I'm trying to do a gradation plotting.
using Plots
using LinearAlgebra
L = 60 #size of a matrix
N = 10000 #number of loops
E = zeros(Complex{Float64},N,L) #set of eigenvalues
IPR = zeros(Complex{Float64},N,L) #indicator for marker_z
Preparing E & IPR
function main()
cnt = 0
for i = 1:N
cnt += 1
H = rand(Complex{Float64},L,L)
eigenvalue,eigenvector = eigen(H)
for j = 1:L
E[cnt,j] = eigenvalue[j]
IPR[cnt,j] = abs2(norm(abs2.(eigenvector[:,j])))/(abs2(norm(eigenvector[:,j])))
end
end
end
Plotting
function main1()
plot(real.(E),imag.(E),marker_z = real.(IPR),st = scatter,markercolors=:cool,markerstrokewidth=0,markersize=1,dpi=300)
plot!(legend=false,xlabel="ReE",ylabel="ImE")
savefig("test.png")
end
#time main1()
358.794885 seconds (94.30 M allocations: 129.882 GiB, 2.05% gc time)
Comparing with a uniform plotting, a gradation plotting takes too much time.
function main2()
plot(real.(E),imag.(E),st = scatter,markercolor=:blue,markerstrokewidth=0,markersize=1,dpi=300)
plot!(legend=false,xlabel="ReE",ylabel="ImE")
savefig("test1.png")
end
#time main2()
8.100609 seconds (10.85 M allocations: 508.054 MiB, 0.47% gc time)
Is there a way of gradation plotting as fast as a uniform plotting?
I solved the problem by myself.
After updating from Julia 1.3.1 to Julia1.6.3, I checked the main1 became faster as Bill's comments.
Related
I have the following coupled system of ODEs (that come from discretizing an integrodifferential PDE):
The xi's are points on an x-grid that I control. I can solve this with the following simple piece of code:
using DifferentialEquations
function ode_syst(du,u,p, t)
N = Int64(p[1])
beta= p[2]
deltax = 1/(N+1)
xs = [deltax*i for i in 1:N]
for j in 1:N
du[j] = -xs[j]^(beta)*u[j]+deltax*sum([u[i]*xs[i]^(beta) for i in 1:N])
end
end
N = 1000
u0 = ones(N)
beta = 2.0
p = [N, beta]
tspan = (0.0, 10^3);
prob = ODEProblem(ode_syst,u0,tspan,p);
sol = solve(prob);
However, as I make my grid finer, i.e. increase N, the computation time grows rapidly (I guess the scaling is quadratic in N). Is there any suggestion on how to implement this using either distributed parallelism or multithreading?
Additional information: I attach the profiling diagram that might be useful to understand where the program spends most of the time
I've looked into your code and found a few problems such as an accidentally introduced O(N^2) behavior due to recalculating the sum term.
My improved version uses the Tullio package to get further speed up from vectorization. Tullio also has tuneable parameters which would allow for multi threading if your system becomes big enough. See here what parameters you can tune in the options section. You might also see GPU support there, i didn't test that but it might yield further speed up or break hooribly. I also choose get the length from the acutal array which should make the use more economical and less error prone.
using Tullio
function ode_syst_t(du,u,p, t)
N = length(du)
beta = p[1]
deltax = 1/(N+1)
#tullio s := deltax*(u[i]*(i*deltax)^(beta))
#tullio du[j] = -(j*deltax)^(beta)*u[j] + s
return nothing
end
Your code:
#btime sol = solve(prob);
80.592 s (1349001 allocations: 10.22 GiB)
My code:
prob2 = ODEProblem(ode_syst_t,u0,tspan,[2.0]);
#btime sol2 = solve(prob2);
1.171 s (696 allocations: 18.50 MiB)
and the result more or less agree:
julia> sum(abs2, sol2(1000.0) .- sol(1000.0))
1.079046922815598e-14
Lutz Lehmanns solution i also benchmarked:
prob3 = ODEProblem(ode_syst_lehm,u0,tspan,p);
#btime sol3 = solve(prob3);
1.338 s (3348 allocations: 39.38 MiB)
However as we scale N to 1000000 with a tspan of (0.0, 10.0)
prob2 = ODEProblem(ode_syst_t,u0,tspan,[2.0]);
#time solve(prob2);
2.429239 seconds (280 allocations: 869.768 MiB, 13.60% gc time)
prob3 = ODEProblem(ode_syst_lehm,u0,tspan,p);
#time solve(prob3);
5.961889 seconds (580 allocations: 1.967 GiB, 11.08% gc time)
My code becomes more than twice as fast due to using the 2 cores in my old and rusty machine.
