numpy 中的 RGB 到 HSV [英] RGB to HSV in numpy
问题描述
我正在尝试使用 这里:
def rgb2hsv_opencv(img_rgb):
img_hsv = cv2.cvtColor(img_rgb, cv2.COLOR_RGB2HSV)
return img_hsv
def rgb2hsv_np(img_rgb):
assert img_rgb.dtype == np.float32
height, width, c = img_rgb.shape
r, g, b = img_rgb[:,:,0], img_rgb[:,:,1], img_rgb[:,:,2]
t = np.min(img_rgb, axis=-1)
v = np.max(img_rgb, axis=-1)
s = (v - t) / (v + 1e-6)
s[v==0] = 0
# v==r
hr = 60 * (g - b) / (v - t + 1e-6)
# v==g
hg = 120 + 60 * (b - r) / (v - t + 1e-6)
# v==b
hb = 240 + 60 * (r - g) / (v - t + 1e-6)
h = np.zeros((height, width), np.float32)
h = h.flatten()
hr = hr.flatten()
hg = hg.flatten()
hb = hb.flatten()
h[(v==r).flatten()] = hr[(v==r).flatten()]
h[(v==g).flatten()] = hg[(v==g).flatten()]
h[(v==b).flatten()] = hb[(v==b).flatten()]
h[h<0] += 360
h = h.reshape((height, width))
img_hsv = np.stack([h, s, v], axis=-1)
return img_hsv
img_bgr = cv2.imread('00000.png')
img_rgb = cv2.cvtColor(img_bgr, cv2.COLOR_BGR2RGB)
img_rgb = img_rgb / 255.0
img_rgb = img_rgb.astype(np.float32)
img_hsv1 = rgb2hsv_np(img_rgb)
img_hsv2 = rgb2hsv_opencv(img_rgb)
print('max diff:', np.max(np.fabs(img_hsv1 - img_hsv2)))
print('min diff:', np.min(np.fabs(img_hsv1 - img_hsv2)))
print('mean diff:', np.mean(np.fabs(img_hsv1 - img_hsv2)))
但我得到了很大的不同:
But I get big diff:
max diff: 240.0
min diff: 0.0
mean diff: 0.18085355
我是否遗漏了什么?
还有可能编写更高效的 numpy 代码,例如没有 flatten
吗?
Also maybe it's possible to write numpy code more efficient, for example without flatten
?
我也很难找到 cvtColor
函数的原始 C++ 代码,据我所知,它实际上应该是 C 代码中的 cvCvtColor
函数,但我找不到实际的带公式的源代码.
Also I have hard time finding original C++ code for cvtColor
function, as I understand it should be actually function cvCvtColor
from C code, but I can't find actual source code with formula.
推荐答案
从最大差异正好是 240 的事实来看,我很确定发生的情况是 v==r
, v==g
与 v==b
同时为真,最后执行.
From the fact that the max difference is exactly 240, I'm pretty sure that what's happening is in the case when both or either of v==r
, v==g
are simultaneously true alongside v==b
, which gets executed last.
如果您更改订单:
h[(v==r).flatten()] = hr[(v==r).flatten()]
h[(v==g).flatten()] = hg[(v==g).flatten()]
h[(v==b).flatten()] = hb[(v==b).flatten()]
致:
h[(v==r).flatten()] = hr[(v==r).flatten()]
h[(v==b).flatten()] = hb[(v==b).flatten()]
h[(v==g).flatten()] = hg[(v==g).flatten()]
最大差异可能开始显示为 120,因为在该等式中添加了 120.因此,理想情况下,您希望按照 b->g->r 的顺序执行这三行.那么差异应该可以忽略不计(仍然注意到最大差异为 0.01~,将其归为某个四舍五入的地方).
The max difference may start showing up as 120, because of that added 120 in that equation. So ideally, you would want to execute these three lines in the order b->g->r. The difference should be negligible then (still noticing a max difference of 0.01~, chalking it up to some round off somewhere).
h[(v==b).flatten()] = hb[(v==b).flatten()]
h[(v==g).flatten()] = hg[(v==g).flatten()]
h[(v==r).flatten()] = hr[(v==r).flatten()]
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