I have a data set of $10^4$ entries. I want to plot the last $2000$ data with the same argument at $x$ axis and fit a curve. That means the range of $x$ axis will be $8001$ to $10000$ incremented by $1$.

I wrote code like the following to know the coefficient of fitting curve:

d = Import["ss.dat"];
T = Table[d[[1, i]], {i, 8001, 10000}];
FindFit[T, a*E^(-b*x), {a, b}, x]

FindFit returns the values of $a$ and $b$ considering $x$ to range from $1$ to $2000$ instead of $8001$ to $10000$.

If anyone can tell me how to deal with this problem that will be great help.

I am attaching data in the following link:

data as DAT file in Google Drive

  • 2
    $\begingroup$ You could try T = Table[{i, d[[1, i]]}, {i, 8001, 10000}];. This will allow you to give FindFit 2D data with the specific x values you want. $\endgroup$
    – MassDefect
    Mar 18, 2019 at 16:11
  • 1
    $\begingroup$ Your first problem has been pointed out by @MassDefect. But you should linearize your model with the following: lmf = LinearModelFit[Transpose[{T[[All, 1]], Log[T[[All, 2]]]}], x, x] Show[LogPlot[Exp[lmf[x]], {x, 8001, 10000}, PlotStyle -> {LightGray, Thickness[0.03]}], ListLogPlot[T, PlotStyle -> Red]]. $\endgroup$
    – JimB
    Mar 18, 2019 at 16:33

1 Answer 1


Generally, you don't need nearly so many points to get a nice fit. Regardless, we can use NonlinearModelFit[] to get what you want:

You can use Drop[list,n] to drop the first n elements of a list.

dir = NotebookDirectory[];
file = "ss.dat";

d = Import[file];
T = Drop[Flatten[d], 8000];
St = Drop[Take[T, {1, -1, 1}], 0];
Stt = Table[{i, T[[i]]}, {i, 1, Length[St]}];
e = NonlinearModelFit[Stt, a*E^(-b*x), {a, b}, x]

GraphicsGrid[{{ListPlot[Stt, ImageSize -> Medium, PlotRange -> All, PlotTheme -> "Scientific"], Plot[{Evaluate[a*E^(-b*x) /. e["BestFitParameters"]]}, {x, 0,Length[Stt]}, PlotTheme -> "Scientific", Epilog -> {Red, Point[Stt]}]}}]


Using the code will drop every x elements from your entire list, drastically reducing your plot points. 2000, or even 10000 are just not needed.

St = Drop[Take[T, {1, -1, 1}], 0];

and changing to:

St = Drop[Take[T, {1, -1, 40}], 0];

Gives us:


A nice visual that the points lay directly on our function, and a good fit for a and b.


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