How to find the index of first and last nonzero elements in each column?

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

Answer 1

[Image Analyst]

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?

[Rabia Zulfiqar]

Thankyou so much for your answer dear,I got this beacuse the MAtLab counts the elements column by column for example in first column there are 7 elements but when it goes to column to its first element will be the 8th one according to MAtlab as far as I have understood but may be I am wrong.

[Rabia Zulfiqar]

@Image analyst how can I do the same thing in 3D which you have explained in 2D matrix?

[Rabia Zulfiqar]

Like here I have a matrix of 7x4 but if I have a matrix of 7x4x20 then how can I do this?

[Image Analyst]

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.

Answer 2

[Stephen23]

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

[Stephen23]

Rabia Zulfiqar's "Answer" moved here:

Stephen Cobeldick Thanks a lot it works perfectly but can you please explain this to me?

[Stephen23]

Lets look at a simple example using one vector (to save space I will use a row vector, but with your data this applies to each column). How can we identify the first non-zero? Start by identifying all non-zeros:

>> V = [0,0,1,2,3,0,0]
V =
  0  0  1  2  3  0  0
>> X = V~=0 % all nonzeros.
X =
  0  0  1  1  1  0  0

Note that only the first non-zero has a cummulative sum of 1:

>> cumsum(X)
ans =
  0  0  1  2  3  3  3
>> X & cumsum(X)==1
ans =
  0  0  1  0  0  0  0

We must include the X & ... because if there is only one non-zero in that vector then all of the trailing zeros will also have a cummulative sum = 1, but we can simply exclude them because they are zero.

To get the last non-zero we do exactly the same, but first flip the columns upside down... and then afterwards flip the result back the right way. That gives the last non-zero in each column.

That gives a simple logical array where each true element is either the first or last non-zero in each column. You can use find on that logical array to trivially return the linear indices of those elements.

[Rabia Zulfiqar]

Thankyou so much for your explanation:)

[Nicole Wan]

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]

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]

A = 7×4

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)

X = 7×4 logical array
   0   0   0   0
   0   0   0   0
   0   0   1   1
   0   0   0   0
   0   1   0   1
   0   1   1   0
   0   0   0   0

B = reshape(find(X),2,[])

B = 2×3

12 17 24 13 20 26

Answer 3

[Matt J]

A=[0    0     0      0
     0     0     0      0
     1     0     4      0
     2     0     5      0
     3     1      6     0
     0      2      7     0
     0      0      0      0];

a=logical(A);
[~,first]=max(a,[],1,'linear');
[~,last]=min( cummax(a,1,'rev'), [],1,'linear');

B=[first;last-1]

B = 2×4

3 12 17 22 5 13 20 21

B(:,~any(A,1))=nan

B = 2×4

3 12 17 NaN 5 13 20 NaN

[Bruno Luong]

I believe this method fails if a column contains only 0s.

[Matt J]

It's easily tweaked to accomodate that case (I've now done so).

Answer 4

[Bruno Luong]

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]

A = 7×4

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)

first = 4×1

3 5 3 3

last=accumarray(c(:),r(:),[size(A,2) 1], @max, 0)

last = 4×1

5 6 6 5

Answer 5

[Bruno Luong]

Some C-mex solution for speed:

https://fr.mathworks.com/matlabcentral/fileexchange/24641-vectorized-find-with-first-option?s_tid=srchtitle