3

I have a piecewise function and I'm trying to extract the different functions for different values of the parameter. I use: 

First/@(InitialEnvelope[\[Beta]]//PiecewiseExpand)[[1]]

This gives be back the different functions except the last piece which is the function when the else or True condition holds. How can I extract the function at True?

Rby
  • 159
  • 11

1 Answers1

3

In Mathematica everything is specified via patterns. So are, of course, Piecewise functions. To obtain a standardized form for nested Piecewise functions you were right to apply PiecewiseExpand first. So let's take a look at an example of a nested Piecewise function:

(*definition*)
pw = Piecewise[{{g[x],x > 5}, {Piecewise[{{h1[x],x < 1}, 
{h2[x],x > 2}}, h3[x]], x < 3}}, f[x]];
(*expansion*)
pwe = PiecewiseExpand[pw];

Remember my remark about patterns before. Let's take a look at the underlying pattern (form) of the expanded Piecewise function:

pwe // FullForm

This gives you a clue on how to extract all the functions via using the Part operator, here is one possibility to do it:

Flatten[{pwe[[1, ;; , 1]], pwe[[2]]}]

{f[x], g[x], h1[x], h2[x], h3[x]}

Have fun playing around with this.

Wizard
  • 2,720
  • 17
  • 28
  • Thank you. I also tried to do it with Union, but it takes more time than Flatten. Any idea why? – Rby Sep 17 '16 at 14:21
  • Union should be slower because it creates a new object, while Flatten does not do that. Also important to note is that according to Mathematica help Union "gives a sorted list of all the distinct elements", so additional sorting is applied, which costs time. For lists containing only a few elements you should not really notice a difference, of course this looks different for large lists. – Wizard Sep 17 '16 at 14:28