CODEWITHSUNDEEP
arraysreplacejuliaprobability
Jim Maas
Jim Maas

Reputation: 1729

Julia, perform substitution with on a probability basis

I'm working with an array from a 3d array, and with lots of previous help here, got a very nice simple elegant solution :

## Reset 44 and 33 to 4 and 3 respectively
replace!(@view(Pop[end, :, 1]), 33=>3, 44=>4)

Now if I wanted to do the replacement on a probability basis, where I would supply the probability, can I do it in one line or do I need to write a function to interrogate the small array, and then loop through and do the substitutions based on the probability? So for example if I wanted to replace 33=>3 with a probability of 0.7, so the sub is made on 70% of occurrences of 33, and replace 44=>4 with a different probability, eg 0.85 how would I write this? Thx. J

Upvotes: 0

Views: 38

Answers (1)

DNF
DNF

Reputation: 12654

Here's one possible solution, which is pretty efficient, only a few times slower than a deterministic replace!

function rreplace!(xs, args...)
    @inbounds for i in eachindex(xs)
        val = xs[i]
        for tup in args
            if val == tup[1][1]
                rand(tup[2]) && (xs[i] = tup[1][2])
                break
            end
        end
    end
    return xs
end

You can call it like this:

rreplace!(xs, (33=>3, Bernoulli(0.7)), (44=>4, Bernoulli(0.85)))

You could of course use other distributions aside from Bernoulli, as long as they return either true or false.

In your particular question, it would be

rreplace!(@view(Pop[end, :, 1]), (33=>3, Bernoulli(0.7)), (44=>4, Bernoulli(0.85)))

Upvotes: 2

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