Suppose, there are two delayed expressions

D1:= RandomVariate[UniformDistribution[{-1, 0}]];
D2:= RandomVariate[ UniformDistribution[{0, 1}]];

D1, D2 work as needed generating a pair of random numbers. How to swap D1, D2 without repeating the full coding

D2:= RandomVariate[UniformDistribution[{-1, 0}]];
D1:= RandomVariate[ UniformDistribution[{0, 1}]];

Clearly, that code {D1,D2}= (or :=) {D2,D1} does not work.


3 Answers 3


There are different usages of SetDelayed. Read SetDelayed documentation, Section Scope > Diffrent Kinds of Values for more information.

This solution works on var := def (like your case) not var[] := def or other kinds.



SetAttributes[swapSetDelays, HoldAll];

swapSetDelays[a_, b_] := With[{temp = OwnValues[a]},
  OwnValues[a] = OwnValues[b] /. HoldPattern[b] :> a;
  OwnValues[b] = temp /. HoldPattern[a] :> b;



r1 := RandomInteger[{1, 10}];
r2 := RandomInteger[{11, 20}];

{r1, r2}
(* Out: {2,15} *)

swapSetDelays[r1, r2];

{r1, r2}
(* Out: {11, 8} *)
  • $\begingroup$ +1 (but strictly, more clever than useful) $\endgroup$
    – mikado
    Jun 4, 2021 at 18:10
  • $\begingroup$ @mikado thanks for your feedback but I should confess that the idea came from OwnValues Neat Examples 🤫. $\endgroup$
    – Ben Izd
    Jun 4, 2021 at 18:21
  • $\begingroup$ This works fine Thanks $\endgroup$
    – Konstantin
    Jun 4, 2021 at 18:29

In the rather unlikely case of you actually wanting to swap two definitions, e.g.

D1 := RandomVariate[UniformDistribution[{-1, 0}]];
D2 := RandomVariate[UniformDistribution[{0, 1}]];

You can do so temporarily

Module[{D1 = D2, D2 = D1}, Print[{D1, D2}]]

If you've already used these definitions e.g.

print := Print[{D1, D2}]

and would like to interchange them temporarily, you can use

Block[{D1 = D2, D2 = D1}, print]

Needless to say, if you interchange them permanently, there is risk of becoming confused about how many times you've done it!


Can't you just define a single function?

d[x_,y_]:=RandomVariate[UniformDistribution[{x, y}]];

Then call it as:

D1 = d[-1,0]


D2 = d[0,1]
  • $\begingroup$ Yeah, this possible, but not very fit to my purpose $\endgroup$
    – Konstantin
    Jun 4, 2021 at 18:20

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