Analyze the formula. Obviously, the atomic terms repeat. So they should only be computed once.
function ode_syst(du,u,p, t)
N = Int64(p[1])
beta= p[2]
deltax = 1/(N+1)
xs = [deltax*i for i in 1:N]
term = [ xs[i]^(beta)*u[i] for i in 1:N]
term_sum = deltax*sum(term)
for j in 1:N
du[j] = -term[j]+term_sum
end
end
This should only linearly increase in N.
In many applications of map(f,X), it helps to create closures that depending on parameters apply different functions f to data X.
I can think of at least the following three ways to do this (note that the second for some reason does not work, bug?)
f0(x,y) = x+y
f1(x,y,p) = x+y^p
function g0(power::Bool,X,y)
if power
f = x -> f1(x,y,2.0)
else
f = x -> f0(x,y)
end
map(f,X)
end
function g1(power::Bool,X,y)
if power
f(x) = f1(x,y,2.0)
else
f(x) = f0(x,y)
end
map(f,X)
end
abstract FunType
abstract PowerFun <: FunType
abstract NoPowerFun <: FunType
function g2{S<:FunType}(T::Type{S},X,y)
f(::Type{PowerFun},x) = f1(x,y,2.0)
f(::Type{NoPowerFun},x) = f0(x,y)
map(x -> f(T,x),X)
end
X = 1.0:1000000.0
burnin0 = g0(true,X,4.0) + g0(false,X,4.0);
burnin1 = g1(true,X,4.0) + g1(false,X,4.0);
burnin2 = g2(PowerFun,X,4.0) + g2(NoPowerFun,X,4.0);
#time r0true = g0(true,X,4.0); #0.019515 seconds (12 allocations: 7.630 MB)
#time r0false = g0(false,X,4.0); #0.002984 seconds (12 allocations: 7.630 MB)
#time r1true = g1(true,X,4.0); # 0.004517 seconds (8 allocations: 7.630 MB, 26.28% gc time)
#time r1false = g1(false,X,4.0); # UndefVarError: f not defined
#time r2true = g2(PowerFun,X,4.0); # 0.085673 seconds (2.00 M allocations: 38.147 MB, 3.90% gc time)
#time r2false = g2(NoPowerFun,X,4.0); # 0.234087 seconds (2.00 M allocations: 38.147 MB, 60.61% gc time)
What is the optimal way to do this in Julia?
There's no need to use map here at all. Using a closure doesn't make things simpler or faster. Just use "dot-broadcasting" to apply the functions directly:
function g3(X,y,power=1)
if power != 1
return f1.(X, y, power) # or simply X .+ y^power
else
return f0.(X, y) # or simply X .+ y
end
end
I have this code(primitive heat transfer):
function heat(first, second, m)
#sync #parallel for d = 2:m - 1
for c = 2:m - 1
#inbounds second[c,d] = (first[c,d] + first[c+1, d] + first[c-1, d] + first[c, d+1] + first[c, d-1]) / 5.0;
end
end
end
m = parse(Int,ARGS[1]) #size of matrix
firstm = SharedArray(Float64, (m,m))
secondm = SharedArray(Float64, (m,m))
for c = 1:m
for d = 1:m
if c == m || d == 1
firstm[c,d] = 100.0
secondm[c,d] = 100.0
else
firstm[c,d] = 0.0
secondm[c,d] = 0.0
end
end
end
#time for i = 0:opak
heat(firstm, secondm, m)
firstm, secondm = secondm, firstm
end
This code give good times when run sequentially, but when I add #parallel it slow down even if I run on one thread. I just need explanation why this is happening? Code only if it doesn't change algorithm of heat function.
