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array generation using logics.

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Sathiya S V
Sathiya S V 2024-5-29
评论： Voss 2024-5-29
x = [0 10 20 30 0 10 20 30 40 0 10 20 30 40 50 0 10]
'x' is the array to be checked.
checked = [0 10 -10 20 0 10 -10 20 -20 0 10 -10 20 -20 30 0 10]
I want the new array like the above mentioned 'checked' array.
The concept is simple to understand in the above given example, if the numbers are countinously increasing in the order of 10 factors in the x array, I need a checked array of incrementing values containing positive and negative number and next should be a next consecutive number in positive sign.
It is simple for me to understand, but for coding I really finding it difficult..I request someone who is more into finding logics to help me with.

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采纳的回答

Voss
Voss 2024-5-29
Ran in:
x = [0 10 20 30 0 10 20 30 40 0 10 20 30 40 50 0 10];
diffx = diff(x(:));
start = find([true; diffx <= 0; true]);
dx = min(diffx(diffx > 0));
checked = zeros(size(x));
sequence_length = diff(start);
for ii = 1:numel(sequence_length)
n = 1:sequence_length(ii);
checked(start(ii):start(ii+1)-1) = ceil((n-1)/2)*dx.*(-1).^n;
end
disp(checked);
0 10 -10 20 0 10 -10 20 -20 0 10 -10 20 -20 30 0 10

更多回答（1 个）

Dyuman Joshi
Dyuman Joshi 2024-5-29
Ran in:
x = [0 10 20 30 0 10 20 30 40 0 10 20 30 40 50 0 10]
x = 1x17
0 10 20 30 0 10 20 30 40 0 10 20 30 40 50 0 10
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y = x;
idx = [1 find(diff(x)~=10)+1 numel(x)]
idx = 1x5
1 5 10 16 17
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for k=1:numel(idx)-1
vec = (1:idx(k+1)-idx(k)-1);
y(vec+idx(k)) = ceil(vec/2).*10.*(-1).^(vec-1);
end
y
y = 1x17
0 10 -10 20 0 10 -10 20 -20 0 10 -10 20 -20 30 0 10
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