Extracting minimum and maximum values from dimensions of a matrix

Question

Faisal Alnahhas 2018-4-12

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/394546-extracting-minimum-and-maximum-values-from-dimensions-of-a-matrix

评论： Faisal Alnahhas 2018-4-15

I have dataset that looks like below (this is just a sample), the last column (with 1, 2,..) are classes. I need to extract the min and max for each column for each class. That means for all columns that have i at end, I need the min and max of all columns. Example:

Data:

5000    0.4600    0.6400    0.3600    0.5000         0    0.4900    0.2200    1
5300    0.5600    0.4900    0.4600    0.5000         0    0.5200    0.2200    1
5200    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    3
3900    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    2
5000    0.4600    0.6400    0.3600    0.5000         0    0.4900    0.2200    2

Output:

Class 1, Dim 1, Min 0.5000, Max 0.5300
Class 1, Dim 2, Min 0.4600, Max 0.5600.

Now I used to calculate mean of such data using the following code (works perfectly):

for i = 1:10
    Mean(i,: ) = mean(file(file(:,end)==i,:));
end

When I try to apply same logic to min, using:

s(i,:) = min(file(file(:,end)==i,:)) (in the same loop)

It works for the first class of data (1), then I get the following error: Subscripted assignment dimension mismatch. How can I fix this, or what is the best way to do what I want to do?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2018-4-12

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/394546-extracting-minimum-and-maximum-values-from-dimensions-of-a-matrix#answer_314837

When you get to i = 4, then file(:,end)==i has no true values and file() indexed there is empty. min() of empty is empty. You cannot assign an empty value to a definite location s(i,:)

9 个评论
显示 7更早的评论隐藏 7更早的评论

Walter Roberson 2018-4-13

在 MATLAB Online 中打开

When file(:,end)==i finds no matches then file(file(:,end)==i,:) is empty.

mean(ARRAY) applied to an empty 0 x N ARRAY is defined as returning nan(1,N), a vector of NaN, which are definite values that can be stored into a vector of locations.

mean(ARRAY) applied to a 1 x N ARRAY (which would be the case if there is exactly one match in the last column) is defined as returning a scalar. If this is being assigned into a vector of locations then the scalar would be copied to each of the locations.

mean(ARRAY) applied to a 2+ x N ARRAY (which would be the case if there are multiple matches in the last column) is defined as returning a 1 x N vector of values of means along each column; these definite values can be stored into a vector of locations.

But

min(ARRAY) applied to an empty 0 x N ARRAY is defined as returning zeros(0, N), an empty vector that cannot be stored into definite locations.

min(ARRAY) applied to a 1 x N ARRAY is defined as returning a scalar, the minimum along the row. If this is being assigned into a vector of locations then the scalar would be copied to each of the locations.

min(ARRAY) applied to a 2+ x N ARRAY (which would be the case if there are multiple matches on the last column) is defined as returning a 1 x N vector of values of minimums along each column; these definite values can be stored into a vector of locations.

I think you also need to give more thought to whether you want the case of a single match to be mean and min along the single row as happens now in your code, or of you instead prefer to treat it as a series of columns each of which only happens to have one value to take the mean() or min() of. To have the per-column mean taken even in the case of only one row, use

mean(THEARRAY, 1)

To have the per-column min taken even in the case of only one row, use

min(THEARRAY, [], 1)

Notice the difference in location of the dimension number. This is because min() has an important second syntax, min(FIRSTARRAY, SECONDARRAY) which returns the element-by-element minimum between the arrays.

Faisal Alnahhas 2018-4-15

编辑：Walter Roberson 2018-4-15

在 MATLAB Online 中打开

function naive_bayes(training_file, type) %, test_file, type, num)
    file_train = load(training_file);
    columns = length(file_train);
    rows = size(file_train, 1);
    %file_test = load(test_file);
    num = 0;
    Mean = zeros(rows, columns);
    Std = zeros(rows, columns);
    s = zeros(rows, columns);
    l  = zeros(rows, columns);
    G = zeros(rows, columns);
    %10 classes 
    % 8 deimensions
      if (strcmp(type, "histograms"))
          for i = 1:rows
              %min(THEARRAY, [], 1)
              s(i,:)= min(file_train(file_train(:, end) ==i,:), [], 1);
              l(i,:) = max(file_train(file_train(:, end) ==i,:), [], 1);
          end
  end

That's the portion of the code that's relevant to my question.

Here's the input:

5000    0.4600    0.6400    0.3600    0.5000         0    0.4900    0.2200    1
3300    0.5600    0.4900    0.4600    0.5000         0    0.5200    0.2200    2
2300    0.5600    0.4900    0.4600    0.5000         0    0.5200    0.2200    1
5200    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    3
5200    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    3
3900    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    2

Here's the output:

s =
      0.2300    0.4600    0.4900    0.3600    0.5000         0    0.4900    0.2200    1.0000
      0.3300    0.5300    0.4900    0.4600    0.5000         0    0.5000    0.2200    2.0000
      0.5200    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    3.0000
           0         0         0         0         0         0         0         0         0
           0         0         0         0         0         0         0         0         0
           0         0         0         0         0         0         0         0         0

Walter Roberson 2018-4-15

编辑：Walter Roberson 2018-4-15

在 MATLAB Online 中打开

Change

rows = size(file_train, 1);

to

      unique_cases = unique(file_train(:,end));
      rows = length(unique_cases);

and change

          for i = 1:rows
              %min(THEARRAY, [], 1)
              s(i,:)= min(file_train(file_train(:, end) ==i,:), [], 1);
              l(i,:) = max(file_train(file_train(:, end) ==i,:), [], 1);
          end

to

          for row_idx = 1:rows
              %min(THEARRAY, [], 1)
              i = unique_cases(row_idx);
              s(row_idx,:)= min(file_train(file_train(:, end) ==i,:), [], 1);
              l(row_idx,:) = max(file_train(file_train(:, end) ==i,:), [], 1);
          end

Faisal Alnahhas 2018-4-15

Thank you!

请先登录，再进行评论。

Extracting minimum and maximum values from dimensions of a matrix

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

9 个评论
显示 7更早的评论隐藏 7更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

Extracting minimum and maximum values from dimensions of a matrix

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

采纳的回答

9 个评论 显示 7更早的评论隐藏 7更早的评论

更多回答（0 个）

另请参阅

类别

标签

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

9 个评论
显示 7更早的评论隐藏 7更早的评论