Separately time different lines in For Loop in Julia?

Question

In Julia, I have a function that times each "kth" iteration of a for loop in the function using the @elapsed macro, and assigns such times to the variable t (and then vector time_vector). However, is there any way to write this function so that different lines within the for loop can be timed separately? I would like to time each line of the for loop, and insert the times into a two-dimensional array time_vector?

function timeloop(k)
    time_vector = zeros(k)
    for x = 1:k
         t = @elapsed 
             L₁=rand(10,k) #time this line with @elapsed?                
             L₂=rand(k,10) #time this line with @elapsed
             L₁*L₂ #time this line with #elapsed
         end
         time_vector[x] = t
     end 
     time_vector  
end

Have you considered profiling? See https://stackoverflow.com/questions/65625496/line-by-line-profiling-of-julia-code-possible and the manual https://docs.julialang.org/en/v1/manual/profile/ or https://www.julia-vscode.org/docs/dev/userguide/profiler/ or https://opensourc.es/blog/benchmarking-and-profiling-julia-code/ — Dan Getz, Aug 29 '22 at 22:20
Note that timing too small a region is hard to do meaningfully on modern out-of-order exec CPUs that are only running full speed when dozens to hundreds of instructions are "in flight" at once. And timing overhead is *at least* 20 clock cycles on modern x86 (for just `rdtsc`), many more for a system-time function using it. Filling a whole array or matrix is something you can meaningfully time if it's large enough, but see [Idiomatic way of performance evaluation?](https://stackoverflow.com/q/60291987) for pitfalls like page fault and CPU-frequency warm-up. — Peter Cordes, Aug 30 '22 at 12:43

score 2 · Accepted Answer · answered Aug 30 '22 at 12:31

You can return a vector of each single @elapsed like.

function timeloop(k)
    time_vector = [zeros(3) for _ in 1:k]
    for x = 1:k
        t = [@elapsed L₁=rand(10,k)
             @elapsed L₂=rand(k,10)
             @elapsed L₁*L₂]
        time_vector[x] = t
    end 
    time_vector  
end

timeloop(5)
#5-element Vector{Vector{Float64}}:
# [1.604e-6, 3.63e-7, 1.1884e-5]
# [5.56e-7, 5.61e-7, 1.193e-6]
# [5.61e-7, 5.22e-7, 1.079e-6]
# [2.471e-6, 5.39e-7, 1.086e-6]
# [5.21e-7, 5.22e-7, 2.742e-6]

Separately time different lines in For Loop in Julia?

1 Answers1