How to preallocate when final numbers of rows in unknown

Hello, I have a for loop that has 3 options like this:

The issue is that the path of the for loop changes depending on user inputs (if they enter 2, then e = 3x and so on). Thing is, I don't know how many rows I'll end up with for each variable as it can change with each cycle of the inputs

a = 1; b = 1; c = 1;
x = 0:1:10
n=5
%preallocating
e = zeros(?,10)
p = zeros(?,10)
o = zeros(?,10)
for i = 1:n
    if x == 2
        e(a,:) = 3x
    elseif y == 15
        p(b,:) = 100/x
    else 
        o(c,:) = 2+x
    end
end 

Answer 1

[Ameer Hamza]

If you don't already know the size of your matrix, then you can make a guess about what will be the maximum size, and you allocate according to that. If fewer numbers of rows are used, then you can delete the extra rows or columns at the end. If more are used, then you can again grow your matrix by a suitable factor until it gets filled again and continue this process.

It can be a bit complicated to implement it in your code; therefore, as a workaround, you can use this FEX submission by John: https://www.mathworks.com/matlabcentral/fileexchange/8334-incremental-growth-of-an-array-revisited. It automatically increases and decreases the size of the array according to your requirement, and you don't need to worry about specifying the maximum matrix size. You can read the description and example on the documentation page of this FEX submission.

[Austin Hernandez]

Hmm, I tried the guessing method but it didn't work out correctly because I don't know to tell MATLAB to "automatically" delete any row containing 0s. for example if I thought that there would be 5 loops, and I guess that there will be 3 rows of option 1, then I could do e = zeros(n-2,10), correct? If I do this, I will end up with a row of 0s if option 1 was only triggered twice. I am trying to take data and make automatic graphs from them, and my system doesn't work if there are a row of 0s in a matrix.

[Walter Roberson]

you keep track of how many the user entered and then e = e(1:n,:) to reduce down to what was used

[Ameer Hamza]

You can create variable 'counter', which get increment by 1 every time the option1 get called. At the end of the loop, you can use the counter to delete extra rows. For example, you created

e = zeros(5,5)

If the option1 runs 3 times, then the counter will be 3. You can delete extra rows like this

e(counter+1:end, :) = []

[Austin Hernandez]

I think this might be the solution!! I am new to matlab so I am still unfamiliar with the syntax. in the statement

e(counter+1:end, : = [ ], this means that for matrix e, take the row after the counter to the end of that row, and then take all columns and delete them with [ ] ?

[Ameer Hamza]

It means that delete all the row numbers starting from counter+1 to the end of the matrix. All the columns of these rows are deleted.

[Austin Hernandez]

Thank you! This fix actually worked for my purposes.

[Ameer Hamza]

I am glad to be of help.