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Fixed#938: Speed up njit-decorated function for sliding dot product [WIP] #939
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aa82e2e
add notebook for njit fft and sliding dot product
NimaSarajpoor 3e16ce8
Add notebook for sliding_dot_product
NimaSarajpoor c2a9b43
add OTFFT notebook
NimaSarajpoor 5d6fc83
add docstrings and put each function in one cell
NimaSarajpoor 2c5dd4d
Add function _fft0
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,289 @@ | ||
| { | ||
| "cells": [ | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "5b88fa48", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "This notebook implements the 6-step / 8-step FFT (fft) algorithm as provided in [OTFFT](http://wwwa.pikara.ne.jp/okojisan/otfft-en/stockham2.html). Accordingly, the inverse FFT (ifft) algorithm will be implemented as well. When the input of FFT, `T` with length `n`, consists of real-valued elements only, we can take advantage of real FFT (rfft) as explained in [2.6.2](https://www.researchgate.net/profile/Christos-Bechlioulis/publication/341270520_FFT_algorithms_are_not_mine_However_I_am_going_to_convince_you_soon_regarding_the_visit_of_RMS_to_our_university_Believe_it_or_not_this_is_me_This_is_us_Univeristy_of_Patras_you_have_chosen_a_quite_wr/links/5fa53ce7299bf10f7328c33b/FFT-algorithms-are-not-mine-However-I-am-going-to-convince-you-soon-regarding-the-visit-of-RMS-to-our-university-Believe-it-or-not-this-is-me-This-is-us-Univeristy-of-Patras-you-have-chosen-a-quite.pdf). " | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": 2, | ||
| "id": "c9a886c2", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [ | ||
| "import math\n", | ||
| "import time\n", | ||
| "\n", | ||
| "import numba\n", | ||
| "import numpy as np\n", | ||
| "import matplotlib.pyplot as plt\n", | ||
| "import scipy\n", | ||
| "\n", | ||
| "from numba import njit, prange\n", | ||
| "import numpy.testing as npt\n", | ||
| "\n", | ||
| "from stumpy import core" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "80765bca", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "Let's start with `rfft`. First, we write the test!" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "8fc07997", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "## rfft" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": 3, | ||
| "id": "7bb4242c", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [ | ||
| "def test_rfft(n_powers_list):\n", | ||
| " seed = 0\n", | ||
| " np.random.seed(seed)\n", | ||
| " for p in n_powers_list:\n", | ||
| " n = 2 ** p\n", | ||
| " T = np.random.rand(n)\n", | ||
| " \n", | ||
| " ref = scipy.fft.rfft(T)\n", | ||
| " comp = rfft(T)\n", | ||
| " \n", | ||
| " npt.assert_almost_equal(ref, comp)" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "5a1c553e", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "We now implement `rfft` function according to the steps provided in [2.6.2](https://www.researchgate.net/profile/Christos-Bechlioulis/publication/341270520_FFT_algorithms_are_not_mine_However_I_am_going_to_convince_you_soon_regarding_the_visit_of_RMS_to_our_university_Believe_it_or_not_this_is_me_This_is_us_Univeristy_of_Patras_you_have_chosen_a_quite_wr/links/5fa53ce7299bf10f7328c33b/FFT-algorithms-are-not-mine-However-I-am-going-to-convince-you-soon-regarding-the-visit-of-RMS-to-our-university-Believe-it-or-not-this-is-me-This-is-us-Univeristy-of-Patras-you-have-chosen-a-quite.pdf)." | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": 4, | ||
| "id": "d0033890", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [ | ||
| "def _rfft(T):\n", | ||
| " \"\"\"\n", | ||
| " For the input `T` with length `n=len(T)`, this function returns its\n", | ||
| " real fast fourier transform (rfft) with length of `(n // 2) + 1`.\n", | ||
| " \n", | ||
| " Parameters\n", | ||
| " ----------\n", | ||
| " T : numpy.ndarray\n", | ||
| " A time series of interest, with real-valued numbers\n", | ||
| " \n", | ||
| " Returns\n", | ||
| " -------\n", | ||
| " out : numpy.ndarray\n", | ||
| " the real fast fourier transform (rfft) of input `T`\n", | ||
| " \"\"\"\n", | ||
| " n = len(T)\n", | ||
| " half_n = int(n // 2)\n", | ||
| " \n", | ||
| " x = T[::2] + 1j * T[1::2]\n", | ||
| " x[:] = scipy.fft.fft(x) # we will implement our fft shortly!\n", | ||
| " \n", | ||
| " out = np.empty(half_n + 1, dtype=np.complex_)\n", | ||
| " out[0] = x[0].real + x[0].imag\n", | ||
| " out[half_n] = x[0].real - x[0].imag\n", | ||
| " out[n // 4] = x[n // 4].conjugate()\n", | ||
| " \n", | ||
| " theta0 = 2 * math.pi / n\n", | ||
| " for k in range(1, n // 4):\n", | ||
| " theta = theta0 * k\n", | ||
