Rearranging array based on number of steps.

Question

Santos García Rosado 2021-6-1

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/845060-rearranging-array-based-on-number-of-steps

评论： Santos García Rosado 2021-6-1

采纳的回答： Stephen23

在 MATLAB Online 中打开

Hi everybody,

Let's say I have a vector such as:

A = (1:24)

And I'd like to separate this array into a 2x12 matrix where the first row will show the first 3 positions skipping the next three and so on... The second row will do the same thing but starting from the fourth position. This is the expexted output:

Output = [1 2 3 7 8 9 13 14 15 19 20 21; 4 5 6 10 11 12 16 17 18 22 23 24]

I could do this manually, but some times I'd like it to take 2 or 4 steps instead of 3, as it is in the example above.

Any ideas of how to to achieve this?

Thank you in advanced,

Santos

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Answer 1

Stephen23 2021-6-1

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/845060-rearranging-array-based-on-number-of-steps#answer_714230

编辑：Stephen23 2021-6-1

在 MATLAB Online 中打开

https://www.mathworks.com/help/matlab/ref/reshape.html

https://www.mathworks.com/help/matlab/ref/permute.html

A = 1:24;
N = 3; % number of "steps"
R = 2; % number of rows
M = reshape(permute(reshape(A,N,R,[]),[2,1,3]),R,[])
M = 2×12
     1     2     3     7     8     9    13    14    15    19    20    21
     4     5     6    10    11    12    16    17    18    22    23    24
[1,2,3,7,8,9,13,14,15,19,20,21;4,5,6,10,11,12,16,17,18,22,23,24]
ans = 2×12
     1     2     3     7     8     9    13    14    15    19    20    21
     4     5     6    10    11    12    16    17    18    22    23    24