Julia returns unexpected values-CodePudding

I am a frequent user of Python for ML and am interested in converting to Julia for its promising improvements in speed. To get a feel for this I have just written a simple loop in both Python and Julia in order to compare the runtime for each one, but I seem to be getting unexpected results in the Julia implementation.

Here are both versions:

In Julia:

B = Array{Float64, 2}(undef, 10000,10000)

function f(x)
    return x^2
end

function g(x)
    return x^3
end

function h(x)
    return x^4
end

function compose(f,g,h,x)
    return f(g(h(x)))
end

for i in 1:10000
    for j in 1:10000
        B[i,j] = compose(f,g,h,i j)
    end
end

println(sum(B))

In Python:

import numpy as np

B = np.zeros((10000, 10000), dtype=np.float64)

def f(x):
    return x**2

def g(x):
    return x**3

def h(x):
    return x**4

def compute(x):
    return f(g(h(x)))

for i in range(10000):
    for j in range(10000):
        B[i, j] = compute(i j 2)

print(np.sum(B))

EDIT: The Julia version returns ~8.3e22 while Python returns ~1.0e109 (which is what I expect).

While the Julia version runs extremely fast compared to the Python version, its results aren't what I expect them to be. As mentioned above, I am using Julia for the first time here, but I can't see where my mistake might be. Is this a result of my array/datatype handling? Are my functions not doing what I think they are?

CodePudding user response：

I'm not sure how you compared the two codes, but on my machine Julia is simply 1000X faster.

f(x) = x^2
g(x) = x^3
h(x) = (x^2)^2
compute(x) = f(g(h(x)))

function compute_sum(B)
    for j in 1:10000
        for i in 1:10000
            B[i,j] = compute(i j 0.0)
        end 
    end 
    sum(B)
end

B = zeros(10000, 10000)
@time compute_sum(B)
1.0337869071875959e109
  0.126618 seconds (1 allocation: 16 bytes)

And Python:

from time import time

t0 = time()
for i in range(10000):
    for j in range(10000):
        B[i, j] = compute(i j 2)

s = np.sum(B)
print('Time:', time()-t0, 'sec')
print(s)
Time: 126.33235883712769 sec
1.0337869071875948e 109

CodePudding user response：

I think one can even get that a little faster by omitting the B and using the iterator feature of the sum

f(x) = x^2
g(x) = x^3
h(x) = (x^2)^2
compute(x) = f(g(h(x)))

large_sum(n) = sum(compute(i j 0.0) for i=1:n, j=1:n)
    
@time large_sum(10000)

yields

  0.115042 seconds
1.0337869071880225e109

and

julia> @allocated large_sum(10000)
0

CodePudding user response：

Your integer values are too large - beyond 64 bits and overflows.

Use Float64 instead (or BigInt):

compose(f,g,h,Float64(i j))