Repetition of elements in a matrix

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Danish Nasir
Danish Nasir 2021 年 9 月 4 日
コメント済み: Walter Roberson 2021 年 9 月 4 日
Suppose i have a matrix A= [25 21 ] . I want to repeat elements by 4 and 2 times respectively. Repeat=[4 2]. However the repeated values will be in proportion i.e. Proportion=[8 17]. The sum of proportion repeated element will be equal to A matrix elements. It means that final matrix will be B= [8 8 8 1 17 4].
(25=8+8+8+1 , 21=17+4 )
Pls suggest code to generate matrix B

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Walter Roberson
Walter Roberson 2021 年 9 月 4 日
A = [25 21 ]
A = 1×2
25 21
Repeat = [4 2]
Repeat = 1×2
4 2
Proportion = [8 17]
Proportion = 1×2
8 17
leftover = A - Proportion .* (Repeat-1);
toRepeat = reshape([Proportion; leftover],1,[]);
repeatCount = reshape([Repeat-1; ones(1,length(Repeat))],1,[]);
B = repelem(toRepeat, repeatCount)
B = 1×6
8 8 8 1 17 4
  2 件のコメント
Danish Nasir
Danish Nasir 2021 年 9 月 4 日
It nearly solved my problem. I just want to add one complexity in the problem. There is a 3rd element in the matrix which i do not want to propotionate
A=[25 21 30]
Repeat=[4 2 1]
Proportion=[8 17 30].
B=[8 8 8 1 17 4 30]
Can you suggest how to generate B?
Walter Roberson
Walter Roberson 2021 年 9 月 4 日
The code already handles that.
A = [25 21 30]
A = 1×3
25 21 30
Repeat = [4 2 1]
Repeat = 1×3
4 2 1
Proportion = [8 17 30]
Proportion = 1×3
8 17 30
leftover = A - Proportion .* (Repeat-1);
toRepeat = reshape([Proportion; leftover],1,[]);
repeatCount = reshape([Repeat-1; ones(1,length(Repeat))],1,[]);
B = repelem(toRepeat, repeatCount)
B = 1×7
8 8 8 1 17 4 30

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