Calculate average between min and max numbers in column 3 depending on values in column 2, for each value in column 1

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Neuro 2022-1-19

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https://ww2.mathworks.cn/matlabcentral/answers/1632230-calculate-average-between-min-and-max-numbers-in-column-3-depending-on-values-in-column-2-for-each

编辑： Voss 2022-1-20

Hello. I was wondering if someone could help me with this:

Let's say you have a 2000 x 3 matrix, where all the elements are integers. In column 1, the values can be between 1 and 16. How can we select one element in column 1 (e.g. 7), then for those rows only, select the rows with the highest 5 values in column 2, and for these rows only, select their elements in column 3 and calculate the average between the min and max values (output).

Thanks!

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Voss 2022-1-19

"select the rows with the highest 5 values in column 2"

Does this mean, "select the 5 rows with the highest value in column 2"?

Or does it mean, "select all rows whose value in column 2 is among the 5 highest unique values in column 2"?

(It's understood that we're talking about only those rows that match the selected element in column 1 already.)

Voss 2022-1-20

编辑：Voss 2022-1-20

在 MATLAB Online 中打开

@Neuro FYI, the accepted answer here selects the 5 rows with the highest value in column 2, whereas the accepted answer in another question you posted selects all rows whose value in column 2 is among the 5 highest unique values in column 2. You should be aware of this distinction because it seems likely that these questions are related.

To illustrate the distinction:

X = [1 2 3 4 5 5 5 6 6 6 6];
maxk(X,5) % 5 highest values
ans = 1×5
     6     6     6     6     5
uX = unique(X);
X(X >= uX(end-4)) % among the 5 highest unique values
ans = 1×10
     2     3     4     5     5     5     6     6     6     6

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Answer 1

Matt J 2022-1-19

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https://ww2.mathworks.cn/matlabcentral/answers/1632230-calculate-average-between-min-and-max-numbers-in-column-3-depending-on-values-in-column-2-for-each#answer_877815

编辑：Matt J 2022-1-19

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A=yourMatrix;
idx=A(:,1)==7;
B=A(idx,:);
[~,idx]=maxk(B(:,2),5);
C=B(idx,3);
result=0.5*(max(C)+min(C))

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Neuro 2022-1-20

Thanks! That's exactly what I needed.

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Answer 2

Image Analyst 2022-1-19

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https://ww2.mathworks.cn/matlabcentral/answers/1632230-calculate-average-between-min-and-max-numbers-in-column-3-depending-on-values-in-column-2-for-each#answer_877820

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Just take it one step at a time. Try this.

% Let's say you have a 2000 x 3 matrix, where all the elements are integers. 
% In column 1, the values can be between 1 and 16. How can we select one element in column 1 (e.g. 7), 
% then for those rows only, select the rows with the highest 5 values in column 2, 
% and for these rows only, select their elements in column 3 and 
% calculate the average between the min and max values (output).
m = randi([1, 16], 2000, 3);
% Find rows with column 1 = 7
rows1Equals7 = m(:, 1) == 7;
% Get values from column 2 in those rows
col2 = m(rows1Equals7, 2)
col2 = 109×1
     7
    11
     6
     3
     5
    12
     7
    12
     8
     1
% Select the rows with the highest 5 values
[highValues, rowsWithHighValues] = maxk(col2, 5)
highValues = 5×1
    16
    16
    16
    16
    16
rowsWithHighValues = 5×1
    33
    35
    41
    44
    46
% Get the values from column 3 in those 5 rows.
col3 = m(rowsWithHighValues, 3)
col3 = 5×1
    11
     4
     5
     4
    15
% Get the average "between" the max and min.  Not sure what "between" means
% so I'm taking the average of the min and max.
max3a = mean([min(col3), max(col3)])
max3a = 9.5000
% Alternative: Get the average of ALL 5 values
max3b = mean(col3)
max3b = 7.8000

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Answer 3

Voss 2022-1-19

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https://ww2.mathworks.cn/matlabcentral/answers/1632230-calculate-average-between-min-and-max-numbers-in-column-3-depending-on-values-in-column-2-for-each#answer_877835

在 MATLAB Online 中打开

matrix = randi([1 16],2000,3);
search_number = 7;
idx_1 = matrix(:,1) == search_number;
col2_vals = unique(matrix(idx_1,2));
idx_2 = idx_1 & matrix(:,2) >= col2_vals(end-4);
col3_min = min(matrix(idx_2,3));
col3_max = max(matrix(idx_2,3));
col3_avg = (col3_min+col3_max)/2;
disp(col3_min);
     1
disp(col3_max);
    16
disp(col3_avg);
    8.5000