How can I do a bitwise-or reduction along an axis of a boolean array in Julia? - julia

Question

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.

Answer 1

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

Answer 2

Try any(x, 3). Just typing a little more here so StackOverflow doesn't nix this response.

Answer 3

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)

Answer 4

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)