1
$\begingroup$

I have the following data:

data = {{66.803, 0.0592517}, {66.836, 0.0591485}, {66.87, 0.058002}, {66.904, 0.0569173}, {66.937, 0.0570737}, {66.97, 0.0570687}, {67.003, 0.0572252}, {67.036, 0.0573801}, {67.068, 0.0577948}, {67.102, 0.0577485}, {67.135, 0.0577799}, {67.168, 0.0581311}, {67.201, 0.058161}, {67.234, 0.0583523}, {67.268, 0.0583076}, {67.301, 0.0584974}, {67.334, 0.0585288}, {67.368, 0.0581627}, {67.4, 0.0584285}, {67.433, 0.0584568}, {67.467, 0.0582475}, {67.5, 0.0582757}, {67.532, 0.05854}, {67.565, 0.0585698}, {67.598, 0.0585997}, {67.631, 0.0584712}, {67.664, 0.0586625}, {67.698, 0.0584563}, {67.731, 0.0582883}, {67.764, 0.0580917}, {67.796, 0.0582942}, {67.829, 0.0580977}, {67.863, 0.0578235}, {67.896, 0.0577852}, {67.93, 0.0573464}, {67.963, 0.0573065}, {67.996, 0.0572651}, {68.029, 0.0570654}, {68.062, 0.0571538}, {68.095, 0.0573135}, {68.128, 0.0573148}, {68.161, 0.0574761}, {68.194, 0.0576357}, {68.227, 0.0577004}, {68.261, 0.0574547}, {68.293, 0.0576841}, {68.326, 0.0576759}, {68.359, 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0.131062}, {89.267, 0.132783}, {89.3, 0.134823}, {89.333, 0.136063}, {89.367, 0.137123}, {89.4, 0.13862}, {89.433, 0.140276}, {89.466, 0.142734}, {89.499, 0.14407}, {89.532, 0.145794}, {89.565, 0.148355}, {89.598, 0.150275}, {89.631, 0.152037}, {89.665, 0.153565}, {89.697, 0.155611}, {89.731, 0.157232}, {89.763, 0.159315}, {89.796, 0.161167}, {89.829, 0.163179}, {89.862, 0.165369}, {89.896, 0.168621}, {89.929, 0.170983}, {89.962, 0.173187}, {89.994, 0.175938}, {90.027, 0.178621}, {90.06, 0.181144}, {90.093, 0.183669}, {90.127, 0.186124}, {90.16, 0.189289}, {90.194, 0.192127}, {90.227, 0.197628}, {90.259, 0.201274}, {90.292, 0.205174}, {90.325, 0.209235}, {90.358, 0.212996}, {90.392, 0.217023}, {90.425, 0.224796}, {90.457, 0.230713}, {90.491, 0.239061}, {90.523, 0.246099}, {90.556, 0.255632}, {90.59, 0.262701}, {90.622, 0.274059}, {90.655, 0.282792}, {90.688, 0.292805}, {90.721, 0.305059}, {90.753, 0.317377}, {90.786, 0.330112}, {90.819, 0.341404}, {90.852, 0.352818}, {90.885, 0.365167}, {90.918, 0.378155}, {90.951, 0.391464}, {90.983, 0.40884}, {91.016, 0.422687}, {91.048, 0.438718}, {91.081, 0.458206}, {91.115, 0.481631}, {91.148, 0.505117}, {91.182, 0.530002}, {91.214, 0.564786}, {91.247, 0.608147}, {91.28, 0.663512}, {91.313, 0.7368}, {91.347, 0.834546}, {91.381, 0.963205}, {91.414, 1.12633}, {91.446, 1.3012}, {91.479, 1.4384}, {91.511, 1.48419}, {91.545, 1.43025}, {91.578, 