Put elements into corresponding locations of upper triangular matrix

Hi all,

Imagine I have a vector:

inpt = (1:6)';

Now I'd like to put elements of inpt in the upper triangular part of a 3 by 3 matrix otpt, so I have:

otpt =
     1     2     4
     0     3     5
     0     0     6

What's the best way to do it? Thanks!

[Jan]

Is this a homework question? If so, please mention it, because then a different type of answers is required.

Answer 1

[Jan]

Start with nested loops:

v = 1:6;
n = round((sqrt(8 * numel(v) + 1) - 1) / 2);
M = zeros(n, n);
c = 0;
for i2 = 1:n
  for i1 = 1:i2
    c         = c + 1;
    M(i1, i2) = v(c);
  end
end

In the next step you can vectorize the inner loop: Move the loop index inside the assignment:

n = round((sqrt(8 * numel(v) + 1) - 1) / 2);
M = zeros(n, n);
a = 1;
for k = 1:n
  b         = a + k - 1;
  M(1:k, k) = v(a:b);
  a         = b + 1;
end

Is this nicer? Questionable, but maybe faster.

Now use a built-in function:

n = round((sqrt(8 * numel(v) + 1) - 1) / 2);
M(triu(ones(n)) == 1) = v;

or better:

M(triu(true(n))) = v;

[EDITED] Some timings - what did you expect?

v = 1:5050;
tic; for k = 1:10000; y = SerialTriU(v); end, toc

Elapsed time is 0.772492 seconds.   % Two loops
Elapsed time is 2.448738 seconds.   % Inner loop vectorized
Elapsed time is 1.029641 seconds.   % TRIU(ONES)
Elapsed time is 0.659360 seconds.   % TRIU(TRUE)

[Xh Du]

Jan you are a legend!

[Jan]

Thanks. I admit, that the timings have less power, if you consider that they are taken for 10'000 loops. If the total time is 0.00024 or 0.000065 seconds is not such important usually. As long as this code is not repeated millions of time and the bottleneck of the code, optimizing it might be very accademically or even a waste of (life) time.

But in the forum I try to share the methods, how efficient code can be developped. Perhaps these methods are useful for other problems also.

[Soumyanil Banerjee]

Hi Jan,

Very good answer. I just had a question. How about if we want to do the following:

v = 1:6;

we want:

M =

1 2 3

0 4 5

0 0 6

Any help would be appreciated.

[Raphael]

This should do the trick:

A=1:6
B=tril(ones(3))
B(B==1)=A
B'

[Anna Iatckova]

A=1:6
B=triu(ones(3))
B(B==1)=A

saves you a line.

Answer 2

[Roger Stafford]

Let vector ‘inpt’ have size = n*(n+1)/2,1.

otpt = zeros(n);
otpt(triu(ones(n),0)==1) = inpt;

[Prabhjot Dhami]

Thanks for this!

[warnerchang]

Brilliant! it's actually the sum formula for arithmetic sequence! very helpful for understanding.

[KUMAR TRIPATHY]

Absolutely brilliant, concise and crisp!