I was a bit curious if anyone knows of any compact way to use Set and/or SetDelayed if we want to, say, generate Fit functions and index them as well.

To demonstrate what I had in mind, consider the following data set for a minimal working example:

data = { {{1, 2}, {2, 4}, {3, 6}} , {{1, 1}, {2, 2}, {3, 3}} };

Now it's apparent that fitting data[[1]] and data[[2]] should provide the functions f[x]=2x and f[x]=x, respectively. But I was hoping to combine Set (=) and/or SetDelayed (:=) in the simplest manner, so that I may define a single compact function that can provide both the fit functions and the numerics for each fit.


fitdata[i_, x_] := Fit[data[[i]], {1, x}, x]

works at the indexing level, where I can input a value for i and evaluate fitdata[1,x] and fitdata[2,x] to return either functional dependences on x. However, trying to input a value for both i and x yields an error, and I was wondering if anyone knows the easiest way to define a function that can both return the ith Fit function AND return a number corresponding to an input x for this function.

I'm aware that there are a number of simple ways we can both index a set of Fit functions and obtain numerical values of the fit, including but not limited to:

  • storing each function into a list that may be indexed, and using a rule to pass an x argument for the desired element
  • removing the x-dependence of the function and using a rule to cast an x-value for fitdata[i_] := Fit[data[[i]], {1, x}, x].

However, I'd like to avoid adding rules or defining new variables if I can, since I'm worried my code will get a bit messy or over-complicated whenever I start iterating over these functions to calculate things here or make some plots there, etc.

The function that I had in mind above would probably be the cleanest way for me to implement things, so I was just wondering if anyone knows of a way that some fitdata[i,x] can be defined to act in this manner I originally envisioned. I suspect there may be some way to use Set and/or SetDelay so that the expression is evaluated in the way that I hope, but I am not sure.

I tried looking some solutions up on the forum but most answers were a bit more complicated than the alternatives I laid out above, or they discussed memoization which doesn't seem to be necessary in the case I've described here.


1 Answer 1


You could use LinearModelFit (or NonlinearModelFit according to your needs):

fitn[dataset_, i_] := LinearModelFit[dataset[[i]], {1, x}, x]

fitn[data, 2]

(* Out: FittedModel[0. + 1. x] *)

When you want a specific calculated value, you can use:

fitn[data, 1][2.5]

(* Out: 5. *)
  • $\begingroup$ works great, thanks for the quick reply $\endgroup$ Apr 2, 2021 at 3:55

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