polyfit with time vector.

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captainmariah
captainmariah 2016-7-28
评论： Steven Lord 2016-7-28
Hi, Im encountered some problems while detrending some data. I have data for example:
15:02:13 0.24
15:02:14 0.25
15:02:16 0.27
and I am trying to find the linear fit to this data to remove the drift.
But when using polyfit and then polyval I get NaN values in polyval and Warning: Polynomial is badly conditioned. Add points with distinct X values, reduce the degree of the polynomial, or try centering and scaling as described in HELP POLYFIT.
My code:
%[x y] = ginput(5) % get points with a click 5 points
dxp = [ 7.3633
7.3633
7.3633
7.3633
7.3633]*1.0e+05;
dyp =[0.1380
0.1380
0.1344
0.1344
0.1292];
degree = 1; % to be adjusted
[pol, S, mu] = polyfit(dxp, dyp, degree);
xp = polyval(pol, t1, [], mu);
dI1 = I1 - xp; % detrended phase
The time vector is the the form above with 10^5 something. I think this is causing the problem. The question is how I can find the slope of the curve to remove the drift .. and plot the detrended curve? I don't know how to solve this. Any suggestions?

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John D'Errico
John D'Errico 2016-7-28
编辑：John D'Errico 2016-7-28
You think you have 5 points.
In reality, you have ONE point, with 5 replicates.
dxp = [ 7.3633
7.3633
7.3633
7.3633
7.3633]*1.0e+05;
ONE point. One single value for x.
How many points determine a line? TWO points, at least. With only one point, no line is possible. EVER. PERIOD.
Yes, I know, but datenum gave you those numbers!!!!
datenum('15:02:13 0.2')
ans =
7.3633e+05
datenum('15:02:14 0.25')
ans =
7.3633e+05
Did it give you those numbers? REALLY?
format long g
datenum('15:02:13 0.2')
ans =
736330.626539352
datenum('15:02:14 0.25')
ans =
736330.626550926
They ARE different. You just chose to supply the wrong numbers when you used polyfit.
You needed to save the EXACT results of datenum in a vector.
By the way, if you tried to use a higher order polyfit than 1 (linear), polyfit will start to fail here again. But that is a completely different story.
Garbage in, garbage out.
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John D'Errico
John D'Errico 2016-7-28
Again, you need to use a scheme that gives you sufficient precision to distinguish the numbers as different.
Steven Lord
Steven Lord 2016-7-28
I would avoid using datenum in this case.
start = datetime('now');
d = randi([1 180], 1, 10);
sampledata = start + cumsum([0, seconds(d)]);
timeBetweenPoints = minutes(sampledata-start)
You wouldn't know what d is in your problem; what you would know is start (the earliest time in your time vector; use min to compute it if your time vector is not sorted) and sampledata (the whole time vector.) Compute timeBetweenPoints (number of minutes since the start -- if your data has a different granularity you may want to use seconds, hours, days, etc.)
Alternately, you could subtract off the mean or median time instead of the min and convert that result to your desired units to use in polyfit.

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