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Finding similar elements in a single matrix

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Md. Nazrul Islam Siddique
Md. Nazrul Islam Siddique 2022 年 12 月 2 日
コメント済み: Star Strider 2022 年 12 月 2 日
I have one matrix,
x = [
2 0 0
3 3 0
4 4 0
5 5 5
6 6 6]
How can I find the common elements between columns in the matrix and count them? for example, here 3 and 4 repeated two times.
  4 件のコメント
Walter Roberson
Walter Roberson 2022 年 12 月 2 日
So you just need to know the total number of non-zero elements? You can use nnz for that.
Md. Nazrul Islam Siddique
Md. Nazrul Islam Siddique 2022 年 12 月 2 日
That's easy. what about the non zero elements between two metrices?

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Star Strider
Star Strider 2022 年 12 月 2 日
One approach using accumarray
x = [2 0 0
3 3 0
4 4 0
5 5 5
6 6 6];
[Aunique,~,idx] = unique(x(:));
Counts = accumarray(idx, 1);
Result = table(Aunique, Counts)
Result = 6×2 table
Aunique Counts _______ ______ 0 4 2 1 3 2 4 2 5 3 6 3
.
  9 件のコメント
Md. Nazrul Islam Siddique
Md. Nazrul Islam Siddique 2022 年 12 月 2 日
I want to know the total number which is repeated. For example, 1 and 2 repeated two times, 5 repeated three times. Other numbers are repeated only one. The total count is: 12.
Star Strider
Star Strider 2022 年 12 月 2 日
Just sum the ‘Count’ results —
x = [1 2 3
4 5 6];
y = [1 5 7
2 5 8];
xy = [x; y]
xy = 4×3
1 2 3 4 5 6 1 5 7 2 5 8
xy = xy(xy~=0);
[XYunique,~,idx] = unique(xy(:));
Counts = accumarray(idx, 1);
Result = table(XYunique, Counts)
Result = 8×2 table
XYunique Counts ________ ______ 1 2 2 2 3 1 4 1 5 3 6 1 7 1 8 1
TotalCounts = sum(Counts)
TotalCounts = 12
If the matrices are not conformable to concatenation (they are conformable here), then reshape both to be column vectors using the column vector operator ‘(:)’ and then vertically concatenate those. Then use unique and accumarray.
.

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