I'm having trouble with a simple linear fit model. Here is my code

x = {23.0, -5.0, 1.5, 1.5, 10.0, 0.0, 0.5}
num = Length[x]
absx = Abs[x]
r = (Ordering@Ordering@# + Reverse@Ordering@Ordering@Reverse@#)/2 &@absx // N
rstar = (r - 0.5)/num // N
p = (rstar + 1)/2
v = Sqrt[2]*InverseErf[2 p - 1]
(* create a table and sort data by r - the 3rd column *)

SortBy[Transpose[{x, absx, r, rstar, p, v}], #[[3]] &] // TableForm
dataset = SortBy[Transpose[{x, absx, r, rstar, p, v}], #[[3]] &]
(* collect the sorted absx variable *)
oabsx := dataset[[All, 2]]
(* collect the sorted v variable *)
ov = dataset[[All, 6]]
(* create the pairs of data to plot *)
xy = Transpose[{oabsx, ov}]
(* construct a linear model for the data *)
model = LinearModelFit[xy, x, x]
(* Plot the data and the linear model *)
Show[ListPlot[xy], Plot[model["BestFit"], {x, 0, Length[xy]}]]

This ListPlot shows up, but the fit does not; I get two errors in return

LinearModelFit::dpbss: The number of unique basis functions is 1 fewer than the number of basis functions specified in {23.,-5.,1.5,1.5,10.,0.,0.5}. Duplicated basis functions will be removed. >>

LinearModelFit::ivar: {23.,-5.,1.5,1.5,10.,0.,0.5} is not a valid variable. >> *)

Any suggestions ?

  • 1
    $\begingroup$ Hi ! You can head to the help centre and read more about proper code formatting practices. Aside from that try executing model = LinearModelFit[xy, x, x] and work from there. $\endgroup$
    – Sektor
    Jan 12, 2015 at 21:03

1 Answer 1


Your problem is that you are using x as the variable to represent the explanators in your model, but you have already defined x to be the list of numbers at the top.

Just change this line to be something other than x, like q:

model = LinearModelFit[xy, q, q] 

And don't forget to make the matching change in the Plot command:

Show[ListPlot[xy], Plot[model["BestFit"], {q, 0, Length[xy]}]]

enter image description here


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