How to find the index of first and last nonzero elements in each column?
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Rabia Zulfiqar
on 22 May 2020
Commented: Stephen23
on 25 Mar 2022
Hi,Is there any way to find the index of first and last nonzero element in a matrix?
I have this matrix
A=[0 0 0 0
0 0 0 0
1 0 4 8
2 0 5 9
3 1 6 7
0 2 7 0
0 0 0 0]
as the first non zero element in 1st column is 3 and and the last is 5. In the second column the the 12th element is the first non zero and the last one is 13th etc
I want to store these values in a matrix in which each row represents the index of first non zero element and the 2nd row shows the index of last non zero elements.
the answer matrix should be like this:
B=[3 12 17 24
5 13 21 26]
How can I do this??
Thankyou for time and consideration.I really appreaciate your help.Kindly guide me
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Accepted Answer
Image Analyst
on 22 May 2020
You could probably use
A = [0 0 0 0
0 0 0 0
1 0 4 8
2 0 5 9
3 1 6 7
0 2 7 0
0 0 0 0]
[rows, columns] = size(A)
B = zeros(2, columns)
for col = 1 : size(A, 2)
B(1, col) = find(A(:, col), 1, 'first');
B(2, col) = find(A(:, col), 1, 'last');
end
This gives
B =
3 5 3 3
5 6 6 5
which makes sense to me but I'm puzzled as to how you get
B=[3 12 17 24
5 13 21 26]
for your example. Can you explain?
4 Comments
Image Analyst
on 23 May 2020
Oh, you wanted the "linear index" rather than the row index within each column (as I had assumed). If you'd said "linear index" (the proper MATLAB lingo) I would have gotten it as :
A = [0 0 0 0
0 0 0 0
1 0 4 8
2 0 5 9
3 1 6 7
0 2 7 0
0 0 0 0]
[rows, columns] = size(A)
B = zeros(2, columns)
for col = 1 : size(A, 2)
B(1, col) = find(A(:, col), 1, 'first') + rows * (col - 1);
B(2, col) = find(A(:, col), 1, 'last') + rows * (col - 1);
end
B
but it looks like Stephen figured out, and it even works for higher dimension arrays, so just use his answer.
More Answers (4)
Stephen23
on 22 May 2020
Edited: Stephen23
on 22 May 2020
This works for any array, 2D, 3D, etc., and returns the requested linear indices:
>> A = [0,0,0,0;0,0,0,0;1,0,4,8;2,0,5,9;3,1,6,7;0,2,7,0;0,0,0,0]
A =
0 0 0 0
0 0 0 0
1 0 4 8
2 0 5 9
3 1 6 7
0 2 7 0
0 0 0 0
>> S = size(A);
>> S(1) = 2;
>> X = A~=0;
>> X = X & (cumsum(X,1)==1 | flipud(cumsum(flipud(X),1))==1);
>> B = reshape(find(X),S)
B =
3 12 17 24
5 13 20 26
5 Comments
Nicole Wan
on 25 Mar 2022
What if there are some columns before and after this data set that are columns of zero but the positions in the matrix need to be kept.
Stephen23
on 25 Mar 2022
A = [0,0,0,0;0,0,0,0;0,0,4,8;0,0,5,9;0,1,6,7;0,2,7,0;0,0,0,0]
X = A~=0;
X = X & (cumsum(X,1)==1 | flipud(cumsum(flipud(X),1))==1)
B = reshape(find(X),2,[])
Bruno Luong
on 25 Mar 2022
Let you do conversion to linear index
A=[0 0 0 0
0 0 0 0
1 0 4 8
2 0 5 9
3 1 6 7
0 2 7 0
0 0 0 0]
[r,c]=find(A);
first=accumarray(c(:),r(:),[size(A,2) 1], @min, 0)
last=accumarray(c(:),r(:),[size(A,2) 1], @max, 0)
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