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是否可以转换此代码的一部分以提高速度?

numba cython cuda python

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Alexandre Bensi · 技术社区 · 3 年前

我是新来的,

请,我需要帮助将此代码的一部分转换为任何东西,以提高速度,并且仍然与python程序兼容以使用此函数。有可能用这个来提高速度吗?

在我的系统中,一个224x224颜色的图像需要360毫秒才能完成。

#!/usr/bin/env python3
# coding: utf-8
# original code https://github.com/verhovsky/squircle/blob/master/squircle.py

import cv2
import math
import time
import numpy

_epsilon = 0.0000000001

def _sgn(x):
    if x == 0.0:
        return 0.0
    if x < 0:
        return -1.0
    return 1.0

def _pixel_coordinates_to_unit(coordinate, max_value):
    return coordinate / max_value * 2 - 1

def _one_coordinates_to_pixels(coordinate, max_value):
    return (coordinate + 1) / 2 * max_value

def _stretch_square_to_disc(x, y):
    if (abs(x) < _epsilon) or (abs(y) < _epsilon):
        return x, y

    x2 = x * x
    y2 = y * y
    hypotenuse_squared = x * x + y * y

    reciprocal_hypotenuse = 1.0 / math.sqrt(hypotenuse_squared)

    multiplier = 1.0

    if x2 > y2:
        multiplier = _sgn(x) * x * reciprocal_hypotenuse
    else:
        multiplier = _sgn(y) * y * reciprocal_hypotenuse

    return x * multiplier, y * multiplier

def _transform(inp):
    result = numpy.zeros_like(inp)
    for x, row in enumerate(inp):

        unit_x = _pixel_coordinates_to_unit(x, len(inp))

        for y, _ in enumerate(row):
            unit_y = _pixel_coordinates_to_unit(y, len(row))

            try:
                uv = _stretch_square_to_disc(unit_x, unit_y)
                if uv is None:
                    continue
                u, v = uv

                u = _one_coordinates_to_pixels(u, len(inp))
                v = _one_coordinates_to_pixels(v, len(row))

                result[x][y] = inp[math.floor(u)][math.floor(v)]
            except IndexError:
                pass

    return result

# -- load and test

img = cv2.imread('circle.png')

elapsed = round(time.time() * 1000)
squareImage = _transform(img[0:224, 0:224])
print(str(round(time.time() * 1000) - elapsed)+' ms to squareImage')

cv2.imshow('square', squareImage)
key = cv2.waitKey(0)

cv2.destroyAllWindows()

我本想尝试将这段代码或其中的一部分转换为更快速的代码,可能是CUDA直接在GPU上运行,也可能是NUMBA、Cython、lib等。。。

1 回复 | 直到 3 年前

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2

Andrej Kesely 3 年前

我用过 numba 装饰功能(删除 _sgn 并使用 np.sign ,删除了 try..except -需要吗?):

import math
import time

import cv2
import numpy
from numba import njit

_epsilon = 0.0000000001


@njit
def _pixel_coordinates_to_unit(coordinate, max_value):
    return coordinate / max_value * 2 - 1


@njit
def _one_coordinates_to_pixels(coordinate, max_value):
    return (coordinate + 1) / 2 * max_value


@njit
def _stretch_square_to_disc(x, y):
    if (abs(x) < _epsilon) or (abs(y) < _epsilon):
        return x, y

    x2 = x * x
    y2 = y * y
    hypotenuse_squared = x * x + y * y

    reciprocal_hypotenuse = 1.0 / np.sqrt(hypotenuse_squared)

    multiplier = 1.0

    if x2 > y2:
        multiplier = np.sign(x) * x * reciprocal_hypotenuse
    else:
        multiplier = np.sign(y) * y * reciprocal_hypotenuse

    return x * multiplier, y * multiplier


@njit
def _transform(inp):
    result = numpy.zeros_like(inp)
    for x, row in enumerate(inp):
        unit_x = _pixel_coordinates_to_unit(x, len(inp))

        for y, _ in enumerate(row):
            unit_y = _pixel_coordinates_to_unit(y, len(row))

            uv = _stretch_square_to_disc(unit_x, unit_y)
            if uv is None:
                continue

            u, v = uv

            u = _one_coordinates_to_pixels(u, len(inp))
            v = _one_coordinates_to_pixels(v, len(row))

            result[x][y] = inp[math.floor(u)][math.floor(v)]

    return result


# -- load and test

img = cv2.imread("circle.png")

# warm jit
# this is needed to let numba do the JIT optimizations
# if you run the the function "cold", the running time will be larger

# you can use compile-ahead-of-time
# https://numba.pydata.org/numba-doc/dev/user/pycc.html

squareImage = _transform(img[0:224, 0:224])

elapsed = time.perf_counter_ns()
squareImage = _transform(img[0:224, 0:224])
print(str((time.perf_counter_ns() - elapsed) / 1000) + " us to squareImage")

cv2.imwrite("shashed.png", squareImage)

在我的电脑(AMD 5700X)上,它打印:

528.596 us to squareImage

# without using numba:
# 47928.774 us to squareImage

使用的图像:

是否可以转换此代码的一部分以提高速度?

`circle.png`

结果