=matrix(0,n,n)
    M[j <= n-i+1]=v[(k+j)[j <= n-i+1]]

    Not so compact, but I like to be clear….

  10. 10 mjm on February 5, 2012 said:

    How do you do RREF on a matrix in R? I think matlab does it out of the box. In R, it seems that you’ve to write your own function!


  1. 1 The DO Loop
  2. 2 Programming Languages: R versus Matlab for quantitative finance? - Quora
  3. 3 Filling an Upper Triangular Matrix from a Vector - The DO Loop
« Progress bars in R (part II) – a wrapper for apply functions
Playing with the ‘playwith’ package »
%d bloggers like this: