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How to find the min(max) value and its index in a multidimensional matrix?

조회 수: 10 (최근 30일)
The problem is that the min(array(:)) returns the index of the min(max) as a number, while I am searching for the indices of it.

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Konstantinos Fragkakis
Konstantinos Fragkakis 2018년 8월 27일
편집: Konstantinos Fragkakis 2018년 8월 27일
Function to calculate the minimum value and its indices, in a multidimensional array - In order to find the max, just replace the min(array(:)) statement with max(array(:)).
function [ minimum,index ] = minmat( array )
% Function: Calculate the minimum value and its indices in a multidimensional array
% -------- Logic description --------
% First of all, identify the Matlab convention for numbering the elements of a multi-dimensional array.
% First are all the elements for the first dimension
% Then the ones for the second and so on
% In each iteration, divide the number that identifies the minimum with the dimension under investigation
% The remainder is the Index for this dimension (check for special cases below)
% The integer is the "New number" that identifies the minimum, to be used for the next loop
% Repeat the steps as many times as the number of dimensions (e.g for a 2-by-3-by-4-by-5 table, repeat 4 times)
neldim = size(array); % Length of each dimension
ndim = length(neldim); % Number of dimensions
[minimum,I] = min(array(:));
remaining = 1; % Counter to evaluate the end of dimensions
index = []; % Initialize index
while remaining~=ndim+1 % Break after the loop for the last dimension has been evaluated
% Divide the integer with the the value of each dimension --> Identify at which group the integer belongs
r = rem(I,neldim(remaining)); % The remainder identifies the index for the dimension under evaluation
int = fix(I/neldim(remaining)); % The integer is the number that has to be used for the next iteration
if r == 0 % Compensate for being the last element of a "group" --> It index is equal to the dimension under evaluation
new_index = neldim(remaining);
else % Compensate for the number of group --> Increase by 1 (e.g if remainder 8/3 = 2 and integer = 2, it means that you are at the 2+1 group in the 2nd position)
int = int+1;
new_index = r;
end
I = int; % Adjust the new number for the division. This is the group th
index = [index new_index]; % Append the current index at the end
remaining = remaining + 1;
end
end

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