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elmar.a
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Follow up: How can I merge two different tables using the first column in common?

elmar.a
さんによって質問されました 2019 年 5 月 21 日
最新アクティビティ Jos (10584)
さんによって コメントされました 2019 年 5 月 22 日
This question is related to How can I merge two different tables using the first column in common? but the accepted answer does not fully solve my issue.
I have a case where there are more than two arrays, some of which have the same identifier in the first column, such as
A = [1 7;
3 15]
B = [2 9;
5 10]
C = [2 5;
3 4]
From this I'd like to get
[1 7 0 0;
2 0 9 5;
3 15 0 4;
4 0 0 0;
5 0 10 0]
that means if identifiers are the same (as for row 3) the values of A, B, and C should appear in the same row.

  2 件のコメント

Not clear, more explanation needed.
Column 1 is just a row index number.
Column 2 are the number is col 2 of "A". Column 3 are the numbers in col 2 of "B". Column 4 are the numbers in col 2 of "C".

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3 件の回答

回答者: Jos (10584)
2019 年 5 月 21 日
 採用された回答

% data, (showing the drawback of storing relates things in different variables)
A = [1 7;
3 15]
B = [2 9;
5 10]
C = [2 5;
3 4]
% simple indexing engine
A(:,3) = 2, B(:,3) = 3, C(:,3) = 4 % add column numbers to input
D = cat(1,A,B,C)
sz = [max(D(:,1)), D(end,3)]
m = zeros(sz)
m(D(:,1), 1) = D(:,1)
m(sub2ind(sz, D(:,1), D(:,3))) = D(:,2)

  1 件のコメント

Thanks, that does the job. I like this short solution!

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回答者: Adam Danz
2019 年 5 月 21 日
編集済み: Adam Danz
2019 年 5 月 21 日

I find it easier to first combine the matrices into a 3D array. This should work with any number of matrices as long as they are the same size. "m" is your final matrix.
ABC = cat(3,A,B,C);
% Create final matrix (all 0s except first column)
m = zeros(max(ABC(:,1,:),[],'all'),size(ABC,3)+1); %prior to r2018b: zeros(max(max(squeeze(ABC(:,1,:)))),size(ABC,3)+1)
m(:,1) = 1:size(m,1);
% Find the column and row indices of M for the elements in ABC
colID = reshape(repelem(2:size(ABC,3)+1,size(ABC,2),size(ABC,2)-1,1),[],1);
[~, rowID] = ismember(reshape(ABC(:,1,:),[],1),m(:,1));
% fill in the rest of the m matrix
m(sub2ind(size(m),rowID,colID)) = ABC(:,2:end,:);
Result
m =
1 7 0 0
2 0 9 5
3 15 0 4
4 0 0 0
5 0 10 0

  5 件のコメント

Note that this solution assumes that all matrices have the same size.
Yeah, I mentioned that in the answer (2nd sentence).
Jos' answer is more fleixible because it allows for varying number of rows.
Ah, my mistake, Adam, I overlooked that in your answer :-)

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回答者: Guillaume
2019 年 5 月 21 日

The question is a bit confusing. Tables are mentioned in the title, but the examples are matrices. Two tables is mentioned in the table, but the example has 3 inputs.
If you were dealing with tables, you'd use outerjoin for what you want:
A = [1 7;
3 15]
B = [2 9;
5 10]
C = [2 5;
3 4]
%cell array of tables
t{1} = array2table(A, 'VariableNames', {'ID', 'A'});
t{2} = array2table(B, 'VariableNames', {'ID', 'B'});
t{3} = array2table(C, 'VariableNames', {'ID', 'C'});
result = outerjoin(t{1}, t{2}, 'MergeKeys', true); %outer join the first two
for tidx = 3:numel(t) %loop over the rest (works with any number of tables)
result = outerjoin(result, t{tidx}, 'MergeKeys', true);
end
For matrices, I'd use Adam's answer.

  1 件のコメント

There are tables mentioned in the title since that question is so similar to my problem. I'm working with matrices though. Thanks anyways for your answer, it works well for tables.

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