Probablity weighted ReplaceAll

Question

I'm looking for an elegant way of doing a probability weighted replacement rule,

something like this

f[x_] :=  x /.  {1,2,3,4} -> {50%{ 4,3,2,1} ,50%{1,3,2,4}}

Using an If statement and generating a RandomReal will be very clumsy as I want different probabilities depending on the pattern I want replaced.

Thanks

Answer 1

You can use RandomChoice to randomly select an item from a list.

You can even randomly select an operator and apply it. For example:

f[x_List] := RandomChoice[{Reverse, Identity}][x]

Table[f[{1, 2, 3, 4}], {10}]

returns:

{{4, 3, 2, 1}, {4, 3, 2, 1}, {1, 2, 3, 4}, {1, 2, 3, 4}, {1, 2, 3, 
  4}, {1, 2, 3, 4}, {4, 3, 2, 1}, {1, 2, 3, 4}, {4, 3, 2, 1}, {1, 2, 
  3, 4}}

As pointed by @WReach you can also assign a weight to each choice (weights are automatically normalized, i.e. they not need to sum to 1). For example:

f[x_List] := RandomChoice[{3, 1} -> {Reverse, Identity}][x]
Table[f[{1, 2, 3, 4}], {10}]

returns:

{{4, 3, 2, 1}, {4, 3, 2, 1}, {4, 3, 2, 1}, {4, 3, 2, 1}, {4, 3, 2, 
  1}, {1, 2, 3, 4}, {4, 3, 2, 1}, {1, 2, 3, 4}, {4, 3, 2, 1}, {1, 2, 
  3, 4}}

Your specific pseudo-code can be replaced with something like:

f[x_] := x /. {1, 2, 3, 4} :> RandomChoice[{{1, 2, 3, 4}, {4, 3, 2, 1}}]
Table[f[{1, 2, 3, 4}], {10}]

Answer 2

Another option is to use Condition:

Replace[foo, {foo_ /; RandomReal[] < p :> bar, foo_ :> baz}]

where p is the probability you want to give to the first replacement.

This is somewhat similar to using an If, but I find it somewhat more idiomatic. I would also avoid using ReplaceAll, when you actually want Replace as ReplaceAll might recurse down the whole expression.

You could also explore random function definition:

ff[x_] /; RandomReal[] < p := bar
ff[x_] := baz

Not completely unlike Church