Re: Seeking Clarification of Procedure for Averaging Images


Rick Gordon
 

Based on the maxim that "the median is affected by outliers and skewed data than the mean, and is usually the preferred measure of central tendency when the distribution is not symmetrical" <https://www.abs.gov.au/websitedbs/a3121120.nsf/home/statistical+language+-+measures+of+central+tendency#:~:text=In%20a%20distribution%20with%20an,value%20is%20the%20middle%20value.&text=Advantage%20of%20the%20median%3A,the%20distribution%20is%20not%20symmetrical.>, that brings up an interesting note: as to whether doing this with Median as opposed to Mean would mitigate the damage of "one bad apple spoiling the apple cart."

Rick Gordon

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On July 8, 2020 at 2:54:31 PM [-0700], John Gillespie wrote in an email entitled "Re: [colortheory] Seeking Clarification of Procedure for Averaging Images":
Once loaded (which may take a while if you choose a lot of images) select all the layers and then select Layers->Smart Objects->Convert To Smart Object (there is an option to do this on the file selection dialog but it doesn't always seem to work). Then select Layers->Smart Objects->Stack Mode->Mean.
You can also have fun with the other stack modes.
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RICK GORDON
EMERALD VALLEY GRAPHICS AND CONSULTING
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WWW: http://www.shelterpub.com

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