Extracting values from array separated by zeros

Hi,

would like to ask if anybody has a hint on how to approach this situation. Basically, I have a big array of numbers (2000+), in which I have 12 sets of numbers that I am interested in, and the rest of the values are set to 0. What I would preferably like to do is to get each of these sets in a separate array, or at least a different column of a new array.

To show an example of what I mean.. array = 0 0 0 0 0 2 3 4 5 0 0 0 0 0 1 1 2 3 0 0 0 0 0 0 4 5 1 2 0 0 0 0 ....

to get final_array = 2 1 4

3 1 5

4 2 1

5 3 2

Thank you for any input.

Answer 1

[David Hill]

yourArray = [0 0 0 0 0 2 3 4 5 0 0 0 0 0 1 1 2 3 0 0 0 0 0 0 4 5 1 2 0 0 0 0];
newMatrix=reshape(yourArray(yourArray~=0),4,[]);

[Tereza Janatova]

I tried that, however, I get an error in the reshape function as the product of known dimension is not divisible into total number of elements..

[David Hill]

Need a cell array because the number lengths are different sizes.

yourArray = [0 0 0 0 0 2 3 4 5 0 0 0 0 0 1 1 2 3 0 0 0 0 0 0 4 5 1 2 0 0 0 0];
s=num2str(yourArray~=0);
s=s(s~=' ');
b=regexp(s,'1*');
e=regexp(s,'1*','end');
for k=1:length(b)
  yourCell{k}=yourArray(b(k):e(k));
end

Answer 2

[Image Analyst]

If you know for a fact that all sets of numbers are of the same length, you can use regionprops() in the Image Processing Toolbox:

array = [0 0 0 0 0 2 3 4 5 0 0 0 0 0 1 1 2 3 0 0 0 0 0 0 4 5 1 2 0 0 0 0];
props = regionprops(array ~= 0, array, 'PixelValues');
output = vertcat(props.PixelValues)'

output = 4×3

2 1 4 3 1 5 4 2 1 5 3 2

If they have different lengths then you'll have to put them into a cell array or else pad some of the columns with zeros or some other value:

array = [0 0 0 0 0 2 3 4 5 0 0 0 0 0 1 2 3 0 0 0 0 0 0 4 5 1 2 9 0 0 0 0];
props = regionprops(array ~= 0, array, 'PixelValues');
for k = 1 : length(props)
    ca{k} = props(k).PixelValues;
end
celldisp(ca)

 
ca{1} =
 
     2     3     4     5

 
 
ca{2} =
 
     1     2     3

 
 
ca{3} =
 
     4     5     1     2     9

 

Answer 3

[Stephen23]

V = [0,0,0,0,0,2,3,4,5,0,0,0,0,0,1,1,2,3,0,0,0,0,0,0,4,5,1,2,0,0,0,0];
D = diff([true,V==0,true]);
B = find(D<0);
E = find(D>0)-1;
C = arrayfun(@(b,e)V(b:e),B,E,'uni',0)

C = 1×3 cell array
    {[2 3 4 5]}    {[1 1 2 3]}    {[4 5 1 2]}