Creating a new column in a table using if condition and equation

Hi I have a table with daily data for a year: 365 rows x 4 columns. I need to create a new column which is defined by a function that I will show below and takes the data from other columns in the table.

For example lets say the columns of the table are labelled: A,B,C,D and the new column would be E

If D<1.2 then E=1

If D>6.5 then E=0.23

anything else E=1.18exp(-0.18*D)

This is what I have so far but the new column is only doing the last calculation and isnt following my limit values.

T = readtable('Data.csv');

if T.D <1.2
    T.E = 1.0;

elseif T.D >6.5
    T.E = 0.23;

else T.E = 1.18*exp(-0.18*T.D);
end 
disp(T)

Answer 1

[Stef1998]

I actually was able to find a solution for this issue for anyone else with similar problems:

unless there is another solution I missed this is how I was able to solve it-

if statements won't do the entire trick, indexing is required

T{:,'E'} = 1.18*exp(-0.18*T{:,'D'});

idx = T{:,'D'} < 1.2;
T{idx,'E'} = 1.0;

idx2 = T{:,'D'} > 6.5;
T{idx2,'E'} = 0.23;

[Peter Perkins]

That's correct, but this is a bit simpler and easy to read:

T.E = 1.18*exp(-0.18*T.D);
idx = T.D < 1.2;
T.E(idx) = 1.0;
idx2 = T.D> 6.5;
T.E(idx2) = 0.23;

To work with one variable, think dot. To work with more than one var, think braces.

Also: you should think of the things in tables as "variables", not "columns". This will save you a lot of confusion when you get to the point of having variables in a table that themselves have multiple columns.

[Stef1998]

Thank you Peter! When I try to repeat the steps above where 2 functions would be necessary for the new variable I run into an error that the size of the left and right do not match, would you know where I am making a mistake for this? I tried 3 different approaches with no success, For instance:

%Method 1
T.F = 1000 * (2834.1 - 0.29 * T.B - 0.004 * T.B.^(2.0));
idx3 = T.B >= 0;
T.F(idx3) = 1000 * (2501 - 2.36 * T.B);

%Method 2
idx3 = T.B < 0;
T.F(idx3) = 1000 * (2834.1 - 0.29 * T.B - 0.004 * T.B.^(2.0));
idx4 = T.B >= 0;
T.F(idx4) = 1000 * (2501 - 2.36 * T.B);

%Method 3
idx3 = T{:,'B'} < 0;
T{idx3,'F'} = 1000 * (2834.1 - 0.29 * T{:,'B'} - 0.004 * T{:,'B'}.^(2.0));
idx4 = T{:,'B'} >= 0;
T{idx4,'F'} = 1000 * (2501 - 2.36 * T{:,'B'});

[Peter Perkins]

I would think you would want

T.F(idx3) = 1000 * (2501 - 2.36 * T.B(idx3))

If T.F and T.B are in one table, they have the same size, but T.F(idx3) is shorter. You need to assign from the same subset of T.B, I think.