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Vectorization, avoiding loops

3 vues (au cours des 30 derniers jours)
Jarl Bergström
Jarl Bergström le 11 Mar 2020
Commenté : Jarl Bergström le 14 Mar 2020
Hello!
I have a Matrix, A=rand(9,10)
Now, I want to add a new number, to a new row, for every column b=[2,3,4;7 1 6];
I can do this using loops but I want to use Vectorization for speed.
A ( end +1 , b(1) ) = 1
A ( end +1 , b(2) ) = 1
A ( end +1 , b(3) ) = 1
Thank you :)

Réponse acceptée

tmarske
tmarske le 11 Mar 2020
Assuming this:
b=[2,3,4;7 1 6];
is a typo and supposed to read:
b=[2,3,4,7,1,6];
Then you can make the new rows as a matrix of zeros, set appropriate elements = 1, then concatenate:
A = rand(9, 10)
b=[2,3,4,7,1,6];
%make the new rows as a single block
newRows = zeros(length(b), size(A, 2));
bIx = sub2ind(size(newRows), 1:length(b), b);
newRows(bIx) = 1;
%concatenate them
A = [A; newRows]
Example output:
A =
0.5470 0.0811 0.8176 0.5502 0.2259 0.9797 0.2217 0.0292 0.5211 0.8852
0.2963 0.9294 0.7948 0.6225 0.1707 0.4389 0.1174 0.9289 0.2316 0.9133
0.7447 0.7757 0.6443 0.5870 0.2277 0.1111 0.2967 0.7303 0.4889 0.7962
0.1890 0.4868 0.3786 0.2077 0.4357 0.2581 0.3188 0.4886 0.6241 0.0987
0.6868 0.4359 0.8116 0.3012 0.3111 0.4087 0.4242 0.5785 0.6791 0.2619
0.1835 0.4468 0.5328 0.4709 0.9234 0.5949 0.5079 0.2373 0.3955 0.3354
0.3685 0.3063 0.3507 0.2305 0.4302 0.2622 0.0855 0.4588 0.3674 0.6797
0.6256 0.5085 0.9390 0.8443 0.1848 0.6028 0.2625 0.9631 0.9880 0.1366
0.7802 0.5108 0.8759 0.1948 0.9049 0.7112 0.8010 0.5468 0.0377 0.7212
A =
0.5470 0.0811 0.8176 0.5502 0.2259 0.9797 0.2217 0.0292 0.5211 0.8852
0.2963 0.9294 0.7948 0.6225 0.1707 0.4389 0.1174 0.9289 0.2316 0.9133
0.7447 0.7757 0.6443 0.5870 0.2277 0.1111 0.2967 0.7303 0.4889 0.7962
0.1890 0.4868 0.3786 0.2077 0.4357 0.2581 0.3188 0.4886 0.6241 0.0987
0.6868 0.4359 0.8116 0.3012 0.3111 0.4087 0.4242 0.5785 0.6791 0.2619
0.1835 0.4468 0.5328 0.4709 0.9234 0.5949 0.5079 0.2373 0.3955 0.3354
0.3685 0.3063 0.3507 0.2305 0.4302 0.2622 0.0855 0.4588 0.3674 0.6797
0.6256 0.5085 0.9390 0.8443 0.1848 0.6028 0.2625 0.9631 0.9880 0.1366
0.7802 0.5108 0.8759 0.1948 0.9049 0.7112 0.8010 0.5468 0.0377 0.7212
0 1.0000 0 0 0 0 0 0 0 0
0 0 1.0000 0 0 0 0 0 0 0
0 0 0 1.0000 0 0 0 0 0 0
0 0 0 0 0 0 1.0000 0 0 0
1.0000 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1.0000 0 0 0 0
  1 commentaire
Jarl Bergström
Jarl Bergström le 14 Mar 2020
Thank you very much!!

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