1.30693}, {91.61, 1.15087}, {91.643, 0.987705}, {91.675, 0.835225}, {91.708, 0.700161}, {91.742, 0.580875}, {91.775, 0.475575}, {91.808, 0.388037}, {91.842, 0.316299}, {91.874, 0.25654}, {91.908, 0.205297}, {91.941, 0.168518}, {91.974, 0.141022}, {92.006, 0.12249}, {92.04, 0.108595}, {92.072, 0.0868284}, {92.106, 0.078694}, {92.139, 0.0727048}, {92.172, 0.0618365}, {92.206, 0.0584736}, {92.238, 0.0531542}, {92.27, 0.0505546}, {92.303, 0.0475734}, {92.336, 0.0452127}, {92.369, 0.0439602}, {92.402, 0.0427077}, {92.435, 0.0411339}, {92.469, 0.0398195}, {92.501, 0.0392685}, {92.534, 0.0383374}, {92.568, 0.0371813}, {92.6, 0.0366319}, {92.632, 0.0346417}, {92.665, 0.0341507}, {92.699, 0.0334095}, {92.732, 0.0328932}, {92.765, 0.0325367}, {92.797, 0.0322453}, {92.831, 0.0313457}, {92.865, 0.0302863}, {92.897, 0.0299948}, {92.93, 0.0294801}, {92.963, 0.0289653}, {92.995, 0.0287641}, {93.028, 0.0282795}, {93.062, 0.0277266}, {93.096, 0.026854}, {93.129, 0.0263709}, {93.162, 0.0260065}, {93.196, 0.0252289}, {93.229, 0.0241996}, {93.262, 0.0238115}, {93.295, 0.0235833}, {93.327, 0.0229452}, {93.36, 0.0228768}, {93.393, 0.0226486}, {93.426, 0.0224204}, {93.458, 0.022262}, {93.491, 0.0220512}, {93.524, 0.0216947}, {93.557, 0.0213383}, {93.589, 0.0213714}, {93.622, 0.0211748}, {93.655, 0.0209973}, {93.688, 0.0206741}, {93.722, 0.0201228}, {93.756, 0.019733}, {93.788, 0.0197978}, {93.821, 0.019492}, {93.853, 0.0197468}, {93.886, 0.0196135}, {93.919, 0.019322}, {93.953, 0.0186425}, {93.986, 0.0182069}, {94.02, 0.0175606}, {94.052, 0.0173674}, {94.085, 0.016946}, {94.118, 0.0166861}, {94.151, 0.0164263}, {94.185, 0.0157784}, {94.218, 0.0156784}, {94.251, 0.0154186}, {94.283, 0.0153836}, {94.317, 0.0148972}, {94.349, 0.0148639}, {94.382, 0.0146024}, {94.416, 0.014116}, {94.449, 0.0138562}, {94.483, 0.0132289}, {94.516, 0.0130007}, {94.549, 0.0127725}, {94.582, 0.0123844}, {94.615, 0.0121562}, {94.647, 0.0119566}, {94.68, 0.011665}, {94.713, 0.0113735}, {94.746, 0.0110835}, {94.779, 0.0106321}, {94.812, 0.010358}, {94.844, 0.0103214}, {94.877, 0.01006}, {94.909, 0.00986512}, {94.942, 0.00960366}, {94.977, 0.00887182}, {95.009, 0.00880519}, {95.042, 0.00851049}, {95.076, 0.0079893}, {95.108, 0.00792267}, {95.141, 0.00762955}, {95.174, 0.00749791}, {95.207, 0.00720479}, {95.24, 0.00707315}, {95.274, 0.00671343}, {95.308, 0.00633472}, {95.34, 0.00639791}, {95.373, 0.00623302}, {95.406, 0.00606972}, {95.439, 0.00574494}, {95.472, 0.00558005}, {95.505, 0.00541675}, {95.539, 