Assign different values per column using logical indexing

Question

Jose Pratdesaba 2023-9-27

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https://ww2.mathworks.cn/matlabcentral/answers/2026542-assign-different-values-per-column-using-logical-indexing

评论： Dyuman Joshi 2023-9-27

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.

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Answer 1

Voss 2023-9-27

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https://ww2.mathworks.cn/matlabcentral/answers/2026542-assign-different-values-per-column-using-logical-indexing#answer_1320277

编辑：Voss 2023-9-27

在 MATLAB Online 中打开

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

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Voss 2023-9-27

编辑：Voss 2023-9-27

在 MATLAB Online 中打开

The third approach I showed is a general approach:

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

Jose Pratdesaba 2023-9-27

Ahh I see now, thank you very much!

Voss 2023-9-27

You're welcome!

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Answer 2

Dyuman Joshi 2023-9-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2026542-assign-different-values-per-column-using-logical-indexing#answer_1320282

在 MATLAB Online 中打开

A simpler approach for your task would be

matrix = magic(5) % data matrix
matrix = 5×5
  24     1     8    15
   5     7    14    16
   6    13    20    22
  12    19    21     3
  18    25     2     9
matrix(:,2) = max(matrix(:,2),10)
matrix = 5×5
  24     1     8    15
  10     7    14    16
  10    13    20    22
  12    19    21     3
  18    25     2     9
matrix(:,3) = max(matrix(:,3),100)
matrix = 5×5
  24   100     8    15
  10   100    14    16
  10   100    20    22
  12   100    21     3
  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

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Jose Pratdesaba 2023-9-27

Thank you very much!!

Dyuman Joshi 2023-9-27

You are welcome!

请先登录，再进行评论。

Assign different values per column using logical indexing

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Assign different values per column using logical indexing

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