我有一段代碼如下：
import cv2
import numpy as np
import operator
from functools import reduce
image = cv2.imread("<some image path>")
bgr = np.int16(image)
h, w, _ = image.shape
mask = np.zeros((h, w), np.uint8)
# Get all channels
blue = bgr[:,:,0]
green = bgr[:,:,1]
red = bgr[:,:,2]
rules = np.where(reduce(operator.and_, [(red > 100), (red > green), (red > blue)]
# Create mask using above rules
mask[rules] = 255
### Then use cv2.findContours ...
這段代碼沒有像我預期的那樣運行得足夠快。我想我可以通過一一應用所有條件來使其更快，即：
rule_1 = np.where(red > 100)
rule_2 = np.where(red[rule_1] > green)
rule_3 = np.where(red[rule_2] > blue)
mask[rule_3] = 255
上述方法可以加速我的代碼嗎？以及如何做到這一點？非常感謝！
uj5u.com熱心網友回復：
一個很好的方法是（改編自 Cris Luengo 的評論）
mask = 255 * ((red > 100) & (red > green) & (red > blue))
但如果你需要更快，你可以使用Numba
from numba import jit, prange
@jit(nopython=True, parallel=True)
def red_dominates(rgb, mask):
M, N, _ = rgb.shape
for i in prange(M):
for j in prange(N):
r = rgb[i,j,0]
g = rgb[i,j,1]
b = rgb[i,j,2]
mask[i,j] = 255 * ((r > 100) & (r > g) & (r > b))
return mask
請注意，prange使用 代替range。這告訴 Numba 回圈是可并行化的。
在我的電腦上，Numba 版本快了大約 3 倍。
>>> bgr = np.int16(255 * np.random.random((100, 100, 3)))
>>> w = np.ones(bgr.shape[:2], np.uint8)
>>> %timeit red_dominates(bgr, w)
13.7 μs ± 26.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> %timeit 255 * ((bgr[:,:,0] > 100) & (bgr[:,:,0] > bgr[:,:,1]) & (bgr[:,:,0] > bgr[:,:,2]))
46.3 μs ± 208 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
祝你好運！
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