How to interpolate between two adjacent points of time series data, if the difference between any two consecutive points is greater than 25 by using for loop?
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I have a time series data like this;
x = [555 554 100 10 36 20 17 5 51 70 101 40 13];
and independent variables as follow
L = numel(x);
fs = 4;
t=0:1/fs:(L-1)/fs;
What I want is to use spline interpolation between two consecutive point of 'x' means (xi and xi+1) if the difference between two point is greater than 25. for example from the above time series abs(x4-x3)= 10-100= 90 therefore I want to interpolate between x3 and x4 with t interval of 0.05 and return all the new time series data by using a loop. The loop have to continue until the difference between every consecutive data in “x” becomes less than 25.
Thanks for your contribution in advance
댓글 수: 2
J. Alex Lee
2021년 5월 26일
Is there any reason you cannot just live with interpolating the whole curve with the new time interval?
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Jan
2021년 5월 26일
편집: Jan
2021년 5월 26일
x = [555 554 100 10 36 20 17 5 51 70 101 40 13];
L = numel(x);
fs = 4;
t = 0:1/fs:(L-1)/fs;
ready = false;
dt = 0.05;
xx = x;
tt = t;
while ~ready
miss = find(abs(diff(xx)) > 25);
if isempty(miss)
ready = true;
else
% Insert new time values for gaps:
for k = numel(miss):-1:1
tt = [tt(1:miss(k)-1), ...
tt(miss(k)):dt:tt(miss(k)+1), ...
tt(miss(k)+2:end)];
end
xx = interp1(t, x, tt, 'spline'); % Using original x, not xx!
dt = dt / 10; % Decrease time interval for next iteration
end
end
figure;
axes;
plot(t, x, 'ro');
hold('on');
plot(tt, xx, 'b.');
댓글 수: 7
Jan
2021년 5월 28일
interp1 is slow. A griddedInterpolant will be faster.
If the input is large, the iterative growing of tt is a bad idea.
Maybe this solution is a bad strategy. So please explain, which problem you want to solve actually. If all want want to achive is a list of numbers with gaps < 25, there are better methods.
추가 답변 (2개)
J. Alex Lee
2021년 5월 28일
Here's [another approximate] way to do it that will only call spline once (could as well do it with gridded interpolant)
The method I suggested above won't work with spline because the interval between t = 0.5 and t = 0.75 is not monotonic after the global spline fit.
So this method is still only approximate to satisfy dx_min.
x = [555 554 100 10 36 20 17 5 51 70 101 40 13];
L = numel(x);
fs = 4;
t = 0:1/fs:(L-1)/fs;
% do the spline fit to the whole set of points once
pp = spline(t,x)
% define the minimum gap in x you want
dxmin = 25
x_agaps = abs(diff(x))
t_agaps = abs(diff(t))
dx_mult = 1.2; % extra refinement tweak
% find the intervals where your gap is violated
intvls = find(x_agaps > 25);
% for each interval, generate
tc = cell(numel(intvls),1);
xc = cell(numel(intvls),1);
for i = 1:numel(intvls)
% for i = 2
idx = intvls(i);
% split the interval into n points such that
% ROUGHLY minimum gap is satisfied (assuming linear)
n = ceil(x_agaps(idx)/dxmin * dx_mult);
tt = linspace(t(idx),t(idx+1),2+n);
xx = ppval(pp,tt);
tc{i} = tt(2:end-1);
xc{i} = xx(2:end-1);
end
tI = horzcat(t,tc{:});
xI = horzcat(x,xc{:});
[tI,srtIdx] = sort(tI);
xI = xI(srtIdx);
% analyze violations there are...
abs(diff(xI))
idxViol = find(abs(diff(xI))>dxmin)
figure(1); cla; hold on;
plot(t,x,'.','MarkerSize',12)
plot(tI,xI,'o')
fplot(@(t)(ppval(pp,t)),t([1,end]),'-')
plot(tI(idxViol),xI(idxViol),'x','MarkerSize',9,'LineWidth',2)
댓글 수: 1
Jan
2021년 5월 28일
Of course, this is an efficient idea: My code computes the spline interpolation for the same intervals repeatedly, but computing te spline onces and insert only additional interpolation steps is leaner and faster.
J. Alex Lee
2021년 5월 28일
If you just want the points to be "appear" roughly evenly distributed, you could consider spacing in the arc length (although you'd have to re-scale one of the dimensions so that it looks "even" in the plot)
x = [555 554 100 10 36 20 17 5 51 70 101 40 13];
L = numel(x);
fs = 4;
t=0:1/fs:(L-1)/fs;
% interpolate more than you ultimately need
tI = linspace(t(1),t(end),2000)';
xI = interp1(t,x,tI,'spline');
% scale by
% scl = 200;
scl = max(xI)/max(tI)
% parameterize the ultra-refined curve w.r.t. its own arclength s
sI = [0;cumsum(sqrt(scl^2*diff(tI).^2+diff(xI).^2))];
% then interpolate both x and t w.r.t. s
% in the steepest part of the curve, dx ~ 25 corresponds to
se = linspace(sI(1),sI(end),50)';
te = interp1(sI,tI,se);
xe = interp1(sI,xI,se);
figure(1); cla; hold on;
plot(t,x,'x')
plot(te,xe,'o')
plot(tI,xI,'-')
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