Have a look at http://docs.julialang.org/en/release-0.4/manual/performance-tips/ . Contrary to advised, you make use of global variables a lot. They are considered to change types anytime so they have to be boxed and unboxed everytime they are referenced. This question also Julia pi aproximation slow suffers from the same. In order to make your function faster, have global variables as input arguments to the function.
There are some points to consider. One of them is the size of m. If it is small, parallelism would give much overhead for not a big gain:
julia 36967257.jl 4
# Parallel:
0.040434 seconds (4.44 k allocations: 241.606 KB)
# Normal:
0.042141 seconds (29.13 k allocations: 1.308 MB)
For bigger m you could have better results:
julia 36967257.jl 4000
# Parallel:
0.054848 seconds (4.46 k allocations: 241.935 KB)
# Normal:
3.779843 seconds (29.13 k allocations: 1.308 MB)
Plus two remarks:
1/ initialisation could be simplified to:
for c = 1:m, d = 1:m
if c == m || d == 1
firstm[c,d] = 100.0
secondm[c,d] = 100.0
else
firstm[c,d] = 0.0
secondm[c,d] = 0.0
end
end
2/ your finite difference schema does not look stable. Please take a look at Linear multistep method or ADI/Crank Nicolson.
I am using Julia version 0.4.5 and I am experiencing the following issue:
As far as I know, taking inner product between a sparse vector and a dense vector should be as fast as updating the dense vector by a sparse vector. The latter one is much slower.
A = sprand(100000,100000,0.01)
w = rand(100000)
#time for i=1:100000
w += A[:,i]
end
26.304380 seconds (1.30 M allocations: 150.556 GB, 8.16% gc time)
#time for i=1:100000
A[:,i]'*w
end
0.815443 seconds (921.91 k allocations: 1.540 GB, 5.58% gc time)
I created a simple sparse matrix type of my own, and the addition code was ~ the same as the inner product.
Am I doing something wrong? I feel like there should be a special function doing the operation w += A[:,i], but I couldn't find it.
Any help is appreciated.
I asked the same question on GitHub and we came to the following conclusion. The type SparseVector was added as of Julia 0.4 and with it the BLAS function LinAlg.axpy!, which updates in-place a (possibly dense) vector x by a sparse vector y multiplied by a scalar a, i.e. performs x += a*y efficiently. However, in Julia 0.4 it is not implemented properly. It works only in Julia 0.5
#time for i=1:100000
LinAlg.axpy!(1,A[:,i],w)
end
1.041587 seconds (799.49 k allocations: 1.530 GB, 8.01% gc time)
However, this code is still sub-optimal, as it creates the SparseVector A[:,i]. One can get an even faster version with the following function:
function upd!(w,A,i,c)
rowval = A.rowval
nzval = A.nzval
#inbounds for j = nzrange(A,i)
w[rowval[j]] += c* nzval[j]
end
return w
end
#time for i=1:100000
upd!(w,A,i,1)
end
0.500323 seconds (99.49 k allocations: 1.518 MB)
This is exactly what I needed to achieve, after some research we managed to get there, thanks everyone!
Assuming you want to compute w += c * A[:, i], there is an easy way to vectorize it:
>>> A = sprand(100000, 100000, 0.01)
>>> c = rand(100000)
>>> r1 = zeros(100000)
>>> #time for i = 1:100000
>>> r1 += A[:, i] * c[i]
>>> end
29.997412 seconds (1.90 M allocations: 152.077 GB, 12.73% gc time)
>>> #time r2 = sum(A .* c', 2);
1.191850 seconds (50 allocations: 1.493 GB, 0.14% gc time)
>>> all(r1 == r2)
true
First, create a vector c of the constants to multiply with. Then multiplay de columns of A element-wise by the values of c (A .* c', it does broadcasting inside). Last, reduce over the columns of A (the part sum(.., 2)).
I'm trying to find the best way to do a bitwise-or reduction of a 3D boolean array of masks to 2D in Julia.
I can always write a for loop, of course:
x = randbool(3,3,3)
out = copy(x[:,:,1])
for i = 1:3
for j = 1:3
for k = 2:3
out[i,j] |= x[i,j,k]
end
end
end
But I'm wondering if there is a better way to do the reduction.
