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

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?

Answer 1

[Stephen23]

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).

[Salad Box]

Thanks Stephen,

I do prefer Method two. But I don't see why it is trivially looped. :)

[Matt J]

But I don't see why it is trivially looped

Even after having carefully studied my answer?

[Stephen23]

"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.

Answer 2

[madhan ravi]

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

[madhan ravi]

https://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html

[Salad Box]

Hi madhan,

Very much appreciated for your prompt solution. :)

[0 0 0] might not be a representitive example in this case. What about if I would like to replace those indexed rows with a vector [255 100 0] instead of [0 0 0]?

P.s. I have also modified my question as I realized that a vector containing three same numbers is more of an extreme case and can be simplified as a single number to solve the problem.

[madhan ravi]

a(idx,:) = [255 100 0] 

[Stephen23]

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

[madhan ravi]

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

Answer 3

[Matt J]

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

[Salad Box]

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.:)

[Matt J]

My loop is over columns!!!