Subtracting two plots from each other after shifting one

Question

I have the following lists representing energy spectra that need to be subtracted from each other to produce a difference spectrum.

data1 = ToExpression@Import["http://pastebin.com/raw.php?i=VfBcE9L5", "Data"][[1]]

data2 = ToExpression@Import["http://pastebin.com/raw.php?i=dFcxfJ4j", "Data"][[1]]

This is how all the spectra look before the necessary shift (I threw a spline fit through the arrays):

The problem is that I need to preform an energy shift to the black spectrum before the subtraction so that the first peaks line up. I usually construct my difference spectrum array like this:

diffdata = Transpose[{data1[[1,;;,1]], data1[[1,;;,2]] - data2[[1,;;,2]]}]

Where the first column is energy and the second column is intensity. I want to energy shift data2 to line up with data1 (it's about a -0.35 shift), then subtract the intensity of data1 from the intensity of data2

Since the subtraction occurs based on the position of the elements within the list, any shifting I do to column 1 of data2 (the energy) will not matter to my difference array. Furthermore, the data point spacing is not homogenous within the spectrum i.e. the data points are more dense in the peak regions than in other regions; however, the point spacing between data1 and data2 are identical. Is there any way I can get around this?

My immediate attempt was to do something with Interpolation, but I do not know how to subtract the y-values of two interpolated functions from each other. I would then input the energies of data1 (i.e. data1[[1,;;,1]] into the interpolated difference spectrum to get the interpolated intensities of with the corresponding energies.

Any help is appreciated.

Answer 1

I always use TimeSeries for these kinds of problems, since it avoids having to worry about indices too much, gives you good control over resampling and you have a whole load of useful functions

dt = 0.05;
{ts1, ts2} = 
 TimeSeries[TimeSeriesResample[#, dt,
  ResamplingMethod -> {"Interpolation", InterpolationOrder -> 1}]] & /@ {data1, data2};

ListLinePlot[{Quiet@
 TimeSeriesThread[First[#] - Last[#] &, {TimeSeriesShift[ts2, -0.35], ts1}],
 ts1, ts2}, PlotStyle -> {Blue, Green, Black}]

The function TimeSeriesShift performs your shift/alignment:

ListLinePlot[{TimeSeriesShift[ts2, -0.35], ts1}, PlotStyle -> {Green, Black}]

TimeSeriesThread is used to do shifted data2 minus data1. If you want it the other way around, change First[#]-Last[#] to Last[#]-First[#].

Answer 2

Because data1 and data2 have an extra set of curly brackets, it is necessary to remove the extra set before generating interpolating functions.

int1 = Interpolation[Flatten[data1, 1]];
int2 = Interpolation[Flatten[data2, 1]];

The difference then can be computed and plotted:

Plot[{int1[t], int2[t + .35], int1[t] - int2[t + .35]}, {t, 1105, 1134}]

The difference alone is

Addendum

In answer to the OP's question below, the difference between the two curves can be converted into an interpolation function itself, if desired, by

diffdata = FunctionInterpolation[int1[t] - int2[t + .35], {t, 1105, 1133.65}, 
  InterpolationOrder -> 1, InterpolationPoints -> 5 Length[Flatten[data1, 1]]]

Answer 3

Using the same variable names. The separation can be approximated using FindPeaks. I have used on the distinctive three middle peaks.

lp1 = ListPlot[data1, 
   Epilog -> {Red, PointSize[0.03], 
     Point@(p1 = 
        Extract[data1[[1]], 
         List /@ FindPeaks[data1[[1, All, 2]]][[All, 1]]])}, 
   Frame -> True, ImageSize -> 400];
lp2 = ListPlot[data2, 
   Epilog -> {Red, PointSize[0.03], 
     Point@(p2 = 
        Extract[data2[[1]], 
         List /@ FindPeaks[data2[[1, All, 2]]][[All, 1]]])}, 
   Frame -> True, ImageSize -> 400];
diff = Drop[p1, 3] - Drop[p2, 2];
md = Mean[diff[[1 ;; 3]]]
i1 = Interpolation[data1[[1]]];
i2 = Interpolation[data2[[1]]];
pld = Plot[{i1[t], i2[t - md[[1]]], i1[t] - i2[t - md[[1]]]}, {t, 
    1105, 1130}, Frame -> True, ImageSize -> 400];
Column[{lp1, lp2, pld}]

The mean difference in 3 peak locations: {-0.309621, 0.0253712}