[Numpy-discussion] Correlation filter
Keith Goodman
kwgoodman@gmail....
Fri Nov 20 13:28:49 CST 2009
On Fri, Nov 20, 2009 at 11:17 AM, <josef.pktd@gmail.com> wrote:
> On Fri, Nov 20, 2009 at 1:51 PM, Keith Goodman <kwgoodman@gmail.com> wrote:
>> def corr3(x, y):
>> x = x - x.mean()
>> x /= x.std()
>> nx = x.size
>> one = np.ones(nx)
>> xy = lfilter(x, 1, y)
>> sy = lfilter(one, 1, y)
>> sy2 = lfilter(one, 1, y*y)
>> d = xy / np.sqrt(nx * sy2 - sy * sy)
>> return d
>
> Is this correct? xy uses the original y and not a demeaned y.
I wouldn't be surprised if I made mistakes. But I don't think I need
to demean y. One way to write the numerator of the correlation
coefficent is
N*sum(xy) - sum(x)*sum(y)
but sum(x) is zero, so it becomes
N*sum(xy)
So I think I only need to demean x. But I'd better test the code. Plus
I have no idea how lfilter will handle my missing values (NaN).
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