This code is suppose to loop through an excel file and tell if each vaule in the second column is either impaired or not. with 1 being impaired and 0 being fine. However it is only looking at the first value instead of looping through all of them.

clear,clc  
num = xlsread('Hwk1_LDL.xlsx');  
m = num  
t = m(:,2)  
for i = 2:50  
    if t >= 130  
        a = 1  
    else   
        a = 0  
    end  
end

These are the instructions for the program. Can someone tell me what is missing so that it will loop through the whole column?

Make a program that:

1. o Automatically imports the excel file from the from the MatLab folder to a matrix called ‘LDL’
2. o Loops through each of the samples in LDL and checks to see if it is classified as impaired or not.
3. o Using column 3 in LDL (a new column) assign a value of 0 for acceptable LDL and 1 for high LDL.

[Image Analyst]

What happened to the question? It's been deleted!

[Rena Berman]

(Answers Dev) Restored edit

Answer 1

[Geoff Hayes]

Courtney - if you are iterating over each element of the t (which is the second column of your input data) then you need to access each element using indexing (see array indexing for more details). For example,

 for k=1:length(t)
    if t(k) >= 130
       % do something
    else
       % do something else
   end
 end

We iterate over each element of t and access the element using k as t(k).

Answer 2

[Image Analyst]

It's a very poorly worded problem - lots of ambiguous/inconsistent wording. Anyway, instead of

num = xlsread('Hwk1_LDL.xlsx');

you need

LDL = xlsread('Hwk1_LDL.xlsx');

because that's what the instructions directed you to do.

Then you need to loop over all elements in column 2 to see if they're "impaired" but I don't know what that means but I guess it means they're more than 130. You can do it vectorized but because is said to do it via a loop, you must use a for loop. And assuming "not impaired" is "acceptable" and "impaired" is high, then you can assign the third column of LDL.

col2 = LDL(:, 2); % Extract column 2 for convenience
for row = 1 : length(col2)
    if col2(row) > 130
        % This element is impaired.
        LDL(row, 3) = 1;
    else
        % Acceptable
        LDL(row, 3) = 0;
    end
end