Computational gain for the one factor model

[TomMaullin]

For the one factor model, large gains can be made by considering that Z'Z is block diagonal. Ideally it would be nice to account for this and implement speedier computation of D(I+Z'ZD)^(-1) for this use case.

[TomMaullin]

Places in the code this could be undertaken:

• Computation of DinvIplusZtZD could be split into 4 cases:
  • r=1,nraneffs[0]=1 (in which case DinvIplusZtZD is diagonal)
  • r=1, nraneffs[0]>1 (in which case it's block diagonal)
  • q>1000 (in which case the recursive inverse method is probably the best choice)
  • else (just do it with numpy).
• Anywhere where there is multiplication by DinvIplusZtZD in the one factor model can be reduced to a running sum of matrices of much smaller dimension (probably using einsum even). Two cases would be of possible speedup:
  • r=1,nraneffs[0]=1 (using a dot product of the diagonal elements)
  • r=1, nraneffs[0]>1 (using Ddict and an einsum)
• For the one factor model, D need never be constructed in full. This would have to be removed from 3 places:
  • Whereever it is created.
  • Where its used to calculate DinvIplusZtZD
  • Where its used in the log-likelihood.

[TomMaullin]

Important: For this to work we must first ensure that the factor is given in increasing order (else pd.get_dummies in blmm_batch will give Z in a non-diagonal format.).

[TomMaullin]

Update: In PR #45 , the following use case has now been accounted for in the 3D code:

• When r=1 and nraneffs[0]=1, the code does not construct ZtZ, DinvIplusZtZD or D.

The case r=1 and nraneffs[0]>1 has not been considered.