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Generating a vector with a pattern?

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aadi
aadi 2011-2-4
How do I generate the following vector efficiently for large lengths?
A=[1^1 2^2 3^3 4^4 5^5 6^6]

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采纳的回答

Vieniava
Vieniava 2011-2-4
check this out:
A=[1:6].^[1:6];
  1 个评论
Andrew Newell
Andrew Newell 2011-2-4
Stylistically better is
A = (1:6).^(1:6);
because brackets are used to construct arrays and 1:6 already is an array.

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更多回答（2 个）

the cyclist
the cyclist 2011-2-4
>> v = 1:6;
>> A = v.^v;
Stylistically even better? ;-)
  3 个评论
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Walter Roberson
Walter Roberson 2011-2-4
Now how would we do tetration, v^^v?
http://en.wikipedia.org/wiki/Tetration
Walter Roberson
Walter Roberson 2011-2-4
Nevermind; 4^4^4^4 is the largest that fits within realmax.

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Jan
Jan 2011-2-4
This job overflows fast for large lengths and computing 200^200 is a waste of time.
EDITED: With Matt Fig's further improvements:
function x = XpowerX(n)
v = 1:min(n,143);
x = inf(1,n);
x(v) = v.^v;
This much faster than the direct 1:n approach:
tic; for i = 1:15000; y = XpowerX(i); end; toc
% 1.8 sec
tic; for i = 1:15000; v = 1:i; y = v.^v; end; toc
% 60 sec
  2 个评论
Matt Fig
Matt Fig 2011-2-4
How about:
v = 1:min(n,143);
x = inf(1,n);
x(v) = v.^v;
Jan
Jan 2011-2-4
@Matt: Some percent faster. I insert it.

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