how do I add the maximum value of arr into arrSorted within the while loop?

clear; clc
rng(1)
a1D = randi([-20,80], 1, randi(20));
a2D = randi([-50, 30], randi(10), randi(16));
ar1DUp = Sort1DAr(a1D, "up");
ar1Down = Sort1DAr(a1D, "down");
sortedUp = Sort1DAr(a2D, "up");
SortedDown = Sort1DAr(a2D, 'down');
disp(ar1DUp);
disp(ar1Down);
disp(sortedUp);
disp(SortedDown);
function arrSorted = Sort1DAr(arr, sortOrder)
arrSorted = [];

while arr > 0
    [M, I] = max(arr(:)); 
    arrSorted = [arrSorted, M];
    arr(I) = [];
end
if sortOrder == "up"
    fliplr(arrSorted);
end
end

Answer 1

[Voss]

while arr > 0

This iterates while all elements of the array arr are greater than zero. If arr contains elements that are less than or equal to zero (which is the case for these a1D and a2D created after rng(1)), the loop iterates zero times. That is, It is never true (for these a1D and a2D) that all elements of arr are greater than zero.

I think what you intended was to iterate as long as arr has some elements. For that you can use

while ~isempty(arr)

Another problem is that when sortOrder is "up" you need to assign the result from fliplr() to arrSorted, which will be returned from the function.

rng(1)

a1D = randi([-20,80], 1,         randi(20));
a2D = randi([-50,30], randi(10), randi(16));

ar1DUp     = Sort1DAr(a1D, "up");
ar1Down    = Sort1DAr(a1D, "down");
sortedUp   = Sort1DAr(a2D, "up");
SortedDown = Sort1DAr(a2D, 'down');

disp(ar1DUp);

   -20   -11    -6    -2    10    14    20    34    52

disp(ar1Down);

    52    34    20    14    10    -2    -6   -11   -20

disp(sortedUp);

  Columns 1 through 33

   -49   -49   -48   -47   -46   -46   -44   -43   -42   -42   -40   -39   -39   -37   -34   -34   -33   -29   -28   -27   -27   -25   -25   -17   -17   -17   -16   -14   -11    -7    -7    -5    -4

  Columns 34 through 55

    -3     3     4     4     5     6     6     6     6    10    10    13    14    17    20    21    21    22    23    27    28    30

disp(SortedDown);

  Columns 1 through 33

    30    28    27    23    22    21    21    20    17    14    13    10    10     6     6     6     6     5     4     4     3    -3    -4    -5    -7    -7   -11   -14   -16   -17   -17   -17   -25

  Columns 34 through 55

   -25   -27   -27   -28   -29   -33   -34   -34   -37   -39   -39   -40   -42   -42   -43   -44   -46   -46   -47   -48   -49   -49

function arrSorted = Sort1DAr(arr, sortOrder)
arrSorted = [];
while ~isempty(arr)             % Iterate as long as there are some elements remaining in arr
    [M, I] = max(arr(:));       % Get the value and index of the maximum element
    arrSorted = [arrSorted, M]; % Append the maximum element to the sorted vector
    arr(I) = [];                % Remove the maximum element from arr
end
if sortOrder == "up"
    % need to assign the result of fliplr to arrSorted, so that the flipped
    % vector is what is returned from this function
    arrSorted = fliplr(arrSorted);
end
end

[Voss]

Note that if all elements of arr are in fact positive, then the loop as you wrote it works.

arr = 1:5;

arrSorted = [];
while arr > 0
    [M, I] = max(arr(:));       % Get the value and index of the maximum element
    arrSorted = [arrSorted, M]; % Append the maximum element to the sorted vector
    arr(I) = [];                % Remove the maximum element from arr
end
disp(arrSorted)

     5     4     3     2     1

This is because, as elements are removed from arr, it remains true that all elements of arr are positive, until the point when arr has no elements, at which point arr > 0 evaluates to false.

arr = [1 2 3]; % some non-empty array of positive numbers

if arr > 0 % "if" works as good as "while" for this purpose
    disp('all positive')
else
    disp('not all positive')
end

all positive

arr = []; % empty array

if arr > 0
    disp('all positive')
else
    disp('not all positive')
end

not all positive

Answer 2

[Sulaymon Eshkabilov]

Here is how ti can be attained:

clearvars; clc
rng(1)
a1D = randi([-20,80], 1, randi(20));
a2D = randi([-50, 30], randi(10), randi(16));
 ar1DUp = Sort1DAr(a1D, "up");
 ar1Down = Sort1DAr(a1D, "down");
sortedUp = Sort1DAr(a2D, "up");
SortedDown = Sort1DAr(a2D, 'down');
disp(ar1DUp);

    10    14    20    34    52

disp(ar1Down);

    52    34    20    14    10

disp(sortedUp);

     3     4     4     5     6     6     6     6    10    10    13    14    17    20    21    21    22    23    27    28    30

disp(SortedDown);

    30    28    27    23    22    21    21    20    17    14    13    10    10     6     6     6     6     5     4     4     3

function arrSorted = Sort1DAr(arr, sortOrder)
arrSorted = [];
arr = arr(:);
arr = arr(arr>0);
N=numel(arr);
ii=0;
while ii<N+1
    ii=ii+1;
    [M, I] = max(arr); 
    arrSorted = [arrSorted, M];
    arr(I) = [];
end
if sortOrder == "up"
   arrSorted = fliplr(arrSorted);
else
   arrSorted = arrSorted;
end
end

[Dyuman Joshi]

The else statement is redundant.

[Sulaymon Eshkabilov]

You're right! I sometimes tend to do "overkill" in my coding :)