Hello Blue,
As per my understanding, you want to create a "category" vector such that:
a) If there is single combination of “set” and “sp”, then the “category” should be 0
b) If there are more than one combination, then –
- "category" corresponding to the largest "mean" will have value 3.
- "category" corresponding to smallest "mean" will have value 1.
- "category" corresponding to all the "mean" in between will have value 2.
As an example code, you can check this:
set = [1, 1, 1, 2, 2, 3, 3, 1, 1]
sp = [23, 23, 34, 45, 55, 22, 22, 23, 23]
mean = [100, 80, 60, 70, 80, 50, 20, 50, 40]
then the "category" should be -
category = [3, 2, 0, 0, 0, 3, 1, 2, 1]
I would suggest you perform the following steps to create "category" vector:
- Find all the indices for the current combination of "set" and "sp".
- Get the "mean" values for current combination.
- Assign values to “category” vector for given indices according to the rules.
Here is an example showing how you can do it:
% Original table
set = [1, 1, 1, 2, 2, 3, 3, 1, 1]';
sp = [23, 23, 34, 45, 55, 22, 22, 23, 23]';
mean = [100, 80, 60, 70, 80, 50, 20, 50, 40]';
T = table(set, sp, mean);
% Group the table by set and sp
groupedTable = groupsummary(T, {'set', 'sp'}, 'mean');
% Initialize the category array
category = zeros(height(T), 1);
% Assign categories based on the mean values
for i = 1:height(groupedTable)
% Find the indices of the rows with the current combination of set and sp
indices = find(T.set == groupedTable.set(i) & T.sp == groupedTable.sp(i));
% Get the mean values for the current combination
currentMeans = T.mean(indices);
% Assign categories based on the mean values
if numel(currentMeans) == 1
category(indices) = 0;
else
[~, maxIndex] = max(currentMeans);
[~, minIndex] = min(currentMeans);
category(indices(maxIndex)) = 3;
category(indices(minIndex)) = 1;
category(indices(setdiff(1:numel(currentMeans), [maxIndex, minIndex]))) = 2;
end
end
% Display the desired answer
category'
I hope this resolves the issue you are facing.