0.00502379}, {95.571, 0.00524687}, {95.604, 0.00508357}, {95.637, 0.00491868}, {95.67, 0.00491369}, {95.703, 0.00475039}, {95.736, 0.0047454}, {95.769, 0.00474199}, {95.803, 0.00444718}, {95.836, 0.0043472}, {95.869, 0.00424722}, {95.902, 0.00398735}, {95.935, 0.00388579}, {95.969, 0.00348017}, {96.002, 0.00334853}, {96.036, 0.00298881}, {96.069, 0.00285559}, {96.101, 0.00295044}, {96.133, 0.00304528}, {96.166, 0.00291364}, {96.199, 0.00278042}, {96.232, 0.00280867}, {96.265, 0.00267703}, {96.298, 0.00264196}, {96.331, 0.00257206}, {96.364, 0.00250058}, {96.396, 0.00266033}, {96.43, 0.00236078}, {96.462, 0.0025617}, {96.495, 0.00255829}, {96.528, 0.0025533}, {96.561, 0.00238999}, {96.595, 0.00199704}, {96.627, 0.0022407}, {96.66, 0.00210589}, {96.693, 0.00197266}, {96.726, 0.00183944}, {96.759, 0.00186611}, {96.792, 0.00171072}, {96.824, 0.00177549}, {96.857, 0.00161061}, {96.89, 0.00160561}, {96.923, 0.00144231}, {96.956, 0.00127742}, {96.989, 0.00111412}, {97.023, 0.000881057}, {97.056, 0.000716171}, {97.089, 0.000712762}, {97.122, 0.000482968}, {97.154, 0.000611061}, {97.187, 0.000507916}, {97.22, 0.000406355}, {97.253, 0.00030321}, {97.286, 0.000342545}, {97.32, -0.0000187506}, {97.352, 0.000242324}, {97.385, 0.000112266}, {97.418, 0.000142103}, {97.451, 0.0000104619}, {97.484, -0.000121179}, {97.517, -0.00025282}, {97.55, -0.000226149}, {97.583, -0.00035779}, {97.615, -0.000297771}, {97.648, -0.00049432}, {97.681, -0.000689285}, {97.713, -0.000493118}, {97.746, -0.000528189}, {97.779, -0.000583841}, {97.812, -0.000652157}, {97.845, -0.000720474}, {97.878, -0.00078879}, {97.911, -0.000857106}, {97.944, -0.000889011}, {97.978, -0.00112524}, {98.011, -0.00113023}, {98.044, -0.00113364}, {98.077, -0.00113863}, {98.11, -0.00112304}, {98.144, -0.00132761}, {98.176, -0.00106812}, {98.209, -0.00119976}, {98.241, -0.00094027}, {98.275, -0.00114484}, {98.308, -0.00111817}, {98.341, -0.00108991}, {98.374, -0.00122155}, {98.407, -0.00119488}, {98.44, -0.00130753}, {98.473, -0.00124761}, {98.506, -0.0011877}, {98.539, -0.00128767}, {98.572, -0.00122934}, {98.604, -0.00106167}, {98.638, -0.0013264}, {98.671, -0.00136305}, {98.704, -0.0013997}, {98.737, -0.00143478}, {98.77, -0.00141444}, {98.803, -0.001516}, {98.836, -0.00145767}, {98.869, -0.00155923}, {98.902, -0.00166079}, {98.935, -0.00162304}, {98.967, -0.00152819}, {99., -0.001663}, {99.033, -0.00179622}, {99.066, -0.00176955}, {99.099, -0.00184103}, {99.132, -0.0018761}, {99.165, -0.00207265}}

