stopping the loop and save in a cell array

1 次查看(过去 30 天)
Hi,
I have a vector:
a = [1 3 4 2 3 4 5 6 3 4 5 6 8 9 10 9 8 7 6 7 8 9 8 7 5 4 3 2 4 5 3 1 2 4 4 3 2 4 6 5 4 6 7 9 9 10 9 8 7 6 7 8 9 8 7 5 4 3 2 4 5 3 1];
The output I would like to have is one cell: {[3 4 2 3 4 5 6 3 4 5 6 8]; [4 4 3 2 4 6 5 4 6 7]} as underlined in the original vector. The code I have written is shown below. I am stuck with getting a very long vector for the code I wrote.
Could anyone please advise how to correct the code? Thanks a lot!
The purpose is to split general forward and back directions. The values are like positions. for example, I am aware that there are backward tracks in the desired output [3 4 2 3 4 5 6 3 4 5 6 8]. I would like to ignore first. 1,2,9 and 10 are positions at the extremes which are excluded as min(a)+2 and max(a) -2.
a = [1 3 4 2 3 4 5 6 3 4 5 6 8 9 10 9 8 7 6 7 8 9 8 7 5 4 3 2 4 5 3 1 2 4 4 3 2 4 6 5 4 6 7 9 9 10 9 8 7 6 7 8 9 8 7 5 4 3 2 4 5 3 1 ];
start = min(a)+2;
ed = max(a) -2;
m = length(a);
f = 1;
i = 2; k=1;
while i < m
if a(i) >= start && a(i-1) <= start && a(i+1) >= a(i)
fwd1(k) = a(i);
for j = i+1:m-1
if a(j+1) >= ed && a(j+2) >= a(j+1) && a(j) <= a(j+1)
fwd2(k) = a(j);
output{k} = a (i:j);
end
end
i=j+1;
else
i = i+1;
end
k= k+1;
end

回答(1 个)

KSSV
KSSV 2020-5-17
Follow somehting likethis:
a = [1 3 4 2 3 4 5 6 3 4 5 6 8 9 10 9 8 7 6 7 8 9 8 7 5 4 3 2 4 5 3 1 2 4 4 3 2 4 6 5 4 6 7 9 9 10 9 8 7 6 7 8 9 8 7 5 4 3 2 4 5 3 1];
[val0,id0] = min(a) ;
[val1,id1] = max(a) ;
iwant1 = a(id0+1:id1-2) ;
a(id0:id1) = [] ;
% second time
[val0,id0] = min(a) ;
[val1,id1] = max(a) ;
iwant2 = a(id0+1:id1-2) ;
a(id0:id1) = [] ;
  1 个评论
HYZ
HYZ 2020-5-17
Sorry ... the vector I gave was an example. The index of where the output vectors start will not be the same always. It would be good to use loop to check each position by each.
If possible, could you help me advise using loop as I mentioned as above? Thank you.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Loops and Conditional Statements 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by