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?
采纳的回答
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).
更多回答（2 个）
madhan ravi
2019-1-29
idx = [1 3];
a(idx,:) = 0
5 个评论
madhan ravi
2019-1-29
编辑：madhan ravi
2019-1-29
a(idx,:) = [255 100 0]
Stephen23
2019-1-29
madhan ravi
2019-1-29
Ah you are right Stephen I was careless when posting it thnx though!
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
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