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RF: Rearranging matmul order and using hermitian flag in ICC_rep_anova #3453

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Merged
merged 8 commits into from
Apr 21, 2022
Merged

RF: Rearranging matmul order and using hermitian flag in ICC_rep_anova #3453

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e3d08bd
np.dot(X.T, T) is Hermitian
Apr 20, 2022
9f2fd97
Reordering matmul operations
Apr 20, 2022
49b7ed5
Using @ operator instead of np.dot
Apr 21, 2022
1b2eddd
Merge branch 'nipy:master' into linalg-speedup
kristoferm94 Apr 21, 2022
835dfe7
Merge branch 'nipy:master' into linalg-speedup
kristoferm94 Apr 21, 2022
883a84a
revert
kristoferm94 Apr 21, 2022
2aaa8a5
removing unneccesary import
Apr 21, 2022
9cb58c3
Update .pre-commit-config.yaml
kristoferm94 Apr 21, 2022
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4 changes: 2 additions & 2 deletions nipype/algorithms/icc.py
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Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
# -*- coding: utf-8 -*-
import os
import numpy as np
from numpy import ones, kron, mean, eye, hstack, dot, tile
from numpy import ones, kron, mean, eye, hstack, tile
from numpy.linalg import pinv
import nibabel as nb
from ..interfaces.base import (
Expand Down Expand Up @@ -114,7 +114,7 @@ def ICC_rep_anova(Y):
X = hstack([x, x0])

# Sum Square Error
predicted_Y = dot(dot(dot(X, pinv(dot(X.T, X))), X.T), Y.flatten("F"))
predicted_Y = X @ (pinv(X.T @ X, hermitian=True) @ (X.T @ Y.flatten("F")))
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For a simulated Y of shape 50 x 10, I get:

In []: %timeit dot(dot(dot(X, pinv(dot(X.T, X))), X.T), Y.flatten("F"))
1.42 ms ± 128 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

Convert to matmul:

In []: %timeit (X @ pinv(X.T @ X) @ X.T) @ Y.flatten("F")
1.4 ms ± 188 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

No change. Reorder multiplications:

In []: %timeit X @ (pinv(X.T @ X) @ (X.T @ Y.flatten("F")))
838 µs ± 43.3 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

41% speedup. Use hermetian optimization:

In []: %timeit X @ (pinv(X.T @ X, hermitian=True) @ (X.T @ Y.flatten("F")))
785 µs ± 17.8 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

6% relative speedup for total of 45%.

residuals = Y.flatten("F") - predicted_Y
SSE = (residuals**2).sum()

Expand Down