通过计算均方误差，numpy有一个简单而快速的解决方案:

before = np.array(get_picture())
while True:
    now = np.array(get_picture())
    MSE = np.mean((now - before)**2)

    if  MSE > threshold:
        break

    before = now

我特别要解决的问题是如何计算它们是否“足够不同”。我假设你能弄清楚如何一个一个地减去像素。

首先，我将取一堆没有任何变化的图像，并找出任何像素变化的最大量，仅仅是因为捕获的变化、成像系统中的噪声、JPEG压缩工件和照明的每时每刻的变化。也许你会发现，即使没有任何移动，1或2位的差异也是可以预期的。

对于“真实”测试，你需要一个这样的标准:

如果最多P个像素的差异不超过E，则相同。

所以，如果E = 0.02, P = 1000，这可能意味着(大约)如果任何单个像素改变超过5个单位(假设8位图像)，或者如果超过1000个像素有任何错误，这将是“不同的”。

这主要是一种很好的“分类”技术，用于快速识别足够接近而不需要进一步检查的图像。“失败”的图像可能更多的是一种更复杂/昂贵的技术，例如，如果相机抖动，或者对光线变化更健壮，就不会产生假阳性。

I run an open source project, OpenImageIO, that contains a utility called "idiff" that compares differences with thresholds like this (even more elaborate, actually). Even if you don't want to use this software, you may want to look at the source to see how we did it. It's used commercially quite a bit and this thresholding technique was developed so that we could have a test suite for rendering and image processing software, with "reference images" that might have small differences from platform-to-platform or as we made minor tweaks to tha algorithms, so we wanted a "match within tolerance" operation.

如果现在回复太晚，我很抱歉，但因为我一直在做类似的事情，我想我可以在某种程度上做出贡献。

也许在OpenCV中你可以使用模板匹配。假设你用的是摄像头

简化图像(可能是阈值?) 应用模板匹配和检查max_val与minMaxLoc

提示:max_val(或min_val取决于所使用的方法)将为您提供数字，较大的数字。为了获得百分比上的差异，使用与相同图像匹配的模板—结果将是100%。

举例的伪代码:

previous_screenshot = ...
current_screenshot = ...

# simplify both images somehow

# get the 100% corresponding value
res = matchTemplate(previous_screenshot, previous_screenshot, TM_CCOEFF)
_, hundred_p_val, _, _ = minMaxLoc(res)

# hundred_p_val is now the 100%

res = matchTemplate(previous_screenshot, current_screenshot, TM_CCOEFF)
_, max_val, _, _ = minMaxLoc(res)

difference_percentage = max_val / hundred_p_val

# the tolerance is now up to you

希望能有所帮助。

使用SSIM测量两幅图像之间的结构相似指数度量。

两种流行且相对简单的方法是:(a)已经提出的欧几里得距离，或(b)标准化互相关。与简单的互相关相比，归一化互相关对光照变化的影响明显更强。维基百科给出了一个标准化互相关的公式。更复杂的方法也存在，但它们需要更多的工作。

使用numpy-like语法，

dist_euclidean = sqrt(sum((i1 - i2)^2)) / i1.size

dist_manhattan = sum(abs(i1 - i2)) / i1.size

dist_ncc = sum( (i1 - mean(i1)) * (i2 - mean(i2)) ) / (
  (i1.size - 1) * stdev(i1) * stdev(i2) )

假设i1和i2为二维灰度图像阵列。

I have been having a lot of luck with jpg images taken with the same camera on a tripod by (1) simplifying greatly (like going from 3000 pixels wide to 100 pixels wide or even fewer) (2) flattening each jpg array into a single vector (3) pairwise correlating sequential images with a simple correlate algorithm to get correlation coefficient (4) squaring correlation coefficient to get r-square (i.e fraction of variability in one image explained by variation in the next) (5) generally in my application if r-square < 0.9, I say the two images are different and something happened in between.

这是强大的和快速的在我的实现(Mathematica 7)

这是值得玩转的部分，你感兴趣的图像，并通过裁剪所有的图像到那个小区域，否则一个远离相机但重要的变化将被错过。

我不知道如何使用Python，但我确信它也有相关性，不是吗?