Detecting samples greater than a threshold and grouping.

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Anum Ahmed Pirkani
Anum Ahmed Pirkani 2020 年 6 月 29 日
コメント済み: Anum Ahmed Pirkani 2020 年 6 月 29 日
Hi all,
Lets say we have a vector: A = [1 2 3 5 2 3 5 2 5 3 10 8 9 11 14 1 4 3 3 4 5 3 2 3 4 5 6 4 2 2 3 4 5 6 6 2 2 3 4 11 23 12 56 23 12 34 12 1 3 4 1 2 3 4 5 ];
In the vector, I have applied a threshold of A>5 for detection of any samples greater than 5. So I detect all the bold samples.
However, as a second step, I want to group these samples. As in, first bold set goes to group 1 and second bold set goes to group 2 (All the consecutive samples goes to same group). How can I do it in Matlab in a way that their original indices in vector A are preserved?
Kind Regards
Anum
  2 件のコメント
the cyclist
the cyclist 2020 年 6 月 29 日
What specifically would you want the output to be, for that input vector A? "Goes to a group" is not enough information to design an algorithm.
Anum Ahmed Pirkani
Anum Ahmed Pirkani 2020 年 6 月 29 日
It can be something like this
Group1val = [10 8 9 11 14]; %group 1 values greater than threshold
Group1ind = [11 12 13 14 15]; %group 1 indices with values greater than threshold
Group2val = [11 23 12 56 23 12 34 12]; %group 2 values greater than threshold
Group2ind = [40 41 42 43 44 45 46 47]; %group 2 indices with values greater than threshold
Indices can be obtained using the find command i.e.
Indices = find(A>5);
But as a next step, I want to seperate them into (in this case) two groups. This is decided if there are 5 values less than the threshold between the groups (in this case there are 24 values less than the threshold between these two groups).

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Image Analyst
Image Analyst 2020 年 6 月 29 日
What you want is bwlabel(), in the Image Processing Toolbox:
A = [1 2 3 5 2 3 5 2 5 3 10 8 9 11 14 1 4 3 3 4 5 3 2 3 4 5 6 4 2 2 3 4 5 6 6 2 2 3 4 11 23 12 56 23 12 34 12 1 3 4 1 2 3 4 5 ];
[groups, numGroups] = bwlabel(A > 5)
groups =
Columns 1 through 27
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 2
Columns 28 through 54
0 0 0 0 0 0 3 3 0 0 0 0 4 4 4 4 4 4 4 4 0 0 0 0 0 0 0
Column 55
0
numGroups =
4
  1 件のコメント
Anum Ahmed Pirkani
Anum Ahmed Pirkani 2020 年 6 月 29 日
Oh brialliant. Thanks a lot... This is exactly what I needed.

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