用Python计算两个图像之间的绝对差之和的最快方法是什么？

Question

我正在尝试在使用Pillow和（可选）Numpy的Python 3应用程序中比较图像。 出于兼容性原因，我不打算使用其他外部非纯Python软件包。 我在Roseta Code中发现了这种基于枕头的算法，虽然可以达到我的目的，但需要一些时间：

from PIL import Image

def compare_images(img1, img2):
    """Compute percentage of difference between 2 JPEG images of same size
    (using the sum of absolute differences). Alternatively, compare two bitmaps
    as defined in basic bitmap storage. Useful for comparing two JPEG images
    saved with a different compression ratios.

    Adapted from:
    http://rosettacode.org/wiki/Percentage_difference_between_images#Python

    :param img1: an Image object
    :param img2: an Image object
    :return: A float with the percentage of difference, or None if images are
    not directly comparable.
    """

    # Don't compare if images are of different modes or different sizes.
    if (img1.mode != img2.mode) \
            or (img1.size != img2.size) \
            or (img1.getbands() != img2.getbands()):
        return None

    pairs = zip(img1.getdata(), img2.getdata())
    if len(img1.getbands()) == 1:
        # for gray-scale jpegs
        dif = sum(abs(p1 - p2) for p1, p2 in pairs)
    else:
        dif = sum(abs(c1 - c2) for p1, p2 in pairs for c1, c2 in zip(p1, p2))

    ncomponents = img1.size[0] * img1.size[1] * 3
    return (dif / 255.0 * 100) / ncomponents  # Difference (percentage)

尝试寻找替代方法时，我发现可以使用Numpy重写此函数：

import numpy as np    
from PIL import Image

def compare_images_np(img1, img2):
    if (img1.mode != img2.mode) \
            or (img1.size != img2.size) \
            or (img1.getbands() != img2.getbands()):
        return None

    dif = 0
    for band_index, band in enumerate(img1.getbands()):
        m1 = np.array([p[band_index] for p in img1.getdata()]).reshape(*img1.size)
        m2 = np.array([p[band_index] for p in img2.getdata()]).reshape(*img2.size)
        dif += np.sum(np.abs(m1-m2))

    ncomponents = img1.size[0] * img1.size[1] * 3
    return (dif / 255.0 * 100) / ncomponents  # Difference (percentage)

我原本希望处理速度有所提高，但实际上需要更长的时间。 除了基础知识之外，我还没有Numpy的经验，所以我想知道是否有任何方法可以使它更快，例如使用某种不暗示for循环的算法。 有任何想法吗？

Answer 1

我想我知道您要做什么。 我不了解我们两台机器的相对性能，因此也许您可以自己对其进行基准测试。

from PIL import Image
import numpy as np

# Load images, convert to RGB, then to numpy arrays and ravel into long, flat things
a=np.array(Image.open('a.png').convert('RGB')).ravel()
b=np.array(Image.open('b.png').convert('RGB')).ravel()

# Calculate the sum of the absolute differences divided by number of elements
MAE = np.sum(np.abs(np.subtract(a,b,dtype=np.float))) / a.shape[0]

唯一的“棘手”之处是将np.subtract()的结果类型强制为浮点数，以确保我可以存储负数。 可能值得在您的硬件上尝试使用dtype=np.int16来看看它是否更快。

---

基准测试的一种快速方法如下。 启动ipython ，然后输入以下内容：

from PIL import Image
import numpy as np

a=np.array(Image.open('a.png').convert('RGB')).ravel()
b=np.array(Image.open('b.png').convert('RGB')).ravel()

现在，您可以使用以下时间来计时我的代码：

%timeit np.sum(np.abs(np.subtract(a,b,dtype=np.float))) / a.shape[0]
6.72 µs ± 21.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

或者，您可以尝试这样的int16版本：

%timeit np.sum(np.abs(np.subtract(a,b,dtype=np.int16))) / a.shape[0]
6.43 µs ± 30.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

如果要计时代码，请粘贴函数，然后使用：

%timeit compare_images_pil(img1, img2)

Answer 2

仔细研究一下，我发现该存储库采用了另一种方法，该方法更多地基于Pillow本身，并且似乎给出了相似的结果。

from PIL import Image
from PIL import ImageChops, ImageStat


def compare_images_pil(img1, img2):
    '''Calculate the difference between two images of the same size
    by comparing channel values at the pixel level.
    `delete_diff_file`: removes the diff image after ratio found
    `diff_img_file`: filename to store diff image

    Adapted from Nicolas Hahn:
    https://github.com/nicolashahn/diffimg/blob/master/diffimg/__init__.py
    '''

    # Don't compare if images are of different modes or different sizes.
    if (img1.mode != img2.mode) \
            or (img1.size != img2.size) \
            or (img1.getbands() != img2.getbands()):
        return None

    # Generate diff image in memory.
    diff_img = ImageChops.difference(img1, img2)

    # Calculate difference as a ratio.
    stat = ImageStat.Stat(diff_img)

    # Can be [r,g,b] or [r,g,b,a].
    sum_channel_values = sum(stat.mean)
    max_all_channels = len(stat.mean) * 255
    diff_ratio = sum_channel_values / max_all_channels

    return diff_ratio * 100

对于我的测试图像样本，结果似乎是相同的（除了一些较小的float舍入错误），并且其运行速度比我上面的第一个版本快得多。