Consider the following program:

g = 2f;
g[x] (* (2f)[x] *)

Is there any way I can change the above expression to have it return 2 f[x]?

With Hold construct perhaps? Sorry, I'm really not familiar with these Hold* functions.

I know I could easily define g=2f[#]& and it would work fine, but in my case what I'm really trying to achieve is something like that :

Transpose @ MapThread[Standardize[{##}, Mean, 2 StandardDeviation] &, xs]

Unfortunately, (2 StandardDeviation) is taken as a single entity. Having a whole Function would make it much more messy.

  • $\begingroup$ There isn't really a better way, just alternatives with similar level of complexity: For example: g = (2#)&@*f uses Composition. $\endgroup$
    – Jens
    Apr 12, 2015 at 4:53
  • $\begingroup$ Yeah, I think I'll just settle for Function[2 StandardDeviation[##]] $\endgroup$
    – Literal
    Apr 12, 2015 at 4:54
  • $\begingroup$ @Kuba OK, I've added an answer... $\endgroup$
    – Jens
    Dec 16, 2015 at 21:59

1 Answer 1


As mentioned in the comments, you could use Composition as follows:

g = (2 #) &@*f

(* ==> 2 #1 &@*f *)


(* ==> 2 f[x] *)

Your own solution using Function is of course also fine, and one could make it slightly more user-friendly by defining the corresponding scalar multiplication operation something like this:

CenterDot[x_, y_] := Function[z, x y[z]]

g = 2·f

(* ==> Function[z$, 2 f[z$]] *)


(* ==> 2 f[x] *)

This would allow you to use CenterDot to define similar constructs on the fly in a more readable form, by using the keyboard shortcut esc.esc ot enter the scalar multiplication. So as an in-line definition this would look like this:


(* ==> 2 h[x] *)

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