比较2张图片并找出％差异 [英] Compare 2 images and find % difference

问题描述

我想比较两个图像并知道它们之间的百分比差异。我在树莓派和python语言上使用raspbian。我已经找到了PIL和magickimage，但是使用magick image我找不到它的功能，而使用PIL我却得到了奇怪的结果。

I want to compare two images and know the % difference between them. I am using raspbian on raspberry pi and python language. I have found PIL and magickimage, but with magick image I can't find a function for this and with PIL I have strange results.

对于Pil，我使用以下代码：

For Pil i use this code :

h1 = image1.histogram()
h2 = image2.histogram()
rms = math.sqrt(reduce(operator.add,map(lambda a,b: (a-b)**2, h1, h2))/len(h1))

当我用0.5秒的间隔拍摄两张照片（无差异）时，我得到以下结果：rms = 4743.766 ....如果我在两张照片之间移动，则有rms： 4699.288 .....
因此，当我移动两个图像时，这并没有区别：/

When i take two pics ( no difference ) with 0.5 seconds of intervall i have this results : rms = 4743.766.... If i move during between the two pics i have rms : 4699.288..... So it's does not make the difference between the two " sames " images and when i move :/

推荐答案

使用 compare ，它是ImageMagick的一部分。像这样：

Use compare which is part of ImageMagick. Like this:

compare -metric AE image1.png image2.png null:

AE 根据像素差的计数给出绝对误差。您还可以使用 MAE （平均绝对错误）或 PAE （峰值绝对错误）或 RMSE （均方根误差）。您还可以添加模糊因子，以允许像素值略有不同，如下所示：

The AE gives the absolute error, in terms of a count of the number of pixels difference. You can also use MAE (mean absolute error), or PAE (peak absolute error) or RMSE (root mean square error). You can also add a fuzz factor to allow slight differences in pixel values like this:

compare -fuzz 10% -metric AE image1.png image2.png null:

如果您想要答案在shell变量中，例如 ndiff ，则可以执行以下操作：

If you want the answer in a shell variable, say ndiff, you can do this:

ndiff=`compare -fuzz 10% -metric AE image1.png image2.png null: `
echo $ndiff

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