使用 PIL 将 RGBA PNG 转换为 RGB

Question

我正在使用 PIL 将使用 Django 上传的透明 PNG 图像转换为 JPG 文件。 输出看起来坏了。

源文件

代码

Image.open(object.logo.path).save('/tmp/output.jpg', 'JPEG')

或者

Image.open(object.logo.path).convert('RGB').save('/tmp/output.png')

结果

两种方式，生成的图像如下所示：

有没有办法来解决这个问题？ 我想要在透明背景曾经是白色背景的地方。

---

解决方案

感谢出色的答案，我想出了以下函数集合：

import Image
import numpy as np


def alpha_to_color(image, color=(255, 255, 255)):
    """Set all fully transparent pixels of an RGBA image to the specified color.
    This is a very simple solution that might leave over some ugly edges, due
    to semi-transparent areas. You should use alpha_composite_with color instead.

    Source: http://stackoverflow.com/a/9166671/284318

    Keyword Arguments:
    image -- PIL RGBA Image object
    color -- Tuple r, g, b (default 255, 255, 255)

    """ 
    x = np.array(image)
    r, g, b, a = np.rollaxis(x, axis=-1)
    r[a == 0] = color[0]
    g[a == 0] = color[1]
    b[a == 0] = color[2] 
    x = np.dstack([r, g, b, a])
    return Image.fromarray(x, 'RGBA')


def alpha_composite(front, back):
    """Alpha composite two RGBA images.

    Source: http://stackoverflow.com/a/9166671/284318

    Keyword Arguments:
    front -- PIL RGBA Image object
    back -- PIL RGBA Image object

    """
    front = np.asarray(front)
    back = np.asarray(back)
    result = np.empty(front.shape, dtype='float')
    alpha = np.index_exp[:, :, 3:]
    rgb = np.index_exp[:, :, :3]
    falpha = front[alpha] / 255.0
    balpha = back[alpha] / 255.0
    result[alpha] = falpha + balpha * (1 - falpha)
    old_setting = np.seterr(invalid='ignore')
    result[rgb] = (front[rgb] * falpha + back[rgb] * balpha * (1 - falpha)) / result[alpha]
    np.seterr(**old_setting)
    result[alpha] *= 255
    np.clip(result, 0, 255)
    # astype('uint8') maps np.nan and np.inf to 0
    result = result.astype('uint8')
    result = Image.fromarray(result, 'RGBA')
    return result


def alpha_composite_with_color(image, color=(255, 255, 255)):
    """Alpha composite an RGBA image with a single color image of the
    specified color and the same size as the original image.

    Keyword Arguments:
    image -- PIL RGBA Image object
    color -- Tuple r, g, b (default 255, 255, 255)

    """
    back = Image.new('RGBA', size=image.size, color=color + (255,))
    return alpha_composite(image, back)


def pure_pil_alpha_to_color_v1(image, color=(255, 255, 255)):
    """Alpha composite an RGBA Image with a specified color.

    NOTE: This version is much slower than the
    alpha_composite_with_color solution. Use it only if
    numpy is not available.

    Source: http://stackoverflow.com/a/9168169/284318

    Keyword Arguments:
    image -- PIL RGBA Image object
    color -- Tuple r, g, b (default 255, 255, 255)

    """ 
    def blend_value(back, front, a):
        return (front * a + back * (255 - a)) / 255

    def blend_rgba(back, front):
        result = [blend_value(back[i], front[i], front[3]) for i in (0, 1, 2)]
        return tuple(result + [255])

    im = image.copy()  # don't edit the reference directly
    p = im.load()  # load pixel array
    for y in range(im.size[1]):
        for x in range(im.size[0]):
            p[x, y] = blend_rgba(color + (255,), p[x, y])

    return im

def pure_pil_alpha_to_color_v2(image, color=(255, 255, 255)):
    """Alpha composite an RGBA Image with a specified color.

    Simpler, faster version than the solutions above.

