Compute probability of each element in each column of a m x n matrix.

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hello_world
hello_world 2014 年 11 月 24 日
回答済み: Sukumar GV 2018 年 7 月 29 日
Hello Friends,
I have a m x n matrix. I want to compute the probability of each element occurring in a column. I want to do it for each column.
For instance, suppose we have a 3 x 4 matrix:
A = [1 1 3 2; 2 3 0 2; 3 1 0 2];
Then, I would like to have the following matrix of probabilities: _P(A) = [ 1/3 2/3 1/3 1; 1/3 1/3 2/3 1; 1/3 2/3 2/3 1];_
Here, entries in P(A) are the probabilities of each element within the column.
For the 1st column: p(1) = p(2) = p(3) = 1/3
For the 2nd column: p(1) = 2/3, p(3) = 1/3
For the 3rd column: p(3) = 1/3, p(0) = 2/3
For the 4th column: p(2) = 3/3 = 1.
Please advise.

採用された回答

hello_world
hello_world 2014 年 11 月 25 日
I found the solution of my own problem posted above. Though, thanks for your kind help. I will still appreciate different ways to solve it specially with hist(), and histc(). Here is my solution to this problem:
[r, c] = size(A);
y = zeros(r,c);
p = zeros(r,c);
for i = 1:c %Looping through the column.
for j = 1:r %Looping through the row in that that column.
y(j,i) = sum(A(:,[i]) == A(j,i)); %Compare each column of the matrix with the entries in that column.
p(j,i) = y(j,i)/r; %Probability of each entry of a column within that column.
end
end

その他の回答 (2 件)

Sukumar GV
Sukumar GV 2018 年 7 月 29 日
You can use the following.
Suppose Matrix is X
[a, b] = hist (X, unique(X))
a/size(X,1)
each row represents item and each column represents the dimension

Yu Jiang
Yu Jiang 2014 年 11 月 24 日
編集済み: Yu Jiang 2014 年 11 月 24 日
One possible but most likely not optimal way is to use the function "find" to count the number of occurrence of a particular number. For example,
[r,c] = size(A); % get the column of A
nr = 0:3; %range of the numbers
P = zeros(r,c); % create a matrix to store the probabilities
for j = 1:c % check each row
for i = 1:length(nr) % check each particular number
P(i,j) = length(find(A(:,j) == nr(i)))/r; % count the occurrence
end
end
  3 件のコメント
Hugo
Hugo 2014 年 11 月 24 日
The range of the numbers refers to the minimum and the maximum number in your matrix. You can just replace the line nr=0:3; with nr=min(A(:)):max(A(:));
This will only work, of course, if A contains only integers. If it doesn't, things get a little more complicated. Hope this helps.
hello_world
hello_world 2014 年 11 月 24 日
編集済み: hello_world 2014 年 11 月 24 日
Thanks for help, though entries in my matrix are not integers. They are real number.
By making your suggested improvement, it gives me a 6 x 5 matrix of probabilities for a 3 x 5 size of matrix A. How is it possible? Should not I get a 3 x 5 size probability matrix instead?

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