I need to repeat numbers in an array with a certain number of repetitions for each value without(repelem or repmat)

% Example
list_1 = [2;3;5;6];
list_2 = [1;4;3;1];
% One possible solution without using repelem/repmat
C = arrayfun(@(x,y) x*ones(y, 1), list_1, list_2, 'UniformOutput', false);
list_3 = cell2mat(C);
% Show the result
disp(list_3)
     2
     3
     3
     3
     3
     5
     5
     5
     6

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Ezzaddin Al-Soufi 2022-8-2

Thanks a lot

for me it doesn't matter if i use repelem/repmat, if i used repelem or repmat i have the problem that i can't have the solution as array or vector. I tried this

n=1:1:4;
list_1 = [2;3;5;6];
list_2 = [1;4;3;1];
    for i=1:1:length(n)
        b=repelem(list_1(i),list_2(i),1)
    end
b = 2
b = 4×1
     3
     3
     3
     3
b = 3×1
     5
     5
     5
b = 6

Stephen23 2022-8-2

编辑：Stephen23 2022-8-2

"for me it doesn't matter if i use repelem/repmat"

Your question title states "without(repelem or repmat)". If it does not matter, why tell us not to use them?

"if i used repelem or repmat i have the problem that i can't have the solution as array or vector."

REPELEM gives exactly the same output as Akira Agata's answer:

list_1 = [2;3;5;6];
list_2 = [1;4;3;1];
list_3 = repelem(list_1,list_2)
list_3 = 9×1
     2
     3
     3
     3
     3
     5
     5
     5
     6

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Answer 2

Bruno Luong 2022-8-2

编辑：Bruno Luong 2022-8-2

Why prefer a simple method when one can do in a complicated manner:

list_1 = [2;3;5;6]
list_1 = 4×1
     2
     3
     5
     6
list_2 = [1;4;3;1]
list_2 = 4×1
     1
     4
     3
     1
idx=cumsum(accumarray(cumsum([1; list_2(:)]),1));
list_1(idx(1:end-1))
ans = 9×1
     2
     3
     3
     3
     3
     5
     5
     5
     6

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Ezzaddin Al-Soufi 2022-8-2

thanks a lot

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Answer 3

Bruno Luong 2022-8-2

编辑：Bruno Luong 2022-8-2

The old for-loop

list_1 = [2;3;5;6];
list_2 = [1;4;3;1];
r = zeros(sum(list_2),1);
start = 0;
for k = 1:length(list_2)
    n = list_2(k);
    r(start+1:start+n)) = list_1(k);
    start = start + n;
end
r
r = 9×1
     2
     3
     3
     3
     3
     5
     5
     5
     6

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Bruno Luong 2022-8-2

Some timing, for-loop seems to be the fatest

list_1 = randi(1000,1000,1);
list_2 = randi(2000,1000,1);
tic
r = zeros(sum(list_2),1);
start = 0;
for k = 1:length(list_2)
    n = list_2(k);
    r(start+1:start+n) = list_1(k);
    start = start + n;
end
toc
Elapsed time is 0.008149 seconds.
tic
C = arrayfun(@(x,y) x*ones(y, 1), list_1, list_2, 'UniformOutput', false);
list_3 = cell2mat(C);
toc
Elapsed time is 0.016814 seconds.
tic
idx=cumsum(accumarray(cumsum([1; list_2(:)]),1));
r=list_1(idx(1:end-1));
toc
Elapsed time is 0.014714 seconds.