Put elements into corresponding locations of upper triangular matrix
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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!
1 Comment
Jan
on 26 Jul 2017
Is this a homework question? If so, please mention it, because then a different type of answers is required.
Accepted Answer
Jan
on 26 Jul 2017
Edited: Jan
on 26 Jul 2017
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)
More Answers (1)
Roger Stafford
on 26 Jul 2017
Let vector ‘inpt’ have size = n*(n+1)/2,1.
otpt = zeros(n);
otpt(triu(ones(n),0)==1) = inpt;
3 Comments
warnerchang
on 4 Jun 2021
Brilliant! it's actually the sum formula for arithmetic sequence! very helpful for understanding.
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