Setting up vector function

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K O
K O 2017-11-9
编辑： K O 2017-11-10
I write a function SetUp to set up three vectors from an input vector k. Vector c = (-k(2); -k(3); ..... -k(n)) where I have n-1 elements. Vector d = (k(1)+k(2); k(2)+k(3); ...... k(n-1)+k(n); k(n)) where there are n elements And vector e = vector c.
Here is the code I wrote:
function [c,e,d] = SetUp
k = input('Enter k vector');
n = size(k,2);
for k1 = 1:n
c = -k(k1)*ones(n-1,1);
d = (k(k1)+k(k1+1))*ones(n,1); d(n) = k(n);
e = c;
end
end
By running this code I get an error: Index exceeds matrix dimensions.
Problem is in this line: d = (k(k1)+k(k1+1))*ones(n,1); d(n) = k(n);
In my opinion the problem is in indexing as the last term when Matlab runs d = (k(k1)+k(k1+1))*ones(n,1); is d = k(n)+k(n+1), where the program does not recognize k(n+1) as there is no such term in vector k.
So, the question is, how to overcome this problem as I don't even need the last term as the last term is d(n) = k(n)???
Thanks in advance!
  1 个评论
Guillaume
Guillaume 2017-11-9
You have to ask yourself several questions:
  • What is the purpose of the loop in your code? Note that at each iteration, you're completely overwriting the previous value of c, d and e.
  • Is a loop required to do what you want? (Hint: No)
  • How can you have a different equation for the last term? (Hint: calculate all the other terms from 1 to n-1 instead of up to n)
  • What happens with your code if the user enters a column vector instead of a row vector? (Hint: use numel instead of size(x, 2))

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采纳的回答

Eric
Eric 2017-11-9
编辑：Eric 2017-11-9
In your for loop, you are using k(k1+1). if k1 is the last element of k, then k(k1+1) will be exceeding the dimension of k. If you dont care about the last term, use
for k1 = 1:n-1
... Stuff ...
end
d(n) = k(n);
or if you need the other vectors to go to n, stick it in a conditional
for k1 = 1:n
... Other Stuff ...
if k1~=n
d = (k(k1)+k(k1+1))*ones(n,1);
end
end
  1 个评论
K O
K O 2017-11-9
I see what I was missing.
if k1~=n
d = (k(k1)+k(k1+1))*ones(n,1);
end
Thanks man!

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更多回答（1 个）

KL
KL 2017-11-9
编辑：KL 2017-11-9
c = -k(2:end); %EDITED
d = k+[k(2:end);0];
e = c;
in case, k is row vector,
c = -k(2:end); %EDITED
d = k+[k(2:end),0];
e = c
that's all you need. No need for a loop!
  3 个评论
显示 1更早的评论
K O
K O 2017-11-9
I'll try without the for loop too.
Ty KL and Eric for discussion.
KL
KL 2017-11-9
Bam! I totally overlooked it! Thanks, I'll edit the answer.

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