How to search for a previous element that is equal in prior loop iteration

1 次查看(过去 30 天)
I have a Matrix called A as below
[1 1 3;
1 2 2;
1 3 3;
2 1 1;
2 2 3;
2 3 2;
2 4 4]
I want to produce a column vector that counts the number of times a value occurs in the 2nd column, I have tried the following function :
function count_stars(A)
total = 0;
for i = 1:size(A,1)
total = total + (A(i,2)==A(:,2));
end
total
This counts the number of matching values in the entire column e.g.
2
2
2
2
2
2
1
What I am wanting is to match to values between 1:i only rather than the entire column so that the output looks like this
1
1
1
2
2
2
1
Any insights you are able to share are greatly appreciated!
  2 个评论
Image Analyst
Image Analyst 2015-12-28
It's too late at night for me to work on this, but I'm thinking it might involve accumarray(), cumsum() or histcounts(). A question though, will the numbers always be integers? Or might they be floating point numbers with fractional parts?
Tim Baker
Tim Baker 2015-12-28
Thanks for your response, yes the numbers are always integers. Am interested in discovering other ways to solve the problem if it can be done more efficiently. Harjeet's solution works well.

请先登录,再进行评论。

采纳的回答

harjeet singh
harjeet singh 2015-12-28
hello tim try this code
a=[1 1 3;
1 2 2;
1 3 3;
2 1 1;
2 2 3;
2 3 2;
2 4 4];
for i=1:size(a,1)
[r,c]=find(a(1:i,2)==a(i,2));
b(i,1)=length(r);
end
b

更多回答(1 个)

Guillaume
Guillaume 2015-12-29
For small arrays, it does not matter but this is likely to be more efficient than harjeet's answer for large arrays:
a = [1 1 3;
1 2 2;
1 3 3;
2 1 1;
2 2 3;
2 3 2;
2 4 4];
[values] = unique(a(:, 3));
counts = zeros(size(a, 1), 1);
for val = values';
occurences = cumsum(a(:, 3) == val);
counts(a(:, 3) == val) = occurences(a(:, 3) == val);
end
Rather than looping over all the rows of a, it only loops over the unique values of a (4 iterations instead of 7 in this example).

类别

Help CenterFile Exchange 中查找有关 Data Type Conversion 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by