Local maxima in a matrix

10 vues (au cours des 30 derniers jours)
Ryan
Ryan le 29 Nov 2014
I need to locate the peaks and valleys of numbers given in a large matrix. I need to compare a value in the matrix with the 8 surrounding values to determine if it is either a peak (maximum) or a valley (minimum). The matrix that I am using is 100X100 and the coordinates of the value will then need to be displayed in a table. For example: [3 6 2 5; 4 9 1 5; 7 3 5 4] has a peak of 9 at (2,2) and a valley of 1 at (3,2). Any ideas would be greatly appreciated.

Réponse acceptée

Image Analyst
Image Analyst le 29 Nov 2014
If you have the IMage Processing Toolbox, you are really in luck because there are two functions meant for exactly this situation: imregionalmax() and imregionalmin(). Here is how to use them:
grayImage = [3 6 2 5; 4 9 1 5; 7 3 5 4]
regionalMaxima = imregionalmax(grayImage)
regionalMinima = imregionalmin(grayImage)
In the command window, observe the results:
grayImage =
3 6 2 5
4 9 1 5
7 3 5 4
regionalMaxima =
0 0 0 0
0 1 0 0
0 0 0 0
regionalMinima =
1 0 0 0
0 0 1 0
0 0 0 0
If you want rows and columns of the max or mins, just use find():
[rows, columns] = find(regionalMaxima)
rows =
2
columns =
2

Plus de réponses (1)

Azzi Abdelmalek
Azzi Abdelmalek le 29 Nov 2014
Modifié(e) : Azzi Abdelmalek le 29 Nov 2014
a=[3 6 2 5; 4 9 1 5; 7 3 5 4]
n=size(a);
[mina,idx1]=min(a(:))
[ii1,jj1]=ind2sub(n,idx1)
[maxa,idx2]=max(a(:))
[ii2,jj2]=ind2sub(n,idx2)
  1 commentaire
Mohammad Abouali
Mohammad Abouali le 29 Nov 2014
Modifié(e) : Mohammad Abouali le 29 Nov 2014
This is not the answer. This gives the global min/max. He is asking for local one.

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