Assign different values per column using logical indexing

I am new to logical indexing and am having some trouble with a specific issue. I want to use logical indexing to assign a certain value to all values in a column where the condition is true, and then a different value to the next column where another condition is true at the same time. E.g.:

matrix = magic(5); % data matrix
boundary_conditions = [10, 100]; % boundary conditions


matrix(matrix(:, [2,3]) < boundary_conditions) = boundary_conditions; % use logical indexing to set values of time columns

Unable to perform assignment because the left and right sides have a different number of elements.

Essentially, I want to find which values in column 2 are below 10, and which values in column 3 are below 100. Then, those that are in column 2, set them to 10, and those that are in column 3, set to 100. And have that reflected in the original matrix, so only columns 2 and 3 are changed.

For some context:

I am working on a molecular dynamics simulation, so I want to check after each timestep whether a particle has passed the boundary in the x- or y- direction.

Because of this, performance is a necessity, since I am working with ~3,000,000 time iterations. So I want to maximize my efficiency.

Answer 1

[Voss]

matrix = magic(5); % data matrix
boundary_conditions = [10, 100]; % boundary conditions

disp(matrix);

    17    24     1     8    15
    23     5     7    14    16
     4     6    13    20    22
    10    12    19    21     3
    11    18    25     2     9

% using logical indexing:

idx = matrix(:,2) < boundary_conditions(1);
matrix(idx,2) = boundary_conditions(1);

idx = matrix(:,3) < boundary_conditions(2);
matrix(idx,3) = boundary_conditions(2);

disp(matrix);

    17    24   100     8    15
    23    10   100    14    16
     4    10   100    20    22
    10    12   100    21     3
    11    18   100     2     9

% another approach using max() function:

matrix = magic(5); % restore original data matrix

matrix(:,2) = max(matrix(:,2),boundary_conditions(1));
matrix(:,3) = max(matrix(:,3),boundary_conditions(2));

disp(matrix);

    17    24   100     8    15
    23    10   100    14    16
     4    10   100    20    22
    10    12   100    21     3
    11    18   100     2     9

% another approach using max() function:

matrix = magic(5); % restore original data matrix

matrix(:,[2 3]) = max(matrix(:,[2 3]),repmat(boundary_conditions,size(matrix,1),1));

disp(matrix);

    17    24   100     8    15
    23    10   100    14    16
     4    10   100    20    22
    10    12   100    21     3
    11    18   100     2     9

[Jose Pratdesaba]

Thank you for the answer!

One question (more so out of curiosity): so there's no way to generalize this process for any number of columns and boundary conditions?

[Voss]

The third approach I showed is a general approach:

matrix = magic(10); % data matrix
disp(matrix);

    92    99     1     8    15    67    74    51    58    40
    98    80     7    14    16    73    55    57    64    41
     4    81    88    20    22    54    56    63    70    47
    85    87    19    21     3    60    62    69    71    28
    86    93    25     2     9    61    68    75    52    34
    17    24    76    83    90    42    49    26    33    65
    23     5    82    89    91    48    30    32    39    66
    79     6    13    95    97    29    31    38    45    72
    10    12    94    96    78    35    37    44    46    53
    11    18   100    77    84    36    43    50    27    59

columns = [2,4,5,7,9];
boundary_conditions = [10,20,40,60,50]; % boundary conditions

matrix(:,columns) = max(matrix(:,columns),repmat(boundary_conditions,size(matrix,1),1));
disp(matrix);

    92    99     1    20    40    67    74    51    58    40
    98    80     7    20    40    73    60    57    64    41
     4    81    88    20    40    54    60    63    70    47
    85    87    19    21    40    60    62    69    71    28
    86    93    25    20    40    61    68    75    52    34
    17    24    76    83    90    42    60    26    50    65
    23    10    82    89    91    48    60    32    50    66
    79    10    13    95    97    29    60    38    50    72
    10    12    94    96    78    35    60    44    50    53
    11    18   100    77    84    36    60    50    50    59

[Jose Pratdesaba]

Ahh I see now, thank you very much!

[Voss]

You're welcome!

Answer 2

[Dyuman Joshi]

A simpler approach for your task would be

matrix = magic(5) % data matrix

matrix = 5×5

17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 10 12 19 21 3 11 18 25 2 9

matrix(:,2) = max(matrix(:,2),10)

matrix = 5×5

17 24 1 8 15 23 10 7 14 16 4 10 13 20 22 10 12 19 21 3 11 18 25 2 9

matrix(:,3) = max(matrix(:,3),100)

matrix = 5×5

17 24 100 8 15 23 10 100 14 16 4 10 100 20 22 10 12 100 21 3 11 18 100 2 9

If you have to do this for some specific columns or all the columns (for which manually doing that would be a tedious task) you can integrate a for loop.

%For example 
matrix = magic(5)

matrix = 5×5

17 24 1 8 15 23 5 7 14 16 4 6 13 20 22 10 12 19 21 3 11 18 25 2 9

s = size(matrix,2);
boundary_conditions = randi([10 20],1,s)

boundary_conditions = 1×5

16 19 11 11 20

for k=1:s
    matrix(:,k) = max(matrix(:,k),boundary_conditions(k));
end

matrix

matrix = 5×5

17 24 11 11 20 23 19 11 14 20 16 19 13 20 22 16 19 19 21 20 16 19 25 11 20

[Jose Pratdesaba]

Thank you for the answer!

I was wondering, more out of curiosity, is there any way to generalize this for any number of columns and boundary conditions? Or do you have to use a for-loop to do that?

[Dyuman Joshi]

That depends on the data you have.

One approach would be -

arr = magic(9)

arr = 9×9

47 58 69 80 1 12 23 34 45 57 68 79 9 11 22 33 44 46 67 78 8 10 21 32 43 54 56 77 7 18 20 31 42 53 55 66 6 17 19 30 41 52 63 65 76 16 27 29 40 51 62 64 75 5 26 28 39 50 61 72 74 4 15 36 38 49 60 71 73 3 14 25 37 48 59 70 81 2 13 24 35

%Random values for example
x = randi(9)

x = 5

%columns for which values to compare and change
idx = randperm(9,x)

idx = 1×5

8 4 3 7 5

%boundary conditions
val = randi([50 100],1,numel(idx))

val = 1×5

67 73 55 61 89

for k=1:x
    arr(:,idx(k)) = max(arr(:,idx(k)),val(k));
end
arr

arr = 9×9

47 58 69 80 89 12 61 67 45 57 68 79 73 89 22 61 67 46 67 78 55 73 89 32 61 67 56 77 7 55 73 89 42 61 67 66 6 17 55 73 89 52 63 67 76 16 27 55 73 89 62 64 75 5 26 28 55 73 89 72 74 67 15 36 38 55 73 89 73 61 67 25 37 48 59 73 89 2 61 67 35

[Dyuman Joshi]

For the same data, using Voss's vectorized approach -

arr = magic(9);
x = 5;
idx = [8 4 3 7 5];
val = [67 73 55 61 89];

arr(:,idx) = max(arr(:,idx),repmat(val,size(arr,1),1))

arr = 9×9

47 58 69 80 89 12 61 67 45 57 68 79 73 89 22 61 67 46 67 78 55 73 89 32 61 67 56 77 7 55 73 89 42 61 67 66 6 17 55 73 89 52 63 67 76 16 27 55 73 89 62 64 75 5 26 28 55 73 89 72 74 67 15 36 38 55 73 89 73 61 67 25 37 48 59 73 89 2 61 67 35

[Jose Pratdesaba]

Thank you very much!!

[Dyuman Joshi]

You are welcome!