Check whether values of a certain column of a matrix is between two values

%{

I have a BUS matrix with 6 rows, column 12 is the max value and column 13 is the minimum value.

I want to check if column 8 values are between column 12 and 13 values.

Now, I know that the answer of volmax are all greater than 0 but some values of volmin are less than 0.

For some reason, my code display Yes which means that volmax and volmin are greater than or equal to zero.

But I know this is not true. My guess is that my code is comparing only the first row and it ignore the rest of the rows

Can you help me to figure this out?

}%

volmax= BUS(:,12)- BUS(:,8) % we are good if volmax >=0

volmin= BUS(:,8)- BUS(:,13) % we are good if volmin >=0

if volmax<0 | volmin <0

disp('NO')

else

disp('YES')

end

end

Answer 1

[the cyclist]

OK, still trying to understand everything, but I think this does what you want:

% Sample input
BUS(:, 8) = [2;2;2];
BUS(:,12) = [3;3;3];
BUS(:,13) = [1;1;1];

volmax = BUS(:,12)- BUS(:,8); % we are good if volmax >=0
volmin = BUS(:,8)- BUS(:,13); % we are good if volmin >=0

if any((volmin<0)|(volmax<0))
    disp('NO')
else
    disp('YES')
end

Change the inputs around, and see if it is correct. If this is NOT correct, then please

• give an example of inputs that give INCORRECT output, and explain why

[Steven Shaaya]

Thank you very much, yes it did work.

Can you help me with something else, it is related to this question

Now in my if statement im using a matrix with 6 rows, is there away that I can use | with another matrix that has only 3 rows.

for example

if any((volmin<0)|(volmax<0) |(H<0)) %where H is a matrix with 3 rows

%where volmin and/or volmax has 6 rows

disp('NO')

else

disp('YES')

end

% it gives the following error, "Matrix dimensions must agree."

[Steven Shaaya]

Thank you again. I figured out the answer to my question which is adding zeros to the matrix in order to make them of the same size.

If you have better method, I would like to hear from you.

Answer 2

[the cyclist]

I think you intended

volmax<0 & volmin <0

rather than

volmax<0 | volmin <0

because you want to meet both conditions ("AND"), not just either condition ("OR").

Also, be aware that for vector input V, you will only enter the if statement

if V
    ...
end

if all elements of the vector V are true. (It is not some kind of implicit loop that works down the vector.)

Note that the expression

volmax<0 | volmin <0

(not using the if structure at all) will be a logical vector, with TRUE/FALSE entries corresponding to each row.

[Steven Shaaya]

I have to use | logic because if one of them is not meet then I need to display "no", I was not clear in my original post.

%I made this example to see how things works, and everthing worked correctly

xmax = [9 9 9;9 9 9;9 9 9];

xmin = [6 6 6;6 6 6;6 6 6];

ymax = [8 8 8;8 8 8;8 8 8];

ymin = [2 2 2;2 2 2;2 2 2];

for y=1:5

ymin = ymin +1;

max=xmax(:,3) - ymax(:,3)

min=ymin(:,3) - xmin(:,3)

if max<0 | min<0

disp('no')

else

disp('okay')

end

end

%%%%%%Matlab display

max =

1

1

1

min =

-3

-3

-3

no

max =

1

1

1

min =

-2

-2

-2

no

max =

1

1

1

min =

-1

-1

-1

no

max =

1

1

1

min =

0

0

0

okay

max =

1

1

1

min =

1

1

1

okay

%%Now I still can not figure out exactly what do you mean by the second part

%%why is it working down the loop for this example but not in my original code

[Steven Shaaya]

Are you saying my original code is reading only the first row because if that is the case then when Matlab displays YES it is correct because on the first 3 rows all the values are equal or greater than zero.

Can you tell me what should I do exactly? Thanks

[the cyclist]

I misunderstood your logic, and I see why you chose | now. Sorry for the confusion.

In your original code, it is not just checking the first element; it is checking whether ALL the elements of the vector meet the condition. Here is a simple example:

if [true true true]
    disp('All elements are true')
else
    disp('At least one element is false')    
end

All elements are true

if [true false true]
    disp('All elements are true')
else
    disp('At least one element is false')    
end

At least one element is false

The reason your second code example works is that you are using a for loop, checking one value at a time (not trying to check the entire vector at once).

Question: Suppose 2 rows do not meet the condition, and 3 rows do. What do you want for output? Should the output be 5 "YES" / "NO" values (one per row), or do you just want a single "YES" / "NO", based on whether all rows are ok?

[Steven Shaaya]

I want just a single "YES" / "NO", based on whether all rows are ok, all values of column 8 should be between the values of columns 12 & 13.

So based on your assumption, it should display 'no'.

For example, I can try

if volmax(1:6,:) <0 | volmax(2:6,:) <0 |volmax(3:6,:) ...........ect

but I do not see this efficient because I will need about 12 OR only for this example and my next example will have about 30 OR

that why I am trying to find another method

Thank you