필터 지우기
필터 지우기

Replace repeatative values by using interpolation ?

조회 수: 2 (최근 30일)
Koustubh Shirke
Koustubh Shirke 2020년 3월 14일
댓글: Ameer Hamza 2020년 3월 14일
Hi Matlabers ,
I have a vector which has repeatative values from 0 to 1.
I want to relace those repeataive values by using interpolated values in matlab script. As a result, I will have no any repeatative value and the plot/ and curve of that vector will be smooth.
Thanks in advance.

채택된 답변

Ameer Hamza
Ameer Hamza 2020년 3월 14일
Try this
x = [0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1]'; % generate random data
indexes = [1; find(x(2:end) - x(1:end-1))+1];
x_new = zeros(size(x));
for i=1:numel(indexes)-2
x_new(indexes(i):indexes(i+1)) = interp1([0 1], x(indexes([i i+1])), linspace(0, 1, indexes(i+1)-indexes(i)+1));
end
x_new(indexes(end-1):end) = interp1([0 1], x([indexes(end-1) end]), linspace(0, 1, numel(x)-indexes(end-1)+1));
  댓글 수: 2
Koustubh Shirke
Koustubh Shirke 2020년 3월 14일
Hi Ameer,
Thanks for the answe. Its really helpfu.
It seems , it keeps first value of start of any duplicate. Can I also shift it to centre ? So that I will keep all centred value of duplictaes and then interploate remained ?
Ameer Hamza
Ameer Hamza 2020년 3월 14일
try something like this
x = [0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0]'; % generate random data
indexes = find(x(2:end) - x(1:end-1))+1;
indexes = floor(movmean(indexes, 2, 'Endpoints', 'discard'));
x_new = zeros(size(x));
x_new(1:indexes(1)) = interp1([0 1], x([1 indexes(1)]), linspace(0, 1, indexes(1)));
for i=1:numel(indexes)-1
x_new(indexes(i):indexes(i+1)) = interp1([0 1], x(indexes([i i+1])), linspace(0, 1, indexes(i+1)-indexes(i)+1));
end
x_new(indexes(end):end) = interp1([0 1], x([indexes(end) end]), linspace(0, 1, numel(x)-indexes(end)+1));

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Creating and Concatenating Matrices에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by