我有一个形状为(1000, 50, 100, 3)
(class 'numpy.ndarray'
)的Numpy图像阵列,其中包含1000个图像RGB(高度= 50,宽度= 100,通道= 3)。我想先将RGB
值转换为YUV
值,然后重新缩放它们以获得yuv
值。下面给出了像素转换器的原型实现。
我的问题:有没有一种简单的方法可以执行此转换?
def yuv(_pixel):
R, G, B = _pixel[0], _pixel[1], _pixel[2]
Y = 0.299 * R + 0.587 * G + 0.114 * B
y = Y / 127.5 - 1
u = (0.493 * (B - Y)) / 127.5 - 1
v = (0.887 * (R - Y)) / 127.5 - 1
return np.array([y, u, v])
答案 0 :(得分:2)
您可以这样做:
images_yuv = np.apply_along_axis( yuv, -1, images_rgb)
编辑:混淆参数的顺序
答案 1 :(得分:1)
您可以向量化转换,以便使用以下方法同时转换所有R,G和B像素:
import java.net.*;
import java.io.*;
public class bpart
{
public static void main(String[] args) throws Exception {
URL oracle = new URL("http://www.google.com/");
BufferedReader in = new BufferedReader(new InputStreamReader(oracle.openStream()));
OutputStream os = new FileOutputStream("my file location");
String inputLine;
while ((inputLine = in.readLine()) != null)
System.out.println(inputLine);
in.close();
}
}
为提高性能,我将演示问题中显示的yuv函数的简短嵌套循环实现:
def yuv_vec(images):
R, G, B = images[:, :, :, 0], images[:, :, :, 1], images[:, :, :, 2]
y = (0.299 * R + 0.587 * G + 0.114 * B) / 127.5 - 1
u = (0.493 * (B - y)) / 127.5 - 1
v = (0.887 * (R - y)) / 127.5 - 1
yuv_img = np.empty(images.shape)
yuv_img[:, :, :, 0] = y
yuv_img[:, :, :, 1] = u
yuv_img[:, :, :, 2] = v
return yuv_img
一些速度比较:
def yuv(_pixel):
R, G, B = _pixel[0], _pixel[1], _pixel[2]
y = (0.299 * R + 0.587 * G + 0.114 * B) / 127.5 - 1
u = (0.493 * (B - Y)) / 127.5 - 1
v = (0.887 * (R - Y)) / 127.5 - 1
return np.array([y, u, v])
def yuvloop(imgs):
yuvimg = np.empty(imgs.shape)
for n in range(imgs.shape[0]):
for i in range(imgs.shape[1]):
for j in range(imgs.shape[2]):
yuvimg[n, i, j] = yuv(imgs[n, i, j])
return yuvimg
因此,这比循环遍历像素快 256倍。使用imgs = np.random.randint(0, 256, size=(100, 50, 100, 3))
%timeit yuvloop(imgs)
# Out: 8.79 s ± 265 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
% timeit np.apply_along_axis(yuv, -1, imgs)
# Out: 9.92 s ± 360 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit yuv_vec(imgs)
# Out: 34.4 ms ± 385 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
似乎更慢。这三个的结果都相同。
我将测试样本的大小减少到100张图像,否则测试太慢了。