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Faster way of replacing multiple rows with same vector without using a loop?

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Hi,
I have a matrix a (see below). I would like to replace the 1st and 3rd row with the same row vector [255 100 0].
a =
245 255 255
254 252 255
251 250 239
Initially I created an index variable 'idx'
idx = [1;3]; % indexing row number for vector replacement
However
a(idx,:) = [255 100 0 ];
would give error messages
'Unable to perform assignment because the size of the left side is 2-by-3 and the size of the right side is 1-by-3'.
How to do it correctly without do it in a loop?

Akzeptierte Antwort

Stephen23
Stephen23 am 29 Jan. 2019
Bearbeitet: Stephen23 am 29 Jan. 2019
Here are your two main choices for avoiding a loop.
Method one: match the elements. You need to make the LHS and the RHS have the same number of elements (and they need to be in the required order). Here is one easy way to do that:
a(idx,:) = repmat([255,100,0],numel(idx),1)
Tested:
>> idx = [1;3];
>> vec = [255,100,0];
>> a = [245,255,255;254,252,255;251,250,239]
a =
245 255 255
254 252 255
251 250 239
>> a(idx,:) = repmat(vec,numel(idx),1)
a =
255 100 0
254 252 255
255 100 0
Method two: use scalars. Treat each column individually and assign scalar values. This can be a viable solution if you working with a fixed, small number of columns (e.g. RGB channels):
>> a = [245,255,255;254,252,255;251,250,239]
a =
245 255 255
254 252 255
251 250 239
>> idx = [1;3];
>> a(idx,1) = vec(1);
>> a(idx,2) = vec(2);
>> a(idx,3) = vec(3)
a =
255 100 0
254 252 255
255 100 0
This can also be trivially looped ( I am sure that you can see how).
  3 Kommentare
Matt J
Matt J am 29 Jan. 2019
Bearbeitet: Matt J am 29 Jan. 2019
But I don't see why it is trivially looped
Even after having carefully studied my answer?
Stephen23
Stephen23 am 29 Jan. 2019
Bearbeitet: Stephen23 am 29 Jan. 2019
"But I don't see why it is trivially looped. :)"
It isn't looped. I wrote that it can be. Look at the indices: 1, 2, 3.

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Weitere Antworten (2)

madhan ravi
madhan ravi am 29 Jan. 2019
idx = [1 3];
a(idx,:) = 0
  5 Kommentare
Stephen23
Stephen23 am 29 Jan. 2019
@madhan ravi: that just takes you back to the same problem in the original question.
madhan ravi
madhan ravi am 29 Jan. 2019
Ah you are right Stephen I was careless when posting it thnx though!

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Matt J
Matt J am 29 Jan. 2019
Bearbeitet: Matt J am 29 Jan. 2019
I don't think it's worth avoiding a loop here. A loop over columns, because there are so few of them, would be as fast as pretty much whatever else you might do, no matter how many rows a has.
rhs=[255,100,0];
for j=1:3
a(idx,j)=rhs(j);
end
  2 Kommentare
Salad Box
Salad Box am 29 Jan. 2019
Thanks Matt,
I understand for small number of rows as shown in the example it's worth to do it in a loop, however in real case I have a large number of rows (could be 50k rows for one image multiplying hundreds of images)... so that I was trying to avoid do it in a loop.:)

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