Trying to index an array and extract its values with a for loop.

I have an table of probabilities and I want to index the table to determine if there is a probability in any of the three columns that is not 1 or 0, extract the row and put it into a new array. I'm fairly new to working with matlab so I'm sure there is a very simple way to do this but it is evading me. Any help would be great.

i = 1;
for scrs(i,:) ~= 1 || scrs(i,:) ~= 0
    unclassifiable = scrs(i,:);
    i = i+1;
    disp('Unclassifiable nanoparticles detected.')
end

Answer 1

[Cris LaPierre]

See MATLAB Onramp Ch 5: Indexing and Modifying Arrays.

You might also find Ch 12: Programming helpful since it covers for loops.

For loops start with a declaration of a counter, which is automatically incremented each loop, and repeats a fixed number of times. Note that this approach assumes scrs only has 1 column of data. If that is not true, you will need to modify this approach.

for i = 1:size(scrs,2)
    if scrs(i,:) ~= 1 || scrs(i,:) ~= 0
        unclassifiable(i,:) = scrs(i,:);
        disp('Unclassifiable nanoparticles detected.')
    end
end

What you have written looks more like a while loop to me, which continue to loop as long as the conditional expression is true. However, this stops looping the first time scrs is 1 or 0. Note that this approach assumes scrs only has 1 column of data. If that is not true, you will need to modify this approach.

i = 1;
while scrs(i,:) ~= 1 || scrs(i,:) ~= 0
    unclassifiable(i,:) = scrs(i,:);
    i = i+1;
    disp('Unclassifiable nanoparticles detected.')
end  

Not sure which one you want so try them both to understand the difference.

[Raven Buckman]

Thanks so much. I want it to keep looping throughout the entire matrix so this is quite helpful. Thank you.

Answer 2

[Kevin Holly]

Below are two possible ways.

If data is in table,

t=table;
t.column1 = [1 2 3 4 0 3 4 2 3 4 0 3 1]';
t.column2 = [8 7 9 0 3 9 2 1 2 3 1 0 3]';
t.column3 = [9 0 9 7 2 1 8 3 9 2 3 0 1]';
t

t = 13×3 table
    column1    column2    column3
    _______    _______    _______

       1          8          9   
       2          7          0   
       3          9          9   
       4          0          7   
       0          3          2   
       3          9          1   
       4          2          8   
       2          1          3   
       3          2          9   
       4          3          2   
       0          1          3   
       3          0          0   
       1          3          1   

t((t.column1==0),:)=[];
t((t.column2==0),:)=[];
t((t.column3==0),:)=[];
t((t.column1==1),:)=[];
t((t.column2==1),:)=[];
t((t.column3==1),:)=[];
t

t = 4×3 table
    column1    column2    column3
    _______    _______    _______

       3          9          9   
       4          2          8   
       3          2          9   
       4          3          2   

If data is in matrix,

t = [1 2 3 4 0 3 4 2 3 4 0 3 1;8 7 9 0 3 9 2 1 2 3 1 0 3;9 0 9 7 2 1 8 3 9 2 3 0 1]'

t = 13×3

1 8 9 2 7 0 3 9 9 4 0 7 0 3 2 3 9 1 4 2 8 2 1 3 3 2 9 4 3 2

index = (sum((t~=1).*(t~=0),2)==3)

index = 13×1 logical array
   0
   0
   1
   0
   0
   0
   1
   0
   1
   1

t(index,:)

ans = 4×3

3 9 9 4 2 8 3 2 9 4 3 2

[Ive J]

The second approach is not efficient; this is way faster:

index = all(t ~= 0 & t ~= 1, 2);