broadcasting linalg.pinv on a 3D matrix

Question

in the example below, there is a 3d matrix of size (4, 3, 3). how to calculate pinv of each of 4 of those 3*3 matrices without a loop in numpy?

dt = np.dtype(np.float32)

a=[[[12,3,1],
   [2,4,1],
   [2,4,2],],
   [[12,3,3],
   [2,4,4],
   [2,4,5],],
   [[12,3,6],
   [2,4,5],
   [2,4,4],],
   [[12,3,3],
   [2,4,5],
   [2,4,6]]]

a=np.asarray(a,dtype=dt)
print(a.shape)

apinv=np.zeros((4,3,3))
print(np.linalg.pinv(a[0,:,:]).shape)

for i in range(4):
   apinv[i,:,:]=np.linalg.pinv(a[i,:,:])

Please note that linalg.inv broadcasts through the matrix as it is described here:

print(np.linalg.inv(a).shape)

But it doesn't work the same way for pinv.

score 1 · Accepted Answer · answered Dec 03 '15 at 19:18

1

One solution would be using map and lambda functions:

apinv = map(lambda n: np.linalg.pinv(n), a)
apinv = np.asarray(apinv,dtype=dt)

answered Dec 03 '15 at 19:18

Cedric Zoppolo

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thanks! are you aware of any similar solution using theano? – PickleRick Dec 03 '15 at 19:33
Glad it worked. Sorry, but I have no experience with `theano` yet. – Cedric Zoppolo Dec 08 '15 at 16:00
no worries. a solution is provided in here: http://stackoverflow.com/questions/34090846/broadcasting-linalg-pinv-on-a-3d-theano-tensor/34103695#34103695 – PickleRick Dec 09 '15 at 16:58

broadcasting linalg.pinv on a 3D matrix

1 Answers1