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 个评论
Jan
2017-7-26
Is this a homework question? If so, please mention it, because then a different type of answers is required.
采纳的回答
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)
5 个评论
Xh Du
2017-7-27
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
2019-4-8
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
2019-5-1
This should do the trick:
A=1:6
B=tril(ones(3))
B(B==1)=A
B'
Anna Iatckova
2020-7-30
A=1:6
B=triu(ones(3))
B(B==1)=A
saves you a line.
更多回答（1 个）
Roger Stafford
2017-7-26
Let vector ‘inpt’ have size = n*(n+1)/2,1.
otpt = zeros(n);
otpt(triu(ones(n),0)==1) = inpt;
3 个评论
Prabhjot Dhami
2020-4-23
Thanks for this!
warnerchang
2021-6-4
Brilliant! it's actually the sum formula for arithmetic sequence! very helpful for understanding.
KUMAR TRIPATHY
2021-10-3
Absolutely brilliant, concise and crisp!
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