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I am trying to use a code written by @Mr Wizard here: Subtract list with part of the same list to remove the baseline of a curve and have the baseline be at zero. The code is as follows:

data = Import["https://pastebin.com/raw/QCAKwZ2P", "Package"];
dat1 = Select[data, 60 <= First[#] <= 140 &];
dat2 = Select[data, 10 <= First[#] <= 70 &];
dat3 = Select[data, 104 <= First[#] <= 150 &];
conectline = {Last[dat2], First[dat3]};

ts2raw = TimeSeries[dat2];
ts3raw = TimeSeries[dat3];
ts4raw = TimeSeries[conectline];
datglass = Array[{#, ts2raw@#} &, Length@dat1, MinMax[First /@ dat2]];
datliquid = Array[{#, ts3raw@#} &, Length@dat1, MinMax[First /@ dat3]];
datline = 
Array[{#, ts4raw@#} &, Length@dat1, MinMax[First /@ conectline]];

Which plotted using ListPlot[{dat1, datglass, datliquid, datline}, PlotStyle -> {Black, Red, Darker[Green], Purple}, PlotRange -> All] gives:

image

I thought that here simply by subtracting the red, purple and green line from the entire curve such as ListPlot[{dat1 - (datliquid + datglass + datline)}, PlotRange -> All], I should get the plot with baseline of zero but I get something very different.

How can I implement this code to subtract the baseline and have it at zero?

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    $\begingroup$ In this answer to one of your previous questions on this topic I showed you how to obtain a peak area. Part of that process involved interpolating a baseline for each area you selected. Could you not repurpose that code from within that peakArea function to your current needs? $\endgroup$
    – MarcoB
    Commented Jun 23, 2020 at 2:28
  • $\begingroup$ @MarcoB yes, I think I could actually. One of the purpuses I sometimes ask these questions is because I am trying to learn as much as possible Mathematica from different angles. I know this is unorthodox but this provides me with more tools to learn how to program better. I hope that's okay and that's also the reason when I am able to do what I ask I also post the answer to the questions. $\endgroup$
    – John
    Commented Jun 23, 2020 at 2:44

2 Answers 2

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data = Import["https://pastebin.com/raw/QCAKwZ2P", "Package"];

ts = TimeSeries @ data;

windows = {{10, 70}, {104, 150}};

{ts2, ts3} = TimeSeriesWindow[ts, #] & /@ windows;

ts23 = TimeSeries @ Join[ts2 @ "Path", ts3 @ "Path"];

window = {60, 120};

{tsw, tsw23} = TimeSeriesWindow[#, window] & /@ {ts, ts23};

ListLinePlot[{ts, tsw - tsw23}, PlotRange -> All]

enter image description here

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  • $\begingroup$ This should likely be faster than what OP was able to attempt, correct? $\endgroup$ Commented Jun 23, 2020 at 11:48
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    $\begingroup$ @CATrevillian, using TimeSeriesWindow instead of Select[...] is faster. $\endgroup$
    – kglr
    Commented Jun 23, 2020 at 14:43
  • $\begingroup$ Ah! Very good! We should be able to implement actual criteria/conditionals within TimeSeriesWindow if I'm not mistaken. That is likely to slow it back down, though. $\endgroup$ Commented Jun 23, 2020 at 14:52
  • $\begingroup$ Thanks kglr! I have accepted your answer as it it more compact, better and way more elegant than mine! $\endgroup$
    – John
    Commented Jun 23, 2020 at 15:40
  • $\begingroup$ @John, thank you for the accept. $\endgroup$
    – kglr
    Commented Jun 23, 2020 at 20:12
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This is a rather non-efficient way to solve the problem as I have to Interpolate the data and make sure that the interpolation have the same lenght, but this does the job:

dat1 = Select[data, 40 <= First[#] <= 140 &];
dat2 = Select[data, 10 <= First[#] <= 70 &];
dat3 = Select[data, 104 <= First[#] <= 150 &];
conectline = {Last[dat2], First[dat3]};

ts2raw = TimeSeries[dat2];
ts3raw = TimeSeries[dat3];
ts4raw = TimeSeries[conectline];
datglass = Array[{#, ts2raw@#} &, Length@dat1, MinMax[First /@ dat2]];
datliquid = Array[{#, ts3raw@#} &, Length@dat1, MinMax[First /@ dat3]];
datline = 
  Array[{#, ts4raw@#} &, Length@dat1, MinMax[First /@ conectline]];
baseline = DeleteDuplicates[Join[datglass, datline, datliquid]];


ts5raw = TimeSeries[baseline];
datbaseline = 
  Array[{#, ts5raw@#} &, Length@dat1, MinMax[First /@ baseline]];

baseinter = Interpolation[datbaseline];
curve = Interpolation[dat1];

Plot[curve[x] - baseinter[x], {x, 60, 120}, PlotRange -> All]

Which gives:

enter image description here

Also notice that ListPlot[{dat1 - datbaseline}, PlotRange -> All] does not work eventhough dat1 and datbaseline are the same lenght and hence the need to interpolate datbaseline and dat1 at the end.

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  • $\begingroup$ This is what I would have suggested. Honestly though, you’re only interpolating from the one side of the connectline to the other, perhaps you can get something from that? $\endgroup$ Commented Jun 23, 2020 at 3:43

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