for loop/periodic implementation for number or arrays?

Question

Hatim Altarteer 2021-8-31

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1443864-for-loop-periodic-implementation-for-number-or-arrays

评论： Hatim Altarteer 2021-8-31

x0.mat

i have 8 arrays of 100x3 configuration. i want to make a for loop (or another method is also appreciated) to accont for 2 constraints.

if the value of a number in the array is less than 1, replace it with one. also, if a number in the array is greater than nx + 1, put it back to 1. (nx is the upper limit, say nx = 100)

for instance,

array1 = [x0,y0,z0];
array2 = [x0,y0,z1];
array3 = [x0,y1,z0];
array4 = [x0,y1,z1];
array5 = [x1,y0,z0];
array6 = [x1,y0,z1];
array7 = [x1,y1,z0];
array8 = [x1,y1,z1];

where xyz0 and xyz1 are coordinates of a particel in the xyz plane. the attached file is an example of x0 where it includes the location of the "x" coordinates of 100 particels. the same applies for the rest of yz0 and xyz1.

thanks in advance!

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Fabio Freschi 2021-8-31

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1443864-for-loop-periodic-implementation-for-number-or-arrays#answer_777699

在 MATLAB Online 中打开

If you want to force the max and min values of your vectors, you can preprocess them

x0(x0 < 1 & x0 > nx+1) = 1;

If the boundary value is different youy can split the instruction in two

x0(x0 < 1) = 1;
x0(x0 > nx+1) = nx+1; % for example

repeat for y0 z0 x1 y1 z1 and then create your 8 arrays

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Hatim Altarteer 2021-8-31

Thanks Fabio!

请先登录，再进行评论。

Answer 2

DGM 2021-8-31

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1443864-for-loop-periodic-implementation-for-number-or-arrays#answer_777704

在 MATLAB Online 中打开

If I'm reading this right...

Let's say you have an array A:

A = randi([-5 110],5,3) % random test array
A = 5×3
    77    87    95
   110    34    87
    18    27    89
     7    -3     8
    67     2    37
allowedrange = [1 101]; % min and max value allowed
A = min(max(A,allowedrange(1)),allowedrange(2))
A = 5×3
    77    87    95
   101    34    87
    18    27    89
     7     1     8
    67     2    37