    Source: http://stackoverflow.com/a/9459208/284318

    Keyword Arguments:
    image -- PIL RGBA Image object
    color -- Tuple r, g, b (default 255, 255, 255)

    """
    image.load()  # needed for split()
    background = Image.new('RGB', image.size, color)
    background.paste(image, mask=image.split()[3])  # 3 is the alpha channel
    return background

表现

简单的非合成alpha_to_color函数是最快的解决方案，但会留下难看的边框，因为它不处理半透明区域。

纯 PIL 和 numpy 合成解决方案都给出了很好的结果，但alpha_composite_with_color比pure_pil_alpha_to_color (79.6 毫秒) 快得多 (8.93 毫秒)。 如果 numpy 在您的系统上可用，那就是要走的路。 （更新：新的纯 PIL 版本是所有提到的解决方案中最快的。）

$ python -m timeit "import Image; from apps.front import utils; i = Image.open(u'logo.png'); i2 = utils.alpha_to_color(i)"
10 loops, best of 3: 4.67 msec per loop
$ python -m timeit "import Image; from apps.front import utils; i = Image.open(u'logo.png'); i2 = utils.alpha_composite_with_color(i)"
10 loops, best of 3: 8.93 msec per loop
$ python -m timeit "import Image; from apps.front import utils; i = Image.open(u'logo.png'); i2 = utils.pure_pil_alpha_to_color(i)"
10 loops, best of 3: 79.6 msec per loop
$ python -m timeit "import Image; from apps.front import utils; i = Image.open(u'logo.png'); i2 = utils.pure_pil_alpha_to_color_v2(i)"
10 loops, best of 3: 1.1 msec per loop

Answer 1

这是一个更简单的版本 - 不确定它的性能如何。 很大程度上基于我在构建RGBA -> JPG + BG支持 sorl 缩略图时发现的一些 django 片段。

from PIL import Image

png = Image.open(object.logo.path)
png.load() # required for png.split()

background = Image.new("RGB", png.size, (255, 255, 255))
background.paste(png, mask=png.split()[3]) # 3 is the alpha channel

background.save('foo.jpg', 'JPEG', quality=80)

结果@80%

结果@ 50%

Answer 2

通过使用Image.alpha_composite ，Yuji 'Tomita' Tomita 的解决方案变得更简单。 如果 png 没有 alpha 通道，此代码可以避免tuple index out of range错误。

from PIL import Image

png = Image.open(img_path).convert('RGBA')
background = Image.new('RGBA', png.size, (255, 255, 255))

alpha_composite = Image.alpha_composite(background, png)
alpha_composite.save('foo.jpg', 'JPEG', quality=80)

Answer 3

透明部分大多具有 RGBA 值 (0,0,0,0)。 由于 JPG 没有透明度，所以 jpeg 值设置为 (0,0,0)，即黑色。

在圆形图标周围，有非零 RGB 值的像素，其中 A = 0。因此它们在 PNG 中看起来是透明的，但在 JPG 中看起来很有趣。

您可以使用 numpy 将 A == 0 的所有像素设置为 R = G = B = 255，如下所示：

import Image
import numpy as np

FNAME = 'logo.png'
img = Image.open(FNAME).convert('RGBA')
x = np.array(img)
r, g, b, a = np.rollaxis(x, axis = -1)
r[a == 0] = 255
g[a == 0] = 255
b[a == 0] = 255
x = np.dstack([r, g, b, a])
img = Image.fromarray(x, 'RGBA')
img.save('/tmp/out.jpg')

---

请注意，徽标还有一些半透明像素，用于平滑文字和图标周围的边缘。 保存到 jpeg 会忽略半透明度，使生成的 jpeg 看起来很锯齿。

使用 imagemagick 的convert命令可以获得更好的质量结果：

convert logo.png -background white -flatten /tmp/out.jpg

---

要使用 numpy 进行质量更好的混合，您可以使用alpha compositing ：

import Image
import numpy as np

def alpha_composite(src, dst):
    '''
    Return the alpha composite of src and dst.

