Subtracting Vector from a Matrix

115 ビュー (過去 30 日間)
Paul Rogers
Paul Rogers 2019 年 11 月 21 日
編集済み: Adam Danz 2019 年 11 月 21 日
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.

採用された回答

Jan
Jan 2019 年 11 月 21 日
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 件のコメント
Jan
Jan 2019 年 11 月 21 日
@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 2019 年 11 月 21 日
編集済み: Adam Danz 2019 年 11 月 21 日
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).

サインインしてコメントする。

その他の回答 (1 件)

Adam Danz
Adam Danz 2019 年 11 月 21 日
"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 件のコメント
Jan
Jan 2019 年 11 月 21 日
Obviously the sizes are not matching. Trust Matlab. See:
size(a)
size(b)
What do you get?
Adam Danz
Adam Danz 2019 年 11 月 21 日
編集済み: Adam Danz 2019 年 11 月 21 日
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.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeMatrix Indexing についてさらに検索

製品


リリース

R2014b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by