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

list_1=[2;3;5;6] is array 1 or [1,4,5,6]

list_2=[1;4;3;1] Number of repetitions for each value in list_1 or [1,4,3,1]

I need the following output

[2;3;3;3;3;5;5;5;6]

or [2,3,3,3,3,5,5,5,6]

for example only one repetition from value 2 becaus of this first value in list_2 it says only one number from the first value in list_1

4 repetitions from number 3

3 repetitions from number 5

1 repetitions from number 6

i need the solution for huge list

Answer 1

[Akira Agata]

I'm not sure why you do not prefer repelem/repmat...

Anyway, how about the following solution?

% 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

[Ezzaddin Al-Soufi]

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]

"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

Answer 2

[Bruno Luong]

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

[Ezzaddin Al-Soufi]

thanks a lot

Answer 3

[Bruno Luong]

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

[Bruno Luong]

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.

[Ezzaddin Al-Soufi]

thanks a lot