Set a time limitation on algorithm in Julia

Question

I have a Julia code which runs for quite sometime. I want to run the code for 3 hours and then terminate it. Does anyone know what is the most efficient way to do this. Would appreciate any advice. Thanks

Answer 1

I recommend using Distributed to spawn your function as a new process and control it's time (I believe I have been answering similar question but I cannot find the answer).

Here is the code:

using Distributed
function run_with_timeout(timeout::Int,f::Function, wid::Int)
    result = RemoteChannel(()->Channel{Tuple}(1));
    @spawnat wid put!(result, (f(),myid()))
    res = (:timeout, wid)
    time_elapsed = 0.0
    while time_elapsed < timeout && !isready(result)
        sleep(0.5)
        time_elapsed += 0.5
    end
    if !isready(result)
        println("Timeout! at $wid")
    else
        res = take!(result)
    end
    return res
end

You can use it like this (please note how external packages are imported):

wid = addprocs(1)[1]
@everywhere using DataFrames
@everywhere function ff()
    sleep(2)
    #code fir making heavy computations
    #this is the place to write whatever you need
    return DataFrame(x=[1,2],y=[1,3])
end

Now let us run it. Note that the second returned value is the workerid to run computations (you might want to remove it):

julia> run_with_timeout(10,() ->(try;ff();catch ee;dump(ee);end ),wid)
(2×2 DataFrame
│ Row │ x     │ y     │
│     │ Int64 │ Int64 │
├─────┼───────┼───────┤
│ 1   │ 1     │ 1     │
│ 2   │ 2     │ 3     │, 2)

If we give our code only 1 sec to execute it will fail:

julia> run_with_timeout(1,() ->(try;ff();catch ee;dump(ee);end ),wid)
Timeout! at 2
(:timeout, 2)

Finally, since the timeout occured the process should be killed (otherwise it would continue to execute):

rmprocs(wid)