which looks like this when plotted:

image2

I am trying to estimate the areas of both peaks using at least 3 "reasonable" different baselines. Thanks to the suggestion of @JimB here: Fitting two peaks at the same time with NonLinearFit, I am estimating the areas as:

left = Select[data, #[[1]] < 82 &];
baselineleft = Mean[Select[data, #[[1]] < 70 &][[All, 2]]];
widthleft = Max[left[[All, 1]]] - Min[left[[All, 1]]];
arealeft = (Mean[left[[All, 2]]] - baselineleft)*widthleft

dright = Select[data, #[[1]] < 98 &];
right = Select[dright, #[[1]] > 84 &];
baselineright = Mean[Select[right, #[[1]] < 85 &][[All, 2]]];
widthright = Max[right[[All, 1]]] - Min[right[[All, 1]]];
arearight = (Mean[right[[All, 2]]] - baselineright)*widthright

My question is:

1) How can I estimate the area of both peaks using this methodology but using at least 3 "reasonable" baselines as to compare what would be the area of both peaks using different base lines and have and idea about the error of estimating these areas?. One example is given below with excel using two baselines where it is obvious that depending on the baseline the area of both peaks will be different. 2) How can I get this comparison of the areas of the two peaks in a table for each baseline?

enter image description here

$\endgroup$
5
  • $\begingroup$ Isn't there some standard definition of a baseline? (Either theoretical or operational). I can't believe it would be considered as "I'll know it when I see it." or "I'll use the one that I like best." My point is: Isn't this a question for a physics or chemistry forum to define what a baseline is and then back to here to implement that definition? $\endgroup$ – JimB May 27 '20 at 16:42
  • $\begingroup$ @JimB in this case not. The baseline is not at the same level for both peaks. That's the reason I am asking for three different baselines as to have an idea of the error (which of course there is). One possible way is to get one whole baseline using the mean values of the second position for when the data is below 70 (using the left side baseline), another baseline could be using the data in the second position between 80 and 85 (using the baseline considering the center) and the same for the right side using the data above 95 $\endgroup$ – John May 27 '20 at 16:47
  • $\begingroup$ As you've guessed I have little to no experience with this. However, if the measurements drift slowly up and down AND there is some experience with this particular instrument that that is the case, why not use as the baseline line segments that connect the troughs between peaks? In other words, if they do drift, why use a constant baseline? OK. I'll stop expanding on my ignorance. $\endgroup$ – JimB May 27 '20 at 16:53
  • $\begingroup$ @JimB we all are ignorant in something. I am very ignorant in Mathematica for instance. Feel free to post a solution as you think would be best to calculate three different baselines and I will modify the code appropiately. I am pretty sure whataver you post as an answer will not only help me but also other people with similar questions in the future. $\endgroup$ – John May 27 '20 at 16:58
  • $\begingroup$ @JimB I posted an answer which seems to be a little better approach but still the arealeft seems to be wrong. Could you tell me what I am doing wrong there?. I am not sure if I am using NIntegrate correctly $\endgroup$ – John May 27 '20 at 21:38
3
$\begingroup$

Here is a function you can apply to your dataset, specifying the areas you want integrated. The function will:

  • select the relevant portion of your dataset
  • take the first and last points in that region to establish a linear baseline by linear interpolation
  • describe the data in between the specified boundaries by interpolation (no point in fitting, the shape of the peak appears irrelevant to whatever the application at hand is)
  • calculate the area of the peak down to the baseline, by integrating the difference of the peak and baseline interpolating functions;
  • plot the result and report the area.

Here is the function's code:

ClearAll[peakArea]
peakArea[dataset_, {start_, end_}] :=
 Module[{region, peak, baseline, area},
   (*select region within data identified as the peak*)
   region = Select[dataset, start <= #[[1]] <= end &];

   (*generate interpolating functions describing the peak and baseline *)
   peak = Interpolation[region];
   baseline = Interpolation[region[[{1, -1}]], InterpolationOrder -> 1];

   (*calculate the area, subtracting the baseline*)
   area = NIntegrate[peak[x] - baseline[x], Flatten@{x, region[[{1, -1}, 1]]}];

   (*generate the plot*)
   Show[
     ListLinePlot[
       Style[dataset, Gray],
       PlotRange -> All,
       PlotLabel -> Style["Peak area: " ~~ ToString[area], Black]
     ],
     Plot[
       {
        Style[peak[x], Directive[Thick, Red]],
        Style[baseline[x], Black]
       },
       Evaluate@Flatten@{x, region[[{1, -1}, 1]]},
       PlotRange -> All,
       Filling -> {1 -> {2}}, FillingStyle -> Opacity[0.2, Red]
     ]
   ]
 ]

