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Switching the elements of an array

Hans123 님이 질문을 제출함. 25 Mar 2019
최근 활동 madhan ravi 님이 편집함. 25 Mar 2019
hi,
I am plotting data from 2 vectors, the x-axis elements (in a vector called M) needs to be flipped and make the first element 0, the last few elements in M recur like this [1 2 3...500 500 500 500...] and these recurring elements should discarded so it should be [...500 500 500 500 499 498... 3 2 0]
The method I went with is flipping each element by creating a new vector (called N) and using a for loop to switch the order. This however changes the size of my array because it ignores the recurring elements. I want to know how to fix this issue.
My code is
for k=1:max(M)-1
N(max(M)-k,:)=k;
end
N(max(M))=0;
The 600x1 M array, becomes a 500x1 N array
*The array M changes because it nested in another loop so I can't use the line for k=1:500
**and I can't use the line for k=1:length(M)-1, because I get the error - Subscript indices must either be real positive integers or logicals.
(Because max(M) = 500 and the length = 600 and N(-100) is not allowed

  댓글 수: 2

Please stop re-posting the same question. You have been given good answers. Use of flip is simple, easy to remember, and efficient.
Isn’t it straight forward?:
[M(end:-1:2),0]
I am unfamiliar with that line of code, how does it work?
Also, I get the error -
Error using horzcat
Dimensions of matrices being concatenated are not consistent.
size(M) %? what is the size of M ?? is it a vector?
%or try:
MM = M(end:-1:1);
MM(end) = 0
The vector is reversed until the second element and the last element is set as zero (which is the first element in the original vector). You have the code , try it out!
Illustrate with a short 1 by 10 vector of M and N so that it's clear what your desired result should be.
Thanks for that explanation
Still have the error - Error using horzcat
Dimensions of matrices being concatenated are not consistent
M is a varying size matrix, it is in a nested loop so after each iteration the size changes
Also, I couldn't mention this. The element of N that corresponds to M should be swapped too.
I realized this now after playing with the code. I would really appreciate it if you can tell me what is th best approach? Should I swap the N matrix as well.
In case I didn't word my question properly
N = [A B C ... ZZZ]
M = [1 2 3.. 500 500]
should be swapped to
N = [ZZZ... A B C]
and
M = [500 500 500... 3 2 0]
M vector = [1 2 3 4 5 6 7 8 8 8]
N vector = [10 20 30 40 50 60 70 80 90 100]
Desired Result;
M vector (swapped) = [8 8 8 7 6 5 4 3 2 0]
N vector = [100 90 80 70 60 50 40 30 20 10]
Both vectors should swap
M is a nx1 array, where n is in the range of 500 to 1000
M_Swapped = M(end:-1:1);
M_Swapped(end) = 0
N_Swapped = N(end:-1:1)
This might clear up what I am trying to achieve.
I want that graph to follow the best fit line in blue
If I just swap the x-axis the co-ordinates will get messed up, because the y-co-ordinate points pair up with the wrong x-co-ordinates.
My graph should be the same shape as the best fit line
I have used the below code previously as a temporary ffix, but all that does is flip the axes
%reverse axes
set(gca, 'XDir','reverse')
set(gca, 'XDir','reverse')
xt = get(gca, 'XTick');
xtr = linspace(max(xt), min(xt), numel(xt));
set(gca, 'XTick',xt, 'XTickLabel',sprintfc('%.1f', xtr))
Using
M_Swapped = M(end:-1:1);
M_Swapped(end) = 0
N_Swapped = N(end:-1:1)
I end up with the original plot! I feared that would happen.
I need to swap the points such that it gets plotted like this

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답변 수: 1

Alex Mcaulley 님의 답변 25 Mar 2019
 채택된 답변

N = flip(M);
N(end) = 0;

  댓글 수: 2

I couldn't mention this. The element of N that corresponds to M should be swapped too.
I realized this now after playing with the code. I would really appreciate it if you can tell me what is th best approach? Should I swap the N matrix as well.
In case I didn't word my question properly
N = [A B C ... ZZZ]
M = [1 2 3.. 500 500]
should be swapped to
N = [ZZZ... A B C]
and
M = [500 500 500... 3 2 0]
So just swap N AND M. What is the problem? Use of flip will be efficient.

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