A simple answer would be
out = x[:,:,1] | x[:,:,2] | x[:,:,3]
but I did some benchmarking:
function simple(n,x)
out = x[:,:,1] | x[:,:,2]
for k = 3:n
#inbounds out |= x[:,:,k]
end
return out
end
function forloops(n,x)
out = copy(x[:,:,1])
for i = 1:n
for j = 1:n
for k = 2:n
#inbounds out[i,j] |= x[i,j,k]
end
end
end
return out
end
function forloopscolfirst(n,x)
out = copy(x[:,:,1])
for j = 1:n
for i = 1:n
for k = 2:n
#inbounds out[i,j] |= x[i,j,k]
end
end
end
return out
end
shorty(n,x) = |([x[:,:,i] for i in 1:n]...)
timholy(n,x) = any(x,3)
function runtest(n)
x = randbool(n,n,n)
#time out1 = simple(n,x)
#time out2 = forloops(n,x)
#time out3 = forloopscolfirst(n,x)
#time out4 = shorty(n,x)
#time out5 = timholy(n,x)
println(all(out1 .== out2))
println(all(out1 .== out3))
println(all(out1 .== out4))
println(all(out1 .== out5))
end
runtest(3)
runtest(500)
which gave the following results
# For 500
simple: 0.039403016 seconds (39716840 bytes allocated)
forloops: 6.259421683 seconds (77504 bytes allocated)
forloopscolfirst 1.809124505 seconds (77504 bytes allocated)
shorty: elapsed time: 0.050384062 seconds (39464608 bytes allocated)
timholy: 2.396887396 seconds (31784 bytes allocated)
So I'd go with simple or shorty
Try any(x, 3). Just typing a little more here so StackOverflow doesn't nix this response.
There are various standard optimization tricks and hints that can be applied, but the critical observation to make here is that Julia organizes array in column-major rather than row-major order. For small size arrays this is not easily seen but when the arrays grow large it's telling. There is a method reduce provided that is optimized to perform an function on a collection (in this case OR), but it comes at a cost. If the number of combining steps is relatively small then it's better to simply loop. In all cases minimizing the number of memory access is over all better. Below are various attempts at optimization using these 2 things in mind.
Various Attempts and Observations
Initial function
Here's a function that takes your example and generalizes it.
function boolReduce1(x)
out = copy(x[:,:,1])
for i = 1:size(x,1)
for j = 1:size(x,2)
for k = 2:size(x,3)
out[i,j] |= x[i,j,k]
end
end
end
out
end
Creating a fairly large array, we can time it's performance
julia> #time boolReduce1(b);
elapsed time: 42.372058096 seconds (1056704 bytes allocated)
Applying optimizations
Here's another similar version but with the standard type hints, use of #inbounds and inverting the loops.
function boolReduce2(b::BitArray{3})
a = BitArray{2}(size(b)[1:2]...)
for j = 1:size(b,2)
for i = 1:size(b,1)
#inbounds a[i,j] = b[i,j,1]
for k = 2:size(b,3)
#inbounds a[i,j] |= b[i,j,k]
end
end
end
a
end
And take the time
julia> #time boolReduce2(b);
elapsed time: 12.892392891 seconds (500520 bytes allocated)
The insight
The 2nd function is a lot faster, and also less memory is allocated because a temporary array wasn't created. But what if we simply take the first function and invert the array indexing?
function boolReduce3(x)
out = copy(x[:,:,1])
for j = 1:size(x,2)
for i = 1:size(x,1)
for k = 2:size(x,3)
out[i,j] |= x[i,j,k]
end
end
end
out
end
and take the time now
julia> #time boolReduce3(b);
elapsed time: 12.451501749 seconds (1056704 bytes allocated)
That's just as fast as the 2nd function.
Using reduce
There is a function called reduce that we can use to eliminate the 3rd loop. Its function is to repeatedly apply an operation on all of the elements with the result of the previous operation. This is exactly what we want.
function boolReduce4(b)
a = BitArray{2}(size(b)[1:2]...)
for j = 1:size(b,2)
for i = 1:size(b,1)
#inbounds a[i,j] = reduce(|,b[i,j,:])
end
end
a
end
Now take it's time
julia> #time boolReduce4(b);
elapsed time: 15.828273008 seconds (1503092520 bytes allocated, 4.07% gc time)
That's ok, but not even as fast as the simple optimized original. The reason is, take a look at all of the extra memory that was allocated. This is because data has to be copied from all over to produce input for reduce.