    Parameters:
    src -- PIL RGBA Image object
    dst -- PIL RGBA Image object

    The algorithm comes from http://en.wikipedia.org/wiki/Alpha_compositing
    '''
    # http://stackoverflow.com/a/3375291/190597
    # http://stackoverflow.com/a/9166671/190597
    src = np.asarray(src)
    dst = np.asarray(dst)
    out = np.empty(src.shape, dtype = 'float')
    alpha = np.index_exp[:, :, 3:]
    rgb = np.index_exp[:, :, :3]
    src_a = src[alpha]/255.0
    dst_a = dst[alpha]/255.0
    out[alpha] = src_a+dst_a*(1-src_a)
    old_setting = np.seterr(invalid = 'ignore')
    out[rgb] = (src[rgb]*src_a + dst[rgb]*dst_a*(1-src_a))/out[alpha]
    np.seterr(**old_setting)    
    out[alpha] *= 255
    np.clip(out,0,255)
    # astype('uint8') maps np.nan (and np.inf) to 0
    out = out.astype('uint8')
    out = Image.fromarray(out, 'RGBA')
    return out            

FNAME = 'logo.png'
img = Image.open(FNAME).convert('RGBA')
white = Image.new('RGBA', size = img.size, color = (255, 255, 255, 255))
img = alpha_composite(img, white)
img.save('/tmp/out.jpg')

Answer 4

这是纯 PIL 中的解决方案。

def blend_value(under, over, a):
    return (over*a + under*(255-a)) / 255

def blend_rgba(under, over):
    return tuple([blend_value(under[i], over[i], over[3]) for i in (0,1,2)] + [255])

white = (255, 255, 255, 255)

im = Image.open(object.logo.path)
p = im.load()
for y in range(im.size[1]):
    for x in range(im.size[0]):
        p[x,y] = blend_rgba(white, p[x,y])
im.save('/tmp/output.png')

Answer 5

它没有坏。 它完全按照您的指示行事； 这些像素是完全透明的黑色。 您将需要遍历所有像素并将完全透明的像素转换为白色。

Answer 6

import numpy as np
import PIL

def convert_image(image_file):
    image = Image.open(image_file) # this could be a 4D array PNG (RGBA)
    original_width, original_height = image.size

    np_image = np.array(image)
    new_image = np.zeros((np_image.shape[0], np_image.shape[1], 3)) 
    # create 3D array

    for each_channel in range(3):
        new_image[:,:,each_channel] = np_image[:,:,each_channel]  
        # only copy first 3 channels.

    # flushing
    np_image = []
    return new_image

Answer 7

from PIL import Image
 
def fig2img ( fig ):
    """
    @brief Convert a Matplotlib figure to a PIL Image in RGBA format and return it
    @param fig a matplotlib figure
    @return a Python Imaging Library ( PIL ) image
    """
    # put the figure pixmap into a numpy array
    buf = fig2data ( fig )
    w, h, d = buf.shape
    return Image.frombytes( "RGBA", ( w ,h ), buf.tostring( ) )

def fig2data ( fig ):
    """
    @brief Convert a Matplotlib figure to a 4D numpy array with RGBA channels and return it
    @param fig a matplotlib figure
    @return a numpy 3D array of RGBA values
    """
    # draw the renderer
    fig.canvas.draw ( )
 
    # Get the RGBA buffer from the figure
    w,h = fig.canvas.get_width_height()
    buf = np.fromstring ( fig.canvas.tostring_argb(), dtype=np.uint8 )
    buf.shape = ( w, h, 4 )
 
    # canvas.tostring_argb give pixmap in ARGB mode. Roll the ALPHA channel to have it in RGBA mode
    buf = np.roll ( buf, 3, axis = 2 )
    return buf

def rgba2rgb(img, c=(0, 0, 0), path='foo.jpg', is_already_saved=False, if_load=True):
    if not is_already_saved:
        background = Image.new("RGB", img.size, c)
        background.paste(img, mask=img.split()[3]) # 3 is the alpha channel

        background.save(path, 'JPEG', quality=100)   
        is_already_saved = True
    if if_load:
        if is_already_saved:
            im = Image.open(path)
            return np.array(im)
        else:
            raise ValueError('No image to load.')