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

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Raven Buckman
Raven Buckman on 11 Jan 2022
Commented: Ive J on 11 Jan 2022
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

Accepted Answer

Cris LaPierre
Cris LaPierre on 11 Jan 2022
Edited: Cris LaPierre on 11 Jan 2022
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.
  1 Comment
Raven Buckman
Raven Buckman on 11 Jan 2022
Thanks so much. I want it to keep looping throughout the entire matrix so this is quite helpful. Thank you.

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More Answers (1)

Kevin Holly
Kevin Holly on 11 Jan 2022
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
  1 Comment
Ive J
Ive J on 11 Jan 2022
The second approach is not efficient; this is way faster:
index = all(t ~= 0 & t ~= 1, 2);

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