How can I fill a matrix column with dta coming from another matrix according to index
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I need to fill a large matrix with data coming from a smaller matrix. Both matrices contains a first column with an index and a second column with the results of an experiment. I need to combine the results according to their index. For example I have a matrix A:
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
And a matrix B
1 10
2 10
4 10
7 10
I would like to combine them to obtain the matrix C with 3 columns (column 1 for index, column 2 results from A, column 3 results from B)
1 1 10
2 1 10
3 1 NaN
4 1 10
5 1 NaN
6 1 NaN
7 1 10
8 1 NaN
9 1 NaN
I have to do that on matrices with millions of rows, therefore methods limiting calculation time would be welcome. Thank you very much for any help.
Risposta accettata
Più risposte (2)
Shoaibur Rahman
il 28 Gen 2015
A = [(1:9)' ones(9,1)];
B = [1 10; 2 10; 4 10; 7 10];
C = NaN(size(A,1),1);
C(B(:,1)) = B(:,2);
C = [A C]
1 Commento
olivier espeli
il 29 Gen 2015
Image Analyst
il 28 Gen 2015
Try this:
A = [...
1 1
2 1
3 1
4 5
5 1
6 1
7 9
8 1
9 1]
B = [...
1 10
2 20
4 70
7 90]
% Find out what the max index could possibly be.
maxRow = max([A(:,1); B(:,1)])
C = zeros(maxRow, 3); % Preallocate C.
% Assign all possible indices to column 1 of C.
C(:,1)=1:maxRow;
% Assign column 2 of A to column 2 of C.
C(A(:,1),2) = A(:, 2);
% Assign column 2 of B to column 3 of C.
C(B(:,1), 3) = B(:, 2)
In the command window:
C =
1 1 10
2 1 20
3 1 0
4 5 70
5 1 0
6 1 0
7 9 90
8 1 0
9 1 0
It's vectorized so it should be pretty fast. Anyway, a million rows is not that many. Here it is with a million rows and it took 0.03 seconds:
numberOfRows = 1000000;
A = int32(randi(99, [numberOfRows, 2]));
B = int32(randi(99, [numberOfRows, 2]));
tic;
% Find out what the max index could possibly be.
maxRow = max([A(:,1); B(:,1)]);
C = zeros(maxRow, 3); % Preallocate C.
% Assign all possible indices to column 1 of C.
C(:,1)=1:maxRow;
% Assign column 2 of A to column 2 of C.
C(A(:,1),2) = A(:, 2);
% Assign column 2 of B to column 3 of C.
C(B(:,1), 3) = B(:, 2);
toc;
1 Commento
olivier espeli
il 29 Gen 2015
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