Subtracting Vector from a Matrix

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Paul Rogers
Paul Rogers el 21 de Nov. de 2019
Editada: Adam Danz el 21 de Nov. de 2019
I have the matrix a (8x1001) and b (1x1001) as in attached.
a and b can change in size.
How do I substract b from each row of a and put in a new matrix c?
Thanks.

Respuesta aceptada

Jan
Jan el 21 de Nov. de 2019
Matlab can subtract vectors from matrices automatically since R2016b - so called "auto expanding". Do you use an older version? Then:
c = bsxfun(@minus, a, b)
  5 comentarios
Jan
Jan el 21 de Nov. de 2019
@Adam: While repmat is (or at least was) an M-function, you can omit the overhead:
c = a - b(ones(1, 8), :);
Adam Danz
Adam Danz el 21 de Nov. de 2019
Editada: Adam Danz el 21 de Nov. de 2019
Good idea, Jan. I would have expected the ones() method to be faster than repmat(), too. But I timed both methods 500000 times and it turns out that the repmat() is actually faster. I repeated this a few times with consistent results. The boxplots below show the results of the 500k reps with outliers removed.
a = repmat((1:8)',1,12);
b = repmat(2,1,12);
% Method 1 (red)
c = a - b(ones(1, 8), :);
% Method 2 (blue)
d = a - repmat(b,size(a,1),1);
And including the outliers
191121 130508-compareSpeedGUI_results.png
Who knows what's going on in either function (closed source).

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Más respuestas (1)

Adam Danz
Adam Danz el 21 de Nov. de 2019
"How do I substract b from each row of a and put in a new matrix c?"
As long as the number of columns in b matches the number of columns in a
c = a - b;
Demo:
% Create data
a = repmat((1:8)',1,12);
b = repmat(2,1,12);
c = a - b;
  3 comentarios
Jan
Jan el 21 de Nov. de 2019
Obviously the sizes are not matching. Trust Matlab. See:
size(a)
size(b)
What do you get?
Adam Danz
Adam Danz el 21 de Nov. de 2019
Editada: Adam Danz el 21 de Nov. de 2019
See the 2nd line of my answer:
As long as the number of columns in b matches the number of columns in a
This requirement is met for the data you provided in matrix a and matrix b in your mat files.
load('a.mat')
load('b.mat')
size(a) % 8 1001
size(b) % 1 1001
c = a-b;
size(c) % 8 1001
If that requirement is not met, you cannot perform matrix subtraction without defining how you plan to resolve the size mismatch.
[update]
If your matlab release is prior to r2016a, see Jan's answer which avoids implicit expansion. This is why it's always good to include your matlab release in the field where it is requested while writing your question.

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