x = [6 5 3 5 9 10 438 4 1 4 7 0 4 8 4 2];
len = length(x);
minval = x(1);
for n = 1:len
for i = 1:len
if x(i) < minval
minval = x(i);
end
end
sortval(i) = minval;
x(n find(x==minval)) = x(find(x==minval) n);
x(n) = [];
end
disp(sortval)
I am trying to swap the position of n and my minval so I can delete that minval later and not affect the process of sorting later in the for loop.

 채택된 답변

James Tursa
James Tursa 2017년 3월 9일
편집: James Tursa 2017년 3월 9일

0 개 추천

This isn't going to work like you hoped. If you delete an element from the array with this line:
x(n) = [];
Then there are no longer len elements in the x array, so this will eventually yield an error because i will be out of bounds for the shortened length of x (i.e., len is no longer the actual length of x):
for i = 1:len
if x(i) < minval
So first thing you need to do is rewrite your code to fix this problem, then we can worry about how to swap elements (if that step still remains in your fixed code).

추가 답변 (0개)

카테고리

도움말 센터File Exchange에서 Shifting and Sorting Matrices에 대해 자세히 알아보기

태그

아직 태그를 입력하지 않았습니다.

질문:

2017년 3월 9일

편집:

2017년 3월 9일

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by