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Apply mean and standard deviation of an image to another image

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Efstathios Kontolatis
编辑： Adam 2016-10-11
Hi I have an image and I calculate the mean and standard deviation like this
meanint=mean(mean(tmpimg));
stdint=std(std(tmpimg));
Now I have another image and I try to make it have the same mean and standard deviation as the previous image. That's what I do
if (meanint2>meanint(1))
tmpimg = imsubtract(tmpimg,meanint2);
tmpimg = imadd(tmpimg,meanint);
tmpimg = imdivide(tmpimg,stdint2);
tmpimg = immultiply(tmpimg,stdint);
elseif (meanint2<meanint(1))
tmpimg = imadd(tmpimg,meanint);
tmpimg = imsubtract(tmpimg,meanint2);
tmpimg = immultiply(tmpimg,stdint);
tmpimg = imdivide(tmpimg,stdint2);
end
But I don't have the result I want. Does anyone knows why?

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采纳的回答

Guillaume
Guillaume 2016-10-11
编辑：Guillaume 2016-10-11
Really, this can be done with just one operation, with much less error accumulation and chance of truncation (if integer images):
meanreference = mean2(referenceimage);
stdreference = std2(referenceimage);
meantoscale = mean2(imagetoscale);
stdtoscale = std2(imagetoscale);
adjustedimage = imlincomb(stdreference / stdtoscale, imagetoscale, ...
meanreference - meantoscale * stdreference / stdtoscale);
If the image is double I wouldn't even bother with the imlincomb:
adjustedimage = meanreference + (imagetoscale - meantoscale) * stdreference / stdtoscale;
  6 个评论
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Adam
Adam 2016-10-11
编辑：Adam 2016-10-11
The problem of standard deviation not being the same is that you calculate it wrongly in the code in the question. Mean is a separable operation that you can do as you are doing (though there is no need to do so), std is not.
std2 as Guillaume uses works correctly or as I usually do more generically, just
std( tmpimg(:) )
since it is a statistical operation and the structure of the data in a 2d matrix rather than 1d is not relevant.

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更多回答（1 个）

Adam
Adam 2016-10-11
tmpimg = imsubtract(tmpimg,meanint2);
tmpimg = imdivide(tmpimg,stdint2);
tmpimg = immultiply(tmpimg,stdint);
tmpimg = imadd(tmpimg,meanint);
seems a more appropriate ordering and gives me the same mean.
  2 个评论
Efstathios Kontolatis
Efstathios Kontolatis 2016-10-11
编辑：Efstathios Kontolatis 2016-10-11
For which of the two if? Also, the standard deviation is not the same.
Adam
Adam 2016-10-11
I don't see the need for two cases, why is the code different depending whether the mean is greater or smaller?

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