Repeating the rows of an array by a number given by another array
显示 更早的评论
Consider the array:
x=[linspace(1,4,4)' linspace(5,20,4)']
I would like to repeat each row by a number given by another array such as
I=[2 3 2 1]'
so that at the end the first row of x is repeat twice, the second row is repeated 3 times and so on.
Is this possible without using a loop?
采纳的回答
Use repelem, which is the built-in efficient solution for repeating elements:
x = [linspace(1,4,4)' linspace(5,20,4)']
I = [2 3 2 1];
y = repelem(x, I, 1)
A speed comparison:
x = rand(1e4, 2);
I = repmat(1:10, 1, 1000);
tic
for k = 1:100
C = cell(length(I),1) ;
for i = 1:length(C)
C{i} = repmat(x(i,:),I(i),1) ;
end
C = cell2mat(C);
end
toc
tic
for k = 1:100 % Slightly modified for I is a row vector:
idx = arrayfun(@(x,y)x*ones(1,y),1:numel(I),I,'UniformOutput',false);
C = x([idx{:}],:);
end
toc
tic
for k = 1:100
C = repelem(x, I, 1);
end
toc
% R2018b:
% Elapsed time is 3.676185 seconds. loop
% Elapsed time is 6.848004 seconds. arrayfun
% Elapsed time is 0.038391 seconds. repelem
更多回答（1 个）
x=[linspace(1,4,4)' linspace(5,20,4)'] ;
I=[2 3 2 1]' ;
C = cell(length(I),1) ;
for i = 1:length(C)
C{i} = repmat(x(i,:),I(i),1) ;
end
C = cell2mat(C)
4 个评论
Mike Croucher
2022-2-17
That's got a loop though
Saeid
2022-2-17
Mike Croucher
2022-2-17
OK good! There has been a lot of work done by MathWorks in recent years on improving the efficiency of loops. You do not need to be afraid of them any more. In many cases, they are on par with vectorised methods.
If the algorithm can be expressed naturally as a loop, do so. Only worry about other methods if the performance is problematic.
Saeid
2022-2-17
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