2 views (last 30 days)
Sahil khandelwal on 22 May 2022
Commented: DGM on 22 May 2022
th=t0+((max(inp(:))+min(inp(:)))./2);

DGM on 22 May 2022
Without any context, all anyone can give is an answer without context.
min(inp(:)) % this is the global minimum of inp
max(inp(:)) % this is the global maximum of inp
((max(inp(:)) + min(inp(:)))./2 % this is the average of the extrema of inp
So then this is the average of the extrema of inp, plus t0 -- whatever that means
th = t0 + ((max(inp(:)) + min(inp(:)))./2);

Image Analyst on 22 May 2022
The stuff between the parentheses is the midpoint of the histogram. My guess is that they have an image with a narrow histogram and they're trying to threshold it a certain number of gray levels above the mean or peak gray level. It might be better to use a triangle threshold, like the attached utility.
DGM on 22 May 2022

R2022a

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