Avoide imaginary part in cell matrix

1 次查看（过去 30 天）
EldaEbrithil
EldaEbrithil 2020-10-13
评论： Ameer Hamza 2020-10-14
Hi all
i have a cell matrix A like that:
val(:,:,1,1,1) =
Columns 1 through 10
{n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double}
Columns 11 through 20
{n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double} {n×2 double}........
i want to obtain another cell matrix with the same global dimension of the previous one but with only the first column values: so i want to obtain a matrix where each cell has dimension equal to {nx1} . I also want to estrapolate from A only those cells that contain elements with imaginary part less than 0.1 setting to zero all the cells that don't satisfy this condition. Is there an easy way for doing that?
Thank you for the help
Regards

请先登录，再回答此问题。

采纳的回答

Ameer Hamza
Ameer Hamza 2020-10-13
编辑：Ameer Hamza 2020-10-13
Here is the solution to the first part
A; % 1x20 cell array
A_new = cellfun(@(x) {x(:,1)}, A);
All cell in A_new will contain first column of matrices in A.
I didn't understand the 2nd question related to setting the imaginary part to zero.
  7 个评论
显示 5更早的评论
EldaEbrithil
EldaEbrithil 2020-10-13
The first and the second column contain complex values. I would like to perform the following operations in order:
  • stay with one column instead of two
  • take those cells (made up of only one column each) that respect that condition on the imaginary part
Ameer Hamza
Ameer Hamza 2020-10-14
Ok. Following should work in this case
A; % 1x20 cell array
A_new1 = cellfun(@(x) {x(:,1)}, A); % only contain 1st columns
A_new2 = cellfun(@(x) {x.*(imag(x)<0.1)}, A_new1); % set the rows which does not satisfy the imaginary number conditions to 0.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Characters and Strings 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by