Anaconda vs Julia Autocorrelation

Ross Mariano:

I am trying to use Julia to do autocorrelation and compare results in Python. Why do they give different results?

Julia code

using StatsBase

t = range(0, stop=10, length=10)
test_data = sin.(exp.(t.^2))

acf = StatsBase.autocor(test_data)

Give

10-element Array{Float64,1}:
  1.0                   
  0.13254954979179642   
 -0.2030283419321465    
  0.00029587850872956104
 -0.06629381497277881   
  0.031309038331589614  
 -0.16633393452504994   
 -0.08482388975165675   
  0.0006905628640697538 
 -0.1443650483145533

Python code

from statsmodels.tsa.stattools import acf
import numpy as np

t = np.linspace(0,10,10)
test_data = np.sin(np.exp(t**2))

acf_result = acf(test_data)

Give

array([ 1.        ,  0.14589844, -0.10412699,  0.07817509, -0.12916543,
       -0.03469143, -0.129255  , -0.15982435, -0.02067688, -0.14633346])
Jacob Neeson:

This is because you test_dataare different:

python:

array([ 0.84147098, -0.29102733,  0.96323736,  0.75441021, -0.37291918,
        0.85600145,  0.89676529, -0.34006519, -0.75811102, -0.99910501])

Julia:

[0.8414709848078965, -0.2910273263243299, 0.963237364649543, 0.7544102058854344,
 -0.3729191776326039, 0.8560014512776061, 0.9841238290665676, 0.1665709194875013,
 -0.7581110212957692, -0.9991050130774393]

sinThis is because of the huge numbers you are taking . For example, with the number t10 in the past, it exp(10^2)is ~2.7 x 10^43. At this scale, floating point inaccuracy is about 3*10^9. So if even the least significant bit is different for Python and Julia, the sinvalue will be the way to go.

In fact, we can examine the underlying binary value of the initial array t. For example, they differ in the third-to-last value:

Julia:

julia> reinterpret(Int, range(0, stop=10, length=10)[end-2])
4620443017702830535

python:

>>> import struct
>>> s = struct.pack('>d', np.linspace(0,10,10)[-3])
>>> struct.unpack('>q', s)[0]
4620443017702830536

Indeed, we can see that they disagree with exactly one machine precision. If we use Julia to take sinthe value obtained by Python:

julia> sin(exp(reinterpret(Float64, 4620443017702830536)^2))
-0.3400651855865199

We get the same value Python does.

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