How can i select certain elements from a vector and set them to zero?
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I have the curve below. it is about a vector of 128 elements and i have to take just the peak and the rest must be replaced to zero. I think it is simple but i don't how because i am a beginner. I thought, i have to perhaps put the peak in anther vector and refill it with zeros but would it be take more time?
So I write the code below but i didn't get what i want, i am trying to creat a loop to test every element if it is smaller than zero and replace it to zero. The whole code workes without any changes in the curve and i get no error.
for i = 1 : length(Vector); % loop to test the elements of the curve from 1 to 128
VectorCurve = PPW ./ Rp + (C .* CPW); % the equation of the curve of 128 elements
if VectorCurve(i) < 0
VectorCurve (i) == 0;
end
end
Has anyone an idea to help me?
5 个评论
Paolo
2018-7-2
Do you just want the peak value or all positive values? You can remove replace negative values with zeros with logical indexing:
VectorCurve(VectorCurve<0) = 0;
Ruaa
2018-7-2
It is actually rare that you have to use a loop in matlab. Most likely, shifting the peak to 0 can be done in one line of code without a loop.
However, I have no idea what shifting the peak to 0 mean in practice. Please, give a numerical example of input and corresponging desired output.
Ruaa
2018-7-2
Thank you Guillaume for your reply:).
Well at the beginning i solve this Problem individually with the code below, so that i saw the values of the curve on the plot and the i replace the negatives with zeros in the code.
VectorCurve = PPW ./ Rp + (C .* CPW); % the equation of the curve of 128 elements
VectorCurve(1:7) = 0;% Zero-Replace before the Peak
VectorCurve(50:128) = 0;% Zero-Replace after the Peak
VectorCurve = circshift(VektorCurve, [0,-7]);% Shifting the Peak to zero
But this was just for one curve and i have many of curves with differnet peaks. In Image is the final plot what i have to get. So, do you think i don't need to the loop?
Ruaa
2018-7-2
采纳的回答
Guillaume
2018-7-2
It has worked with the Code below and without a loop
This is indeed the way to do it. I would recommend one change though:
FF = find(VectorCurve> 0, 1); %only get the location of the first element > 0
As you wrote it, FF is a vector of all the indices for which your curve is > 0. So, in your earlier example, FF would be [8, 9, 10, 11, ..., 49]. You're then passing that vector to circshift which shift the respective dimensions by 8, 9, 10, 11, etc. position. Thankfully, since dimensions 3 to 44 are all singleton, it doesn't do anything but really, you only meant to pass the 8. Adding the 1 option to find means it only returns that 8 you wanted.
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