How to remove zeros from an array?

I want to remove zeroes from an array. The array has exactly one zero per row. For example: a = [1 4 0 3; 0 1 5 5; 1 0 8 1; 5 4 4 0; 0 1 5 2] Should be turned into a = [1 4 3; 1 5 5; 1 8 1; 5 4 4; 1 5 2] I have tried using the command a(a==0) = []; However, this turns the 2000x50 array into an 1x98000 array instead of an 2000x49 array like I want it. Any ideas?

Answer 1

[Beder]

To remove a single zero from each row of a matrix and rebuild the new matrix of nonzero entries, try the following code: a = [1 4 0 3; 0 1 5 5; 1 0 8 1; 5 4 4 0; 0 1 5 2] v = nonzeros(a'); newmat = reshape(v,3,5)'

[Elvis Somers]

Thanks!

[eloy garcia venegas]

why a' and not just a?

[Trevon McHansen]

@eloy garcia venegas If you give it a try in MATLAB you'll see that getting the appropriate sized output takes a bit of thinking.

Calling nonzeros on the matrix a will return a vector of elements. This is because MATLAB doesn't attempt to "naturally" resize the outputs any other way. (because MATLAB doesn't actually know how many zeros it will omit or what their relationships are to one another).

Since we know that there are exactly 1 '0' elements per row, we know that the output matrix size will be the size of the input matrix with one less column. So the size will go from 5,4 to 5,3.

When you get the vector from nonzeros, the values are considered column-by-column. But we're considering a row-wise size adjustment.

By transposing a in the nonzeros call, you're effectively telling it to treat rows as columns and vice versa, making it behave "as if it were a row-wise operation".

Then when you get your row-wise output, you can reshape to a matrix. Reshape is also a column-wise operation, it will take the vector v, and assign values column-first. So to get around this, Beder first reshaped directly into a 3,5 and then transposed the entire matrix.

To get a more visual take on this. Try to get the right output without using a'.

Answer 2

[saber kazemi]

If we do not know how much of the elements to submit after we remove the zero elements.

a = [is a big matrix]

v = nonzeros(a');

newmat = reshape(v,?,?)'

Any ideas?

[ytzhak goussha]

I have been working on this problme for the some time now,

so far the solution that i have reached works better if you know for certain that the number of zeros in each row is the same, if not, then it must be converted to a cell class.

a=matrix                    %a big matrix with unknown number of zeroes
[sz1,sz2]=size(a);         
new_sz2=zeros(1,sz1)        %This array will hold the size of each row after removing zeros
for i=1:sz1
    new_mat{i}= nonzeros(a(i,:));
    new_row_sz(i)=size(new_mat{i},1)
end
if range(new_row_sz) == 0
    new_mat=cell2mat(new_mat);
end

[Farshad Bolouri]

This is how I have solved this problem:

Z = find(~a);
[x,~] = ind2sub(size(a),Z);
a(x,:) = [];

This worked for my application. Please me know if it also works for you. It is possible to tweek it for different applications.

[Daniel Boateng]

It helped me. Thanks Farshad.

[Patrick Benz]

Both methods don't seem to work for me.

I have an 22762 x 20 double array which contains 19488 zeros. The problem is that I don't know how many zeros are in each row and the second problem is that they aren't equally distributed.

A (1:10)
4707	0	0	    5178	947	    0	0	    1417	0	0	0	    2506	0	0	0	    2974	2037	0	0	    3442
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
963	    0	0	    1433	4693	0	0	    5164	0	0	0	    6258	0	0	0	    6729	5787	0	0	    7200
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
4708	0	0	    5179	948	    0	0	    1418	0	0	0	    2509	0	0	0	    2977	2040	0	0	    3445

A (1287:1293)
0	    0	614	    1084	0	    0	1555	2023	0	0	4186	0	    0	0	7009	0	    0	    0	6538	7480
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	0	    0	    0	    0	0	    0	    0	0	0	    0	    0	0	0	    0	    0	    0	0	    0
0	    0	3893	3427	0	    0	4830	4361	0	0	2785	0	    0	0	5598	0	    0	    0	6069	5127

The uneven spacing for is to show the irregularities.

Is there any chance that I can remove the zeros from the matrix without "loosing" the form for the nonzero elements? so that it would be something like this?

A(1)
4707    5178    947     1417    2506    2974	2037	3442
963	    1433	4693	5164	6258	6729	5787	7200

with the method from ytzhak goussha it just skips the if term, because range(new_row_sz) = 8.

When I try to use cell2mat I get an error

Error using cat
Dimensions of arrays being concatenated are not consistent.

Error in cell2mat (line 75)
m{n} = cat(2,c{n,:});

But to be honest, I never worked with cells until now so I don't really know what the error means.

[Stephen23]

"Both methods don't seem to work for me."

Neither method is suitable for your data. Both methods are fragile and should be avoided.

The simplest approach is to detect the values that you want to remove and then use ANY with its dimension argument (or whatever logic you need for your task) to create a logical index vector of the rows that you need to remove. Then use that logical index to remove the rows.

For example:

A = randi([0,3],7,4)

A = 7×4

2 1 1 3 2 0 0 0 3 0 3 0 1 2 0 0 3 0 0 1 3 3 2 1 0 0 0 0

X = any(A==0,2)

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

A(X,:) = []

A = 2×4

2 1 1 3 3 3 2 1

[Patrick Benz]

Das scheint zumindest schon einmal für die Reihen zu funktionieren, welche nur 0 Elemente enthalten.

Bei den gemischten Reihen funktioniert es aber offensichtlich nicht, da nur die Reihen von A behalten werden, welche keine 0 enthalten.

Wenn ich den Code so auf meine Matrix anwende, bekomme ich einen 0 x 20 double als Ergebnis.

Wäre es einfacher, wenn ich wüsste, wie viele 0 pro Reihe in der Matrix enthalten sind?

Edit:

Auch wenn es sicherlich nicht das schnellste oder eleganteste Ergebnis ist, habe ich es mit geschachtelten Schleifen geschafft.

Elementset_Nodes = 22762 x 21 double

Node_Coord = 7765 x 1 double
    r=1;
    [LIA, LOCB]=ismember(Elementset_Nodes(:,2:end),Node_Coord(:,1));
    for i=1:length(LOCB)
        X=0;
        X=find(LOCB(i,:));
        if sum(X)~=0
            for j=1:length(X)
                new_Mat(r,1)=Elementset_Nodes(i,1);
                new_Mat(r,j+1)=LOCB(i,X(j));
            end
        else
            new_Mat(r,:)=0;
            new_Mat(r,:)=[];
        end
            [r, c]=find(new_Mat);
            r=max(r)+1;
    end
    Elementset_Nodes=horzcat(Elementset_Nodes(:,1),new_Mat);

Answer 3

[Mathieu PEYRE]

Hello,

Rather than remove the zeros of your matrix, you can create an other matrix with the non zeros rows in it.

EXEMPLE: A=[ 1 2 3; 0 0 0; 3 4 5; 0 1 0] and you want B=[1 2 3; 3 4 5; 0 1 0]

(If you want to remove all rows with at least one zero and obtain B=[1 2 3; 3 4 5], you can replace the "||" by "&&" in the "if" condition)

L=1;
for i=1:length(A) 
    if A(i,1)~=0 || A(i,2)~=0 || A(i,3)~=0
        B(L,:)=A(i,:);
        L=L+1;
    end
end