Conditioning on group of functions

Question

I have two sets of functions which I want to implement them in a single function with an option indicating which set of functions should be used.

Suppose that I have two lists like:(updated example)

b = {-1, 1, 0, 2}}
u = {{1, 1, 1, 1}, {1, 0, 0, 1}, {1, 0, 1, 1}, {0, 1, 1, 0}, {1, 0, 1,
1}, {0, 1, 0, 1}, {1, 0, 1, 0}, {1, 0, 1, 1}, {0, 1, 1, 1}, {1, 1,
0, 0}}

and I have this function:

 AbEst[b0_, u0_, method_] := 
 Module[{t, b = b0, u = u0, l, a, pos, theta, dif, se},
  t = Table[i, {i, -4, 4, .01}];
  L[b_, t_, u_] := 
   Product[p[b[[i]], t]^u[[i]]*(1 - p[b[[i]], t])^(1 - u[[i]]), {i, 
     1, Length[u]}];
  p[b_, t_] := 1/(1 + Exp[-1.7*(t - b)]);
  l = L[b, t, #] & /@ u;
  a = PDF[NormalDistribution[0, 1], t];
  pos = Table[a, {Dimensions[u][[1]]}]*l;
  Which[method == "ML",
   theta = 
    t[[Position[l[[#]], Max[l[[#]]]][[1, 1]] & /@ 
      Range[Dimensions[u][[1]]]]],
   method == "MAP",
   theta = 
    t[[Position[pos[[#]], Max[pos[[#]]]][[1, 1]] & /@ 
      Range[Dimensions[u][[1]]]]],
   method == "EAP",
   theta = pos.t/Total[Transpose[pos]]];
  dif = (Table[t, {Dimensions[u][[1]]}] - 
      Partition[Flatten[Table[theta, {Length[t]}]], Length[t]])^2;
  Which[method == "ML",
   se = Sqrt[Total[Transpose[l*dif]]/Total[Transpose[l]]],
   method == "MAP",
   se = Sqrt[Total[Transpose[pos*dif]]/Total[Transpose[pos]]],
   method == "EAP",
   se = Sqrt[Total[Transpose[pos*dif]]/Total[Transpose[pos]]]];
  Return[Transpose[{theta, se}]]]

Now it returns {theta,se} for given method. I want to change it such that it returns theta by default (i.e. without computing se) and computes and returns {theta,se} if it was requested.

1) How can I do that?

2) How can I re-write more efficient function in which fits my purpose?

Answer 1

One possible way is using Switch as follows:

f[a_,b_,opts___?OptionQ]:=
    With[{t=method/.{opts}/.{method->1}},
        Switch[t,
        1,{a+b,a-b},
        2,{a*b,a/b}]]

In this definition, the default option is 1. Is this what you want?

In[7]:= f[a,b,method->2]

Out[7]={{10,30,60,100,150},{1/10,2/15,3/20,4/25,1/6}}

Updated:

You can combine the two Whichs together and add If condition to control whether the output includes se or not. For your case of different behaviors at different methods, I prefer Switch to Which in this case. The following is the outline of the code:

myAbEst[b0_, u0_, method_, opts___?OptionQ] :=
  Module[{t, b = b0, u = u0, l, a, pos, theta, dif, se, L, p, 
         op = Se /. {opts} /. {Se -> False}}, 
(*op is the option that can control whether se should be output.The default is not.*)
  ...(*Your code about the assignments and definitions*)
  Switch[method,
   "ML", theta=(*your code*);If[op,dif=(*your code*);se=(*your code*)],
   "MAP", theta=(*your code*);If[op,dif=(*your code*);se=(*your code*)],
   "EAP", theta=(*your code*);If[op,dif=(*your code*);se=(*your code*)]];
  If[op,Transpose[{theta, se}], theta]]

By the way, functional program always returns the value of the last expression. You don't need to use Return to return values.

myAbEst[b,u,"ML"](*This is the default case, which won't return se.*)
myAbEst[b,u,"ML",Se->True](*This is the case that returns se.*)