I've been struggling to make the Which command more efficient. Below is my original code:

fitmodel[set_, k1_, k2_, k3_, k4_, k5_, m_, n_][t_] := 
 Which[set == 1, Evaluate@model[1][k1, m, n][t], set == 2, 
  Evaluate@model[2][k2, m, n][t], set == 3, 
  Evaluate@model[3][k3, m, n][t], set == 4, 
  Evaluate@model[4][k4, m, n][t], set == 5, 
  Evaluate@model[5][k5, m, n][t]]

I Prepend the set number in front of my data and use this fitmodel to assign different model to different data. Since it's very inconvenient to change this fitmodel whenever I have different numbers of kinetic constants(k1,k2,...kn), I'm trying to make the code easier to handle. I tried the following:

args = Append[
     Table[ToExpression["k" <> ToString[i] <> "_"], {i, 1, 
       Length@allNames}], ToExpression["set_"]], ToExpression["m_"]], 

k = Flatten[{set == #, 
      Evaluate@model[#][ToExpression["k" <> ToString[#]], m, n][t]}&/@Range@5, 2];

fitmodel[args][t_] := Which[Sequence@@k]

But it failed and the error message is:

Which::argctu: Which called with 1 argument. >>

I checked the output of "k" and it's exactly the same as my original code inside Which command. Then I used "Sequence@@k" to get rid of the curly bracket. So I thought it should work. Obviously I'm wrong and I couldn't come up with a solution to this issue. Hope someone could guide me forward. Thank you!!


1 Answer 1


It seems that set is the index you need for your $k_i$ in your model. If this is the case then you don't need Which.

fitmodel[set_, k_List, m_, n_][t_] := model[set][k[[set]], m, n][t]

fitmodel could then be called with as many $k_i$ as needed as they are contained in a list. You just need to add additional $k_i$ to the list.

kVars = {k1, k2, k3, k4, k5};
fitmodel[s, kVars, m, n][t]

Hope this helps.

  • $\begingroup$ Thank you @Edmund ! It's indeed a much better way to deal with the problem. But I'm getting the following error Part::pkspec1: The expression set cannot be used as a part specification Could you please help me a bit more on this issue? Thanks again! $\endgroup$
    – DavidC
    Aug 3, 2016 at 6:03
  • 1
    $\begingroup$ @DavidC It is due to the Evaluate that you have in the function. I don't think you need that as you are forcing it to resolve before it has any values. $\endgroup$
    – Edmund
    Aug 3, 2016 at 10:00

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