Combining things
But what if we max out the insight as best we can. Instead of the last index being reduced, the first one is?
function boolReduceX(b)
a = BitArray{2}(size(b)[2:3]...)
for j = 1:size(b,3)
for i = 1:size(b,2)
#inbounds a[i,j] = reduce(|,b[:,i,j])
end
end
a
end
And now create a similar array and time it.
julia> c = randbool(200,2000,2000);
julia> #time boolReduceX(c);
elapsed time: 1.877547669 seconds (927092520 bytes allocated, 21.66% gc time)
Resulting in a function 20x faster than the original version for large arrays. Pretty good.
But what if medium size?
If the size is very large then the above function appears best, but if the data set size is smaller, the use of reduce doesn't pay enough back and the following is faster. Including a temporary variable speeds things from version 2. Another version of boolReduceX using a loop instead of reduce (not show here) was even faster.
function boolReduce5(b)
a = BitArray{2}(size(b)[1:2]...)
for j = 1:size(b,2)
for i = 1:size(b,1)
#inbounds t = b[i,j,1]
for k = 2:size(b,3)
#inbounds t |= b[i,j,k]
end
#inbounds a[i,j] = t
end
end
a
end
julia> b = randbool(2000,2000,20);
julia> c = randbool(20,2000,2000);
julia> #time boolReduceX(c);
elapsed time: 1.535334322 seconds (799092520 bytes allocated, 23.79% gc time)
julia> #time boolReduce5(b);
elapsed time: 0.491410981 seconds (500520 bytes allocated)
It is faster to devectorize. It's just a matter of how much work you want to put in. The naïve devectorized approach is slow because it's a BitArray: extracting contiguous regions and bitwise OR can both be done a 64-bit chunk at a time, but the naïve devectorized approach operates an element at a time. On top of that, indexing BitArrays is slow, both because there is a sequence of bit operations involved and because it can't presently be inlined due to the bounds check. Here's a strategy that is devectorized but exploits the structure of the BitArray. Most of the code is copy-pasted from copy_chunks! in bitarray.jl and I didn't try to prettify it (sorry!).
function devec(n::Int, x::BitArray)
src = x.chunks
out = falses(n, n)
dest = out.chunks
numbits = n*n
kd0 = 1
ld0 = 0
for j = 1:n
pos_s = (n*n)*(j-1)+1
kd1, ld1 = Base.get_chunks_id(numbits - 1)
ks0, ls0 = Base.get_chunks_id(pos_s)
ks1, ls1 = Base.get_chunks_id(pos_s + numbits - 1)
delta_kd = kd1 - kd0
delta_ks = ks1 - ks0
u = Base._msk64
if delta_kd == 0
msk_d0 = ~(u << ld0) | (u << (ld1+1))
else
msk_d0 = ~(u << ld0)
msk_d1 = (u << (ld1+1))
end
if delta_ks == 0
msk_s0 = (u << ls0) & ~(u << (ls1+1))
else
msk_s0 = (u << ls0)
end
chunk_s0 = Base.glue_src_bitchunks(src, ks0, ks1, msk_s0, ls0)
dest[kd0] |= (dest[kd0] & msk_d0) | ((chunk_s0 << ld0) & ~msk_d0)
delta_kd == 0 && continue
for i = 1 : kd1 - kd0
chunk_s1 = Base.glue_src_bitchunks(src, ks0 + i, ks1, msk_s0, ls0)
chunk_s = (chunk_s0 >>> (64 - ld0)) | (chunk_s1 << ld0)
dest[kd0 + i] |= chunk_s
chunk_s0 = chunk_s1
end
end
out
end
With Iain's benchmarks, this gives me:
simple: 0.051321131 seconds (46356000 bytes allocated, 30.03% gc time)
forloops: 6.226652258 seconds (92976 bytes allocated)
forloopscolfirst: 2.099381939 seconds (89472 bytes allocated)
shorty: 0.060194226 seconds (46387760 bytes allocated, 36.27% gc time)
timholy: 2.464298752 seconds (31784 bytes allocated)
devec: 0.008734413 seconds (31472 bytes allocated)