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

3 views (last 30 days)
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

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.
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.

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

R2021b

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