Using your dataset from the OP (data), here are two usage examples:

peakArea[data, {82, 95}]

rightmost peak in data

peakArea[data, {62, 82}]

leftmost peak in data

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8
  • $\begingroup$ MarcoB thank you so much for always helping me!. I really apreciate it. This is a fantastic answer!. This is also similar to the answer I posted above, which I still do not have very clear why my arealeft is so high. Could you check that so that this post can have two good answers for me and for other people?. Thank you ! $\endgroup$ – John May 27 '20 at 21:54
  • $\begingroup$ @MarcoB I certainly can't think of anything better. But it sure does seem awfully arbitrary. Do machines that spit out hydrocarbon profiles, for example, have fancier algorithms built in? (I vaguely remember those providing areas for peaks.) $\endgroup$ – JimB May 27 '20 at 22:35
  • 1
    $\begingroup$ @JimB Not much more refined than this, no. This is not too far from typical in the analysis of e.g. nuclear magnetic resonance data, in which case peak integration is very important, but peak shape can be any old blob because of superposition. Then a reasonably well-behaved global baseline (often linear, in some cases polynomial) is found, often with user input and then numerical integration is carried out not very differently than here. $\endgroup$ – MarcoB May 27 '20 at 22:43
  • 1
    $\begingroup$ @JimB Spectroscopic results (light absorption, emission, scattering, etc) are very different: in those case the underlying physics dictate the shape of the baseline (and sometimes of the peaks as well). For instance, in light absorption measurements, one expects a flat baseline, as close to zero as possible after appropriate allowances have been made (the predictable absorption of the sample container, of the air the light goes through, the unavoidable scattering at interfaces...). Then, great deviations from the expected shape are typically a clear sign of experimental trouble. $\endgroup$ – MarcoB May 27 '20 at 22:47
  • $\begingroup$ @MarcoB Thanks! $\endgroup$ – JimB May 27 '20 at 22:56
0
$\begingroup$

I think I found one way to do it but the areas do not seem to be correct. Here's an example with one baseline for each curve:

left = Select[data, #[[1]] < 82 &];
baselineleft = 0.056;
ipleft = Interpolation[left];


dright = Select[data, #[[1]] < 98 &];
right = Select[dright, #[[1]] > 84 &];
baselineright = 0.026;
ipright = Interpolation[right];




p1 = ListPlot[data, PlotStyle -> {{Yellow, PointSize[0.01]}}, 
   PlotRange -> {{66, 100}, All}, PlotMarkers -> {Automatic, 5}, 
   PlotLegends -> {"Original data"}, LabelStyle -> {Black, Bold, 14}];
p2 = ListLinePlot[{{65, baselineleft}, {81, baselineleft}}, 
   PlotStyle -> Black, PlotLegends -> {"Baseline 1"}];
p3 = ListLinePlot[{{82, baselineright}, {98, baselineright}}, 
   PlotStyle -> Blue, PlotLegends -> {"Baseline 2"}];
p4 = Plot[ipleft[x], {x, 67, 80}, PlotStyle -> {{Thin, Red}}, 
   PlotRange -> All, PlotLegends -> {"Fitted data"}];
p5 = Plot[ipright[x], {x, 85, 95}, PlotStyle -> {{Thin, Red}}, 
   PlotRange -> All];

Show[p1, p2, p3, p4, p5]

arealeft = NIntegrate[ipleft[x], {x, 67, 80}] - baselineleft
arearight = NIntegrate[ipright[x], {x, 85, 95}] - baselineright

which gives:

![Image

and arealeft=1.59149 and arearight=1.487459, which at least arealeft is obvously a wrong value because it cannot be bigger than arearight.

Can someone tell me what I am doing wrong here for arealeft? This area should be around 0.7-0.8.

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2
  • 2
    $\begingroup$ That's new information. Why should arealeft be around 0.7-0.8 ? Is that from another instrument? Theory? Visual inspection? $\endgroup$ – JimB May 27 '20 at 21:58
  • $\begingroup$ @JimB I did the calculation of the area with origin but it may well be around 0.9-1 as MarcoB found. It certainly is not 1.59149. So, anything around 1 or below should be alright. I just do not understand with this code why arealeft would give that value $\endgroup$ – John May 27 '20 at 22:18

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