Is there any numpy autocorrellation function with standardized output?

Question

I followed the advice of defining the autocorrelation function in another post:

def autocorr(x):
    result = np.correlate(x, x, mode = \'full\')
    maxcorr         


        

Answer 1

So your problem with your initial attempt is that you did not subtract the average from your signal. The following code should work:

timeseries = (your data here)
mean = np.mean(timeseries)
timeseries -= np.mean(timeseries)
autocorr_f = np.correlate(timeseries, timeseries, mode='full')
temp = autocorr_f[autocorr_f.size/2:]/autocorr_f[autocorr_f.size/2]
iact.append(sum(autocorr_f[autocorr_f.size/2:]/autocorr_f[autocorr_f.size/2]))

In my example temp is the variable you are interested in; it is the forward integrated autocorrelation function. If you want the integrated autocorrelation time you are interested in iact.

Answer 2

I'm not sure what the issue is.

The autocorrelation of a vector x has to be 1 at lag 0 since that is just the squared L2 norm divided by itself, i.e., dot(x, x) / dot(x, x) == 1.

In general, for any lags i, j in Z, where i != j the unit-scaled autocorrelation is dot(shift(x, i), shift(x, j)) / dot(x, x) where shift(y, n) is a function that shifts the vector y by n time points and Z is the set of integers since we're talking about the implementation (in theory the lags can be in the set of real numbers).

I get 1.0 as the max with the following code (start on the command line as $ ipython --pylab), as expected:

In[1]: n = 1000
In[2]: x = randn(n)
In[3]: xc = correlate(x, x, mode='full')
In[4]: xc /= xc[xc.argmax()]
In[5]: xchalf = xc[xc.size / 2:]
In[6]: xchalf_max = xchalf.max()
In[7]: print xchalf_max
Out[1]: 1.0

The only time when the lag 0 autocorrelation is not equal to 1 is when x is the zero signal (all zeros).

The answer to your question is: no, there is no NumPy function that automatically performs standardization for you.

Besides, even if it did you would still have to check it against your expected output, and if you're able to say "Yes this performed the standardization correctly", then I would assume that you know how to implement it yourself.

I'm going to suggest that it might be the case that you've implemented their algorithm incorrectly, although I can't be sure since I'm not familiar with it.