Computing powers in matlab
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find A whose ith element is A = ( 2.^i ) X ./ Y .... where X and Y are vectors of equal length
without looping. ie. X = [1 2 3] Y = [4 5 6].
I know how to do X ./ Y but the ith element of (2.^i) is confusing me and I do not know how to do it without looping. Thanks
Edit. Thank you 2.^(1:length(x)) worked. What if it was -2 instead of 2??? I tried fooling around with it but I had no luck
回答 (3 件)
Andreas Goser
2014 年 1 月 16 日
doc power
Mischa Kim
2014 年 1 月 16 日
編集済み: Mischa Kim
2014 年 1 月 16 日
Does this do it?
X = [1 2 3]; Y = [4 5 6];
A = (2.^sort(randperm(length(X)))).*X./Y
A =
0.5000 1.6000 4.0000
1 件のコメント
Walter Roberson
2014 年 1 月 16 日
A little obfuscation to start the day right ;-)
Walter Roberson
2014 年 1 月 16 日
2.^(1:3)
4 件のコメント
Mischa Kim
2014 年 1 月 16 日
編集済み: Mischa Kim
2014 年 1 月 16 日
Am I reading the equation for A incorrectly? Does it not depend on X and Y? And yes, I did it for the general case of arbitrary vector sizes :)
Walter Roberson
2014 年 1 月 16 日
the ith element of (2.^i) is confusing me
so I showed an example of creating that portion. I could have generalized that portion as
2.^(1:length(X))
which is what your code does in an obscure way.
Mischa Kim
2014 年 1 月 16 日
Ok, I understand. For the vector size portion, I have to admit, I tried to come up with a challenge for myself.
Walter Roberson
2014 年 1 月 17 日
What if it was -2, mark asked.
(-2).^(1:length(X))
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