How to add 1 matrix to another matrix ?

6 vues (au cours des 30 derniers jours)
Ahsan Abbas
Ahsan Abbas le 27 Avr 2016
Commenté : Ahsan Abbas le 30 Avr 2016
I have 2 matrices with different size as show below and just want to concatenate both of them. I know 2nd matrix have 1st 6 columns entries that are lying in b/w 1st matrix somewhere, so now i just want to add last 2 column values to 1st matrix of there corresponding rows of 1st matrix. Can someone help me how to write syntax for this ?
  4 commentaires
Azzi Abdelmalek
Azzi Abdelmalek le 27 Avr 2016
This is not clear. post the result to make it clear
Ahsan Abbas
Ahsan Abbas le 27 Avr 2016
x = [1 2 3; 4 5 6; 7 8 9; 10 11 12]; y = [4 5 6 12 15; 10 11 12 25 52]; result would be like z = [1 2 3 0 0; 4 5 6 12 15; 7 8 9 0 0; 10 11 12 25 52];

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Star Strider
Star Strider le 27 Avr 2016
Modifié(e) : Star Strider le 27 Avr 2016
One approach, taking advantage of the fact that both of your matrices are sorted by the first column:
M1 = sortrows(rand(15, 4), 1); % Create 1st Matrix & Sort By Column #1
M2 = [M1(1:5:end,:) rand(3,2)]; % Create 2nd Matrix & Add Columns #5 & #6
idx = find(ismember(M1(:,1), M2(:,1))); % Find Indices Of Matching Column #1 Entries
M1 = [M1 zeros(size(M1,1),2)]; % Add Zeros To End Of ‘M1’
M1(idx,5:6) = M2(:,5:6); % Fill Columns #5 & #6 With Same Values From ‘M2’
Since you are dealing with floating-point numbers, if you have R2015a or later, you would likely best use ismembertol instead of ismember to guarantee that your data match.
  2 commentaires
Ahsan Abbas
Ahsan Abbas le 27 Avr 2016
Thanks Star Strider for helping, i am using matlab 2013b so ismember is the only option.
Star Strider
Star Strider le 27 Avr 2016
My pleasure.
I’m hoping that your matrix entries are exact, so that ismember will work. If not, we will work on a way to use a tolerance with it. Rounding the first column of each matrix to a particular number of decimal places first would likely be easiest.

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Plus de réponses (1)

Azzi Abdelmalek
Azzi Abdelmalek le 27 Avr 2016
[n1,m1]=size(x)
[n2,m2]=size(y)
z=zeros(n1,m2)
idx=ismember(x,y(:,1:m1),'rows')
z(idx,:)=y
z(~idx,1:m1)=x(~idx,:)
  1 commentaire
Ahsan Abbas
Ahsan Abbas le 30 Avr 2016
Thank you Azzi for your response.

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