Optimize based on on max min of columns

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Thomas 2016-5-10

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https://ww2.mathworks.cn/matlabcentral/answers/283487-optimize-based-on-on-max-min-of-columns

评论： Stephen23 2016-5-10

Hi,

I have an array where each row indicates the parameters of an item, I want to set constraints for each column (or parameter) and find the optimum item that closely meets these parameters. Here is an example,

 %
A = [80 7.5 1000 0.5;120 500 3000 1;50 256 500 0.5];

For the first column I want the maximum, for the second I want the minimum, the third I want the max and the fourth I want the max. However as each row represents an item I need the item which closely matches the parameters I set. So far I have been trying to use a logic array that gives a 1 where the max/min value is and a zero where it does not meet the criteria then summing each row and the highest number will give the optimum. I have been trying to use a for loop to go through each column but I am having difficulty selecting both maximum and minimum across the array. So I would end up with an array like this,

%
A1 = [0 1 0 0;1 0 1 1;0 0 0 0]

The logical array indicates that the 2nd column is the optimum. If anyone has any advice on how to proceed, I would be very grateful.

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Stephen23 2016-5-10

"indicates that the 2nd column is the optimum"

presumably you mean row, not column.

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Answer 1

Stephen23 2016-5-10

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https://ww2.mathworks.cn/matlabcentral/answers/283487-optimize-based-on-on-max-min-of-columns#answer_221572

编辑：Stephen23 2016-5-10

在 MATLAB Online 中打开

With MATLAB using a loop to solve this is a waste of time. Try something like this:

>> A = [80 7.5 1000 0.5;120 500 3000 1;50 256 500 0.5]
A =
                    80                   7.5                  1000                   0.5
                   120                   500                  3000                     1
                    50                   256                   500                   0.5
>> vec = max(A,[],1);
>> vec(2) = min(A(:,2))
vec =
                   120                   7.5                  3000                     1
>> idx = bsxfun(@eq,A,vec)
idx =
                     0                     1                     0                     0
                     1                     0                     1                     1
                     0                     0                     0                     0
>> [~,row] = max(sum(idx,2))
row =                    2