| " a = x[half_n - k].conjugate()\n", | ||
| " b = 0.5 * (x[k] - a) * (1.0 + complex(math.sin(theta), math.cos(theta)))\n", | ||
| " out[k] = x[k] - b\n", | ||
| " out[half_n - k] = (a + b).conjugate()\n", | ||
| " \n", | ||
| " return out" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": 5, | ||
| "id": "d5db0eb9", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [ | ||
| "def rfft(T):\n", | ||
| " \"\"\"\n", | ||
| " For the input `T` with length `n=len(T)`, this function returns its\n", | ||
| " real fast fourier transform (rfft) with length of `(n // 2) + 1`.\n", | ||
| " \n", | ||
| " Parameters\n", | ||
| " ----------\n", | ||
| " T : numpy.ndarray\n", | ||
| " A time series of interest, with real-valued numbers\n", | ||
| " \n", | ||
| " Returns\n", | ||
| " -------\n", | ||
| " out : numpy.ndarray\n", | ||
| " the real fast fourier transform (rfft) of input `T`\n", | ||
| " \"\"\"\n", | ||
| " return _rfft(T)" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": 6, | ||
| "id": "ded21f89", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [ | ||
| "n_powers_list = np.arange(2, 11)\n", | ||
| "test_rfft(n_powers_list)" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "6398a98a", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "We now work on implementing 6-step / 8-step FFT algorithm. We then revisit the two functions above. We will replace `scipy.fft.fft(x)` with our FFT function, and then change them accordingly." | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": 133, | ||
| "id": "09e2fb6d", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [ | ||
| "@njit(fastmath=True)\n", | ||
| "def _fft0(n, s, eo, x, y):\n", | ||
| " \"\"\"\n", | ||
| " A recurive function that is being called by six-step / eight-step FFT algorithm, \n", | ||
| " and update `x` in place.\n", | ||
| " \n", | ||
| " Parameters\n", | ||
| " ----------\n", | ||
| " n : int\n", | ||
| " Length of sequence\n", | ||
| " \n", | ||
| " s : int\n", | ||
| " size of striding window\n", | ||
| " \n", | ||
| " eo : bool\n", | ||
| " If False, `x` is output. If True, `y` is the output.\n", | ||
| " \n", | ||
| " x : numpy.ndarray\n", | ||
| " A 1D numpy.ndarray with `np.complex_` dtype\n", | ||
| " \n", | ||
| " y : nummpy.ndarray\n", | ||
| " A 1D numpy.ndarray with same size as `x` and `np.complex_` dtype. \n", | ||
| " \n", | ||
| " Returns\n", | ||
| " -------\n", | ||
| " \n", | ||
| " Notes\n", | ||
| " -----\n", | ||
| " <http://wwwa.pikara.ne.jp/okojisan/otfft-en/sixstepfft.html>`__\n", | ||
| "\n", | ||
| " See function `fft0` provided in \"List-11: Six-Step FFT\"\n", | ||
| " \"\"\"\n", | ||
| " if n == 2:\n", | ||
| " if eo:\n", | ||
| " z = y\n", | ||
| " else:\n", | ||
| " z = x\n", | ||
| " \n", | ||
| " for i in range(s):\n", | ||
| " j = i + s\n", | ||
| " a = x[i]\n", | ||
| " b = x[j]\n", | ||
| " z[i] = a + b\n", | ||
| " z[j] = a - b\n", | ||
| " \n", | ||
| " elif n >= 4:\n", | ||
| " m = n // 2\n", | ||
| " sm = s * m\n", | ||
| " \n", | ||
| " theta = 2 * math.pi / n\n", | ||
| " c = complex(math.cos(theta), -math.sin(theta))\n", | ||
| " \n", | ||
| " twiddle_factor = 1.0\n", | ||
| " for p in range(m):\n", | ||
| " sp = s * p\n", | ||
| " two_sp = 2 * sp\n", | ||
| " for q in range(s):\n", | ||
| " i = sp + q\n", | ||
| " j = i + sm\n", | ||
| " \n", | ||
| " k = two_sp + q\n", | ||
| " y[k] = x[i] + x[j]\n", | ||
| " y[k + s] = (x[i] - x[j]) * twiddle_factor\n", | ||
| " \n", | ||
| " twiddle_factor = twiddle_factor * c\n", | ||
| " \n", | ||
| " _fft0(m, 2 * s, not eo, y, x)\n", | ||
| " \n", | ||
| " else:\n", | ||
| " pass" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": null, | ||
| "id": "9813fb90", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": null, | ||
| "id": "a57bf1ac", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": null, | ||
| "id": "88312d26", | ||
| "metadata": {}, | ||
| "outputs": [], | ||
| "source": [] | ||
| } | ||
| ], | ||
| "metadata": { | ||
| "kernelspec": { | ||
| "display_name": "Python 3 (ipykernel)", | ||
| "language": "python", | ||
| "name": "python3" | ||
| }, | ||
| "language_info": { | ||
| "codemirror_mode": { | ||
| "name": "ipython", | ||
| "version": 3 | ||
| }, | ||
| "file_extension": ".py", | ||
| "mimetype": "text/x-python", | ||
| "name": "python", | ||
| "nbconvert_exporter": "python", | ||
| "pygments_lexer": "ipython3", | ||
| "version": "3.10.12" | ||
| } | ||
| }, | ||
| "nbformat": 4, | ||
| "nbformat_minor": 5 | ||
| } | ||
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We will later work on implementing this
fftfunction using 6-step / 8-step algorithm. We then change this line and the caller function accordingly.Reply via ReviewNB