[Numpy-discussion] Correlation filter
Mon Nov 23 00:32:03 CST 2009
On Fri, Nov 20, 2009 at 4:27 PM, <email@example.com> wrote:
> On Fri, Nov 20, 2009 at 2:28 PM, Keith Goodman <firstname.lastname@example.org> wrote:
>> On Fri, Nov 20, 2009 at 11:17 AM, <email@example.com> wrote:
>>> On Fri, Nov 20, 2009 at 1:51 PM, Keith Goodman <firstname.lastname@example.org> 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
>> 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).
> I think nans fully propagate, trick might be to fill with zero and use
> another convolution/correlation with the non-nan mask to get the
> number of observations included (I think I saw it as a trick by Pierre
> for autocorrelation with missing observations).
Just an update, I used this for moving correlation between x and y
of equal shape, just posted to the scipy-user mailing list.
In the test case it works correctly.
> Here is a moving slope function which estimate the slope coefficient
> for a linear trend if x is arange()
> I compared the ranking with your moving correlation and it is quite different.
> def moving_slope(x,y)
> '''estimate moving slope coefficient of regression of y on x
> filters along axis=1, returns valid observations
> Todo: axis and lag options
> xx = np.column_stack((np.ones(x.shape), x))
> pinvxx = np.linalg.pinv(xx)[1:,:]
> windsize = len(x)
> lead = windsize//2 - 1
> return signal.correlate(y, pinvxx, 'full'
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