Faster way of replacing multiple rows with same vector without using 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 个评论
显示 1更早的评论隐藏 1更早的评论

Matt J 2019-1-29

编辑：Matt J 2019-1-29

But I don't see why it is trivially looped

Even after having carefully studied my answer?

Stephen23 2019-1-29

编辑：Stephen23 2019-1-29

"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.

请先登录，再进行评论。

Answer 2

madhan ravi 2019-1-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/442081-faster-way-of-replacing-multiple-rows-with-same-vector-without-using-a-loop#answer_358530

在 MATLAB Online 中打开

idx = [1 3];
a(idx,:) = 0

5 个评论
显示 3更早的评论隐藏 3更早的评论

Stephen23 2019-1-29

@madhan ravi: that just takes you back to the same problem in the original question.

madhan ravi 2019-1-29

Ah you are right Stephen I was careless when posting it thnx though!

请先登录，再进行评论。

Answer 3

Matt J 2019-1-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/442081-faster-way-of-replacing-multiple-rows-with-same-vector-without-using-a-loop#answer_358534

编辑：Matt J 2019-1-29

在 MATLAB Online 中打开

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 个评论
显示无隐藏无

Salad Box 2019-1-29

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.:)