I have this experimental data set which I know should follow the Complementary error function.

My goal is to find the σ of the associated gaussian distribution.

However, I cannot seem to do it in any way using NonlinearModelFit. Could you please give me some hints?

The data is in a list of lists, e.g.:

data = ToExpression@Import["http://pastebin.com/raw.php?i=BtcXHh9c"];

And when plotted it looks like this:


  • $\begingroup$ where is the data? $\endgroup$ Jul 7, 2014 at 8:15
  • $\begingroup$ Actually, I should have first of all put the data in ordered pairs: {{x1,y1},{x2,y2} ... }. $\endgroup$
    – Danno
    Jul 7, 2014 at 10:03
  • $\begingroup$ I think you mean something like this. $\endgroup$
    – Öskå
    Jul 7, 2014 at 10:04
  • $\begingroup$ To fix data download: data = ReplaceRepeated[ ToExpression[ ToString@Import["http://pastebin.com/raw.php?i=BtcXHh9c"]][[1, 1]], {a___, {b_, Null}, {c_}, d___} :> {a, {b, c}, d}]; ListPlot[data] $\endgroup$ Jul 7, 2014 at 15:34

2 Answers 2


I don't think there is anything unusual here that can't be found in the documentation. Normalization parameters in the x and y dimensions (b, and a, respectively) can be added into the model, as can the x-axis shift (parameter c). Parameter estimates are taken from a look at the original data.

nlm = NonlinearModelFit[data, 
  a Erfc[b (x + c)], {{a, 8 10^-8}, {b, 1/10}, {c, -35000}}, x]
Plot[nlm[x], {x, 35000, 35400}, Epilog -> {Red, Point /@ data}]

Mathematica graphics

  • $\begingroup$ agree with your comment. Originally the data posted was not such a beautiful fit...hence my hamfisted approach to force a function where NonlinearModelFit did not work...timing...but at least it led to provision of correct data...though this fit seems to good to be true $\endgroup$
    – ubpdqn
    Jul 7, 2014 at 20:35
  • $\begingroup$ @ubpdqn +1 for "spending" the time on the bad data! $\endgroup$ Jul 7, 2014 at 21:02
  • $\begingroup$ Thanks. I need to ask, are parameter estimates in NonlinearModelFit necessary? $\endgroup$
    – Danno
    Jul 8, 2014 at 15:50
  • $\begingroup$ @Danno I can't think of a time when knowing the uncertainty of your parameter model is not necessary. Get them with nlm["ParameterTable"] $\endgroup$ Jul 8, 2014 at 16:00
  • $\begingroup$ But, and it really may be a newbie question on many levels, isn't the fitting algorithm's job to find the parameter values? I am talking about the estimates you gave, e.g. the 8 10^-8besides $a$ etc. Notice I may not have fully understood your last comment. $\endgroup$
    – Danno
    Jul 8, 2014 at 17:15

I am a little unsure what the aim is here. Looking at the data only the first third of the plot could be fitted to Erfc like function.

I post this, perhaps, as motivation.

d = Transpose@data;
ListPlot[ds = SortBy[d, #[[1]] &], Joined -> True]

enter image description here

Extracting the part of plot I referred to:

fd = ds[[1 ;; 35]]
fdlp = ListPlot[fd]

enter image description here

Now as a rather ugly approach to trying to fit desired function. This will involve rescaling, removal of troublesome 0 or negative values, logarithmic transform and back transforming. This is NOT ideal and am sure expert users will have better approaches (assuming this is what is desired).

{min, max} = Through[{Min, Max}[fd[[All, 2]]]];
res = Rescale[fd[[All, 2]]];
trans = Transpose[{fd[[All, 1]], res}];
int = Interpolation[trans, InterpolationOrder -> 1];
{xmin, xmax} = {34800.`, 35137.`};
dif[u_] := D[int[x], x] /. x -> u
w = {#, -dif[#]} & @@@ trans;
wlp = ListPlot[w]
peak = Cases[w, {_, Max[w[[All, 2]]]}][[1, 1]]
shift = Cases[{#1 - peak, #2} & @@@ w, {_, _?Positive}];
lg = {#1, Log@#2} & @@@ shift;
lm = LinearModelFit[lg, {1, x^2}, x]
func[u_] := Exp[Normal@lm] /. x -> u
Show[Plot[func[t - peak], {t, xmin, xmax}], wlp]
ex = Rescale[v, {0, 1}, {min, max}]
h[u_] := ex /. v -> u;
soln[t_] := NIntegrate[func[x], {x, -Infinity, t}]
Show[fdlp, Plot[h[1 - soln[x - peak]], {x, xmin, xmax}]]

The derivative of interpolating function (and fit):

enter image description here

The final "fit":

enter image description here

The fit expression (with back transforming and rescaling):

enter image description here

  • $\begingroup$ Thank you very much for your answer, however I didn't notice that the data I published was corrupted. I just edited the question with the actual data, that looks way different, and I am quite sure it fits a complementary error function (apart from some two constant pieces at the beginning and end of the data set). $\endgroup$
    – Danno
    Jul 7, 2014 at 14:03
  • $\begingroup$ @Danno...ok...bobthechemist has provided the very clear answer $\endgroup$
    – ubpdqn
    Jul 7, 2014 at 20:39

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