python - 使用 Numpy 更有效地变换色调

Question

我正在制作一个在线图像编辑器，我正在实现一个色相/周六/明度编辑器。这是我使用给定值更改图像的功能。

def make_edit(pixels, hue, sat, lum):
    shape = pixels.shape
    new = np.empty(shape)
    print(time.time())
    for row_count, row in enumerate(pixels):
        for pixel_count, p in enumerate(row):
            new_hue = p[0] + hue
            if new_hue < 0:
                new_hue += 255
            elif new_hue > 255:
                new_hue -= 255

            new_sat = p[1] + sat
            if new_sat < 0:
                new_sat += 255
            elif new_sat > 255:
                new_sat -= 255

            new_lum = p[2] + lum
            if new_lum < 0:
                new_lum = 0
            elif new_lum > 255:
                new_lum = 255

            new[row_count, pixel_count] = np.array([new_hue, new_sat, new_lum])
    print(time.time())
    return new

该函数采用形状为(高度、宽度、3)的 numpy 数组。我逐个像素地做，然后将色调、饱和度和亮度值添加到每个像素。然而，它需要 13 秒(在 (648, 1152, 3) 形状的阵列上)，显然太长了。是否有一个 numpy 函数可以用我给它的数量来抵消所有值。
附言该功能还不起作用，hue 似乎可以，但是 sat 和 lum 没有提供正确的图像。

Answer 1

由于您还没有让 sat 和 lum 正常工作，因此可能需要根据您的最终代码的样子进行调整，但它确实与您当前过程的结果相匹配，并且速度快了几个数量级:

def getPic():   
    return  np.random.randint(0, 255, 648*1152*3).reshape(648, 1152, 3)

def make_edit(pixels, hue, sat, lum):
    shape = pixels.shape
    new = np.empty(shape)
    #print(time.time())
    for row_count, row in enumerate(pixels):
        for pixel_count, p in enumerate(row):
            new_hue = p[0] + hue
            if new_hue < 0:
                new_hue += 255
            elif new_hue > 255:
                new_hue -= 255

            new_sat = p[1] + sat
            if new_sat < 0:
                new_sat += 255
            elif new_sat > 255:
                new_sat -= 255

            new_lum = p[2] + lum
            if new_lum < 0:
                new_lum = 0
            elif new_lum > 255:
                new_lum = 255

            new[row_count, pixel_count] = np.array([new_hue, new_sat, new_lum])
    #print(time.time())
    return new


def new_make_edit(pixels, hue, sat, lum):
    new = np.empty_like(pixels)
    new[:,:,0] = pixels[:,:,0] + hue
    new[:,:,0][new[:,:,0]<0] += 255
    new[:,:,0][new[:,:,0]>255] -= 255

    new[:,:,1] = pixels[:,:,1] + sat
    new[:,:,1][new[:,:,1]<0] += 255
    new[:,:,1][new[:,:,1]>255] -= 255

    new[:,:,2] = pixels[:,:,2] + lum
    new[:,:,2][new[:,:,2]<0] = 0
    new[:,:,2][new[:,:,2]>255] = 255
    return new

def tEd():
    pic = getPic()
    old = make_edit(pic, 10, -25, 211)
    new = new_make_edit(pic, 10, -25, 211)
    return old, new

def timeOld():
    pic = getPic()
    old = make_edit(pic, 10, -25, 211)
    return old

def timeNew():
    pic = getPic()
    new = new_make_edit(pic, 10, -25, 211)
    return new

在同一图像上执行 old 和 new 并验证输出匹配:

>>> o,n=tEd()
>>> np.all(o==n)
True

性能对比:

>>> timeit.timeit(timeNew, number=10)
0.5608299169980455
>>> timeit.timeit(timeOld, number=10)
58.86368254100671