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Find index of a matrix of values into another matrix

Georgios Tertikas さんによって質問されました 2019 年 4 月 15 日
最新アクティビティ Jon
さんによって コメントされました 2019 年 4 月 15 日
Hello,
I have a matrix A 151x1 double and a matrix B 151x1000 double. Each value of A corresponds to a certain value in the same row of B. For example if A=22;14;53;2, then B will have the value 22 in one of its columns in the first row, the value 14 in one of its columns in the second row etc. Is there a way to have the matrix of the index of all the columns in B that contain the values of A (always in their coresponding row?).

  3 件のコメント

madhan ravi
2019 年 4 月 15 日
Explicitly show how your expected output should look like with a small example of A & B.
if A = 1;5;15;22 and B= 31 23 10 1 9; 221 31 51 5 11; 15 22 33 44 55; 20 21 22 23 24
then I want a matrix that will be C=4;4;1;3 (for each position in B)
madhan ravi
2019 年 4 月 15 日
See my answer below.

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3 件の回答

回答者: madhan ravi
2019 年 4 月 15 日
 採用された回答

Indices = cell(numel(A),1);
for k = 1:numel(A)
Indices{k} = find(B(k,:)==A(k));
end
%celldisp(Indices)
% Indices{1} represents first element of A match in first row of B likewise others

  9 件のコメント

madhan ravi
2019 年 4 月 15 日
Partially correct but try with this and see why it fails :
B= [ 31 23 1 1 9;...
% ^ ^---- see what happens if there is more than one common ground
221 31 51 5 11;...
15 22 33 44 55;...
20 21 22 23 24]
Adam Danz
2019 年 4 月 15 日
Now test it with these inputs and you'll see why the simpler suggestion doesn't work.
A = [1;5;15;22]
B = [1 23 10 1 9; 1 31 51 5 11; 15 22 33 44 55; 20 5 22 22 22]
Jon
2019 年 4 月 15 日
Yes I agree your approach will cover the case where there is more than one matching entry in a row. I did condition my original response with limitation that for the simple approach to work there had to be exactly one match in each row. In the end I'm not sure whether the OP has multiple matching entries in a row. Thanks for the clarification and discussion.

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回答者: Adam Danz
2019 年 4 月 15 日

If I understood you correctly, you have a matrix "B" and a column vector "A" and you want the identify the columns of B that are identical to A (if any). Here's a demo that achieves that:
% Create data
A = randi(100, 151,1);
B = randi(100, 151, 1000);
B(:,55) = A; %column #55 of B equals A
B(:,66) = A; %column #66 of B also equals A
% Identify columns of B that match A
% matchIdx is a logical row vectors containing 'true' for columns of B that match A
matchIdx = ismember(B.', A.', 'rows').';
% Find the column numbers
colNums = find(matchIdx);
colNums =
55 66

  2 件のコメント

So indeed I want to find which column in B has the same value as A but not with A as a whole matrix. I want for the first row (where A has one value and B has 1000) to find which column of B has the same value. And then do this also for the other 151 rows each time specifically for each row.
Adam Danz
2019 年 4 月 15 日
madhan ravi's solution should do what you're describing. If not, you can elaborate.

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回答者: Jon
2019 年 4 月 15 日

This should work assuming you have exactly one match in each row. If some rows have no matches or more than one match you will have to do something more elaborate
% find the row and column indices of all the places in B that match A
[row,col] = find(A==B)
% sort the matching column locations by row to get them in the desired (row) order
idx = col(row)

  3 件のコメント

The two different matrices have different sizes so I cant use find(A==B)
Jon
2019 年 4 月 15 日
Actually, it is ok the == operation apparently operates column wise in this situation and then find with two left hand arguments works perfectly for this situation. Please try it to see.
For more info on A==B with A a column with the same number of rows as B please see https://www.mathworks.com/help/matlab/matlab_prog/compatible-array-sizes-for-basic-operations.html
Adam Danz
2019 年 4 月 15 日
The problem is not that A and B are not the same size. The problem is that this does not address the question. The OP wants to list column numbers where row 'n' of B matches the value of row 'n' of A.

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