Extracting minimum and maximum values from dimensions of a matrix

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:

      0.5000    0.4600    0.6400    0.3600    0.5000         0    0.4900    0.2200    1
      0.5300    0.5600    0.4900    0.4600    0.5000         0    0.5200    0.2200    1
      0.5200    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    3
      0.3900    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    2
      0.5000    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?

Answer 1

[Walter Roberson]

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,:)

[Faisal Alnahhas]

Okay, I think I understand. How can I fix this? Or What is the best way to go about doing this? Thank you

[Walter Roberson]

for i = 1:10
    temp = mean(file(file(:,end)==i,:));
    if ~isempty(temp)
      Mean(i,: ) = temp;
    else
      Mean(i,:) = nan;
    end
end

[Faisal Alnahhas]

Would this work for minimum? because just using mean(file(file(:,end)==i,:)) works for mean. Also can you explain why this solves the problem of getting the minimum value?

[Walter Roberson]

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]

That was a great explanation, thank you. Well I am trying to build histograms for the dataset given using some formula -that depends on min and max values- to define the bin size. I actually don't need the mean for this, I just posted it because I was confused as to why it works for mean and not min (and max). I will try what you suggested and report back.

UPDATE: I tried using: s(i,:) = min(file_train(file_train(:, end) ==i,:), [], 1) in my loop and it seemed to be extracting the correct values, however the s matrix has a bunch of zeros after the first three rows. I am iterating i = 1:3. The total number of lines in the input file is 6 with with 3 classes. Why isn't it not stopping after 3? Does it update the minimum values associated with each class? Is there a way to make it just show those 3 lines not all 6 including the zeros? Thank you a bunch, you've been very helpful.

[Walter Roberson]

I will need to see more code.

[Faisal Alnahhas]

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:

      0.5000    0.4600    0.6400    0.3600    0.5000         0    0.4900    0.2200    1
      0.3300    0.5600    0.4900    0.4600    0.5000         0    0.5200    0.2200    2
      0.2300    0.5600    0.4900    0.4600    0.5000         0    0.5200    0.2200    1
      0.5200    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    3
      0.5200    0.5300    0.5800    0.6900    0.5000         0    0.5000    0.2200    3
      0.3900    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]

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]

Thank you!