CHALLENGER njit serial numpy dot

[seanlaw]

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.

[seanlaw]

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.

[NimaSarajpoor]

@seanlaw
I was looking at the code and the plot provided in the first comment of this PR. The code is similar to njit_sdp.py, and the difference is that np.dot is used here instead of a for-loop. And, similar to njit_sdp, it does not outperform pyfftw when len(T) > len(Q) > 2^6.

So, I was wondering if we should just compare the challenger with njit_sdp and focus on len(Q) <= 2^6 where we know njit_sdp.py can outperform the other methods.

[seanlaw]

Sure