How to replace values of a logical index with values from another matrix if certain conditions are met?

I have a matrix labeled "A" [daily values] with the following values [10, 15, 30]

where 10 is a minimum temperature, 15 is an average temperature, and 30 is a maximum temperature

I also have a matrix labeled "B" [long term values] with the values [9, 12, 18; 11, 16 14; 14, 17, 19; 13, 14, 17] where the first column represents minimum temperatures, the second column represents average temperatures, and the third column represents maximum temperatures.

I am assigned the following task:

Determine if any of the minimum values in the daily term values array are less than any of the minimum values in the long term values array. Create a matrix that lists all the values in a column, that are less than the long term values. The second column should contain the day at which these values occurred. (each row of matrix A and B is a day. Day 1 = row 1).

I have tried everything, but I can not seem to accomplish this task.

[Walter Roberson]

Please show your attempts.

Hint: find()

[Andrew Padilla]

I decided to try to figure it out when I saw your hint because several of the attempts I tried included the find function. I ended up solving it!

I = find(LONGTERM(:,1) > DAILYVALUES(:,1));

C = [DAILYVALUES(Index), B];

C is a new matrix that, in column 1, displays the values where DAILY VALUES are less than LONG TERM values. The second column lists the days when these temperatures happened!

[Walter Roberson]

What is the difference between LONGTERM and B in that code?

Is DAILYVALUES the same as A?

Is it correct that DAILYVALUES has only the three entries? Or does it have three columns, one per day?

At the moment your code looks unlikely to be correct.

[Andrew Padilla]

I added an image for clarification. I will try to explain the variable names above.

B SHOULD BE REPLACED WITH I IN MY PREVIOUS RESPONSE sorry!

OVERALL are the long term values. It is basically matrix "B" which I defined above.

MA3Task1 are the daily term values. It is matrix "A" defined above.

So, ColdDays is a matrix where the first column are the values where overall is greater than the values in the first column from MA3Task1. In this case, overall is greater than the values in MA3Task1 column 1 zero times because the first column of overall has one row with a value of -80 which is not greater than any of the values in the first column of MA3Task1. Since the number of rows determines how many days where Overall is greater than MA3Task1, I used the size function to find the number of rows of "ColdDays" and used that to fill my placeholder in fprintf to determine the number of days where overall is greater. Also, the second row of "ColdDays" is the day at which the overall temperature was greater than the MA3Task1 temperatures occurred.

[Walter Roberson]

Okay, that looks better.

I would suggest, though, that when a problem is worded in terms of "less than" that it makes the most sense to use a < operator than to use a > operator. This would involve switching the two operands involved.

[Andrew Padilla]

Thank you for pointing that out. I will make sure to fix that! I also appreciate the hint! It helped out quite a lot.