[Numpy-discussion] Efficient orthogonalisation with scipy/numpy
Tue Jan 19 14:52:54 CST 2010
On Tue, Jan 19, 2010 at 3:47 PM, Gael Varoquaux
> On Tue, Jan 19, 2010 at 02:22:30PM -0600, Robert Kern wrote:
>> > y = y - np.dot(confounds.T, linalg.lstsq(confounds.T, y))
>> > with np = numpy and linalg = scipy.linalg where scipy calls ATLAS.
>> For clarification, are you trying to find the components of the y
>> vectors that are perpendicular to the space spanned by the 10
>> orthonormal vectors in confounds?
> Yes. Actually, what I am doing is calculating partial correlation between
> x and y conditionally to confounds, with the following code:
> def cond_partial_cor(y, x, confounds=):
> """ Returns the partial correlation of y and x, conditionning on
> # First orthogonalise y and x relative to confounds
> if len(confounds):
> y = y - np.dot(confounds.T, linalg.lstsq(confounds.T, y))
> x = x - np.dot(confounds.T, linalg.lstsq(confounds.T, x))
you can combine x, and y for one call to leastsq, if it makes a difference
linalg.lstsq(confounds.T, [x,y]) #format? columnstack?
I don't see anything else yet
> return np.dot(x, y)/sqrt(np.dot(y, y)*np.dot(x, x))
> I am not sure that what I am doing is optimal.
>> > Most of the time is spent in linalg.lstsq. The length of the vectors is
>> > 810, and there are about 10 confounds.
>> Exactly what are the shapes? y.shape = (810, N); confounds.shape = (810, 10)?
> Sorry, I should have been more precise:
> y.shape = (810, )
> confounds.shape = (10, 810)
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