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CHALLENGER njit serial numpy dot #8

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@seanlaw seanlaw commented Feb 27, 2025 •
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Tested the njit dot function from numpy:

import numpy as np
from numba import njit


def setup(Q, T):
    sliding_dot_product(Q, T)
    return


@njit(fastmath=True)
def sliding_dot_product(Q, T):
    m = Q.shape[0]
    l = T.shape[0] - m + 1
    out = np.empty(l)
    for i in range(l):
        out[i] = np.dot(Q, T[i : i + m])

    return out

Using ./test.py -noheader -pmin 6 -pmax 23 -pdiff 6 challenger >> timing.csv.

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@seanlaw seanlaw changed the title CHALLENGER njit numpy dot CHALLENGER njit serial numpy dot Feb 27, 2025
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seanlaw commented Feb 27, 2025 •
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The results are comparable to the original njit and, possibly, "better" when len(Q) == len(T) and len(T) <= 2^14.

I would consider comparing the lower p values.

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