How to compute the mean of two disjoint region ?

1 Ansicht (letzte 30 Tage)
Makrim
Makrim am 1 Mai 2015
Kommentiert: Image Analyst am 2 Mai 2015
Suppose I have to fragment of an image J : J_out_1 et J_out_2.
J_out_1 = J(1:h,startj:i);
J_out_2 = J(1:h,k:endj);
I would like to compute the mean of the union of those two regions , is it possible ?
m_out = mean2(J_out_1 union J_out_2);
Thank you in advance

Akzeptierte Antwort

Guillaume
Guillaume am 1 Mai 2015
Bearbeitet: Guillaume am 1 Mai 2015
m_out = mean([J_out_1(:); J_out_2(:)])
would be one way to do it assuming the image has only one colour channel. If they are RGB images:
m_out = mean([reshape(J_out_1, 1, [], 3), reshape(J_out_2, 1, [], 3)])
Note that if the two regions are the same size, you could just concatenate them without any reshaping (by colon or reshape).
  1 Kommentar
Makrim
Makrim am 2 Mai 2015
excellent, that's what I am looking for.finally I have done it as follow :
m_out = mean2([J_out_1 , J_out_2])

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Image Analyst
Image Analyst am 1 Mai 2015
Why not just take the weighted mean of the two?
numerator = numel(J_out_1) * mean2(J_out_1) + numel(J_out_2) * mean2(J_out_2)
denominator = numel(J_out_1) + numel(J_out_2)
m_out = numerator / denominator
If you want, you could make a binary image and use that as a mask to extract all the pixels in just the two regions:
binaryImage = false(size(J));
binaryImage(1:h,startj:i) = true;
binaryImage(1:h,k:endj) = true;
m_out = mean(J(binaryImage))
  2 Kommentare
Makrim
Makrim am 2 Mai 2015
that's another way, Yes , thank you. by the way the idea of using Logical is also brilliant.
Image Analyst
Image Analyst am 2 Mai 2015
You're welcome. Those ways will also work even if the two subimages don't have the same number of rows. So "h" could be different for each image and they would still work.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Image Processing Toolbox finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by