numpy.dot() result different with classic computation for large-size arrays

[paul18fr]

I was not fast enough

Try this for example (have a look to "print" structure)

import time
import numpy
import array
 
N=100000000

# # 8 bytes size int
# a = array.array('q')

# for i in range(N):
#         a.append(i);
 
# b = array.array('q')
# for i in range(N, 2*N):
#         b.append(i)

    
a = numpy.arange(N)
b = numpy.arange(N, 2*N)
# classic dot product of vectors implementation
tic = time.process_time()
dot = 0;
 
for i in range(N):
        dot += a[i] * b[i]
 
toc = time.process_time()
 
print(f"dot_product = {dot}")
print(f"Classic Computation time = {1000*(toc - tic )}ms")
 
n_tic = time.process_time()
n_dot_product = numpy.dot(a, b)     ## or n_dot_product = a @ b
n_toc = time.process_time()
 
print(f"\nn_dot_product = {(n_dot_product)}")
print(f"Vec Computation time = {1000*(n_toc - n_tic )} ms")

print(f"Difference on the dot product results = {numpy.abs(n_dot_product - dot)}")