Grouping the Elements of the Array

16 次查看（过去 30 天）
tinkyminky93
tinkyminky93 2022-6-21
评论： Voss 2022-6-21
Hello, I have a double array with the element number of 897880. I want to group them like 30000 30000 but it gives me error because of the 897880 divided by 30000 is not an integer. How can I group it like 30000 30000 and the last grouped part as it is, if it is not the multiple of 30000? Thank you.
  4 个评论
显示 2更早的评论
Image Analyst
Image Analyst 2022-6-21
What form are these numbers in? They can't be in a double array because matrices can't have ragged right edges. Maybe you have a cell array. Or do you have one long vector of 897880 elements and you want to make it square with the last few elements padded with zeros?
If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:
TUTORIAL: How to ask a question (on Answers) and get a fast answer
tinkyminky93
tinkyminky93 2022-6-21
编辑：tinkyminky93 2022-6-21
I have a .mat file and i am importing them. I can see that it is double array and it is 897880x1. I just want to group them like above.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Voss
Voss 2022-6-21
Ran in:
% Taking your example of an array with 17 elements (rather than 897880),
data = 1:17;
% and you want to group them into groups of 5 (rather than 30000).
group_size = 5;
% You can append NaNs to the end of the data as necessary, to make the
% number of elements a multiple of 5:
% number of elements in data
n_data = numel(data);
% number of groups of 5 necessary
n_groups = ceil(n_data/group_size);
% append NaNs off the end of data, storing the result as new_data
new_data = data;
new_data(end+1:n_groups*group_size) = NaN;
% And finally reshape the NaN-appended data set:
new_data = reshape(new_data,group_size,[]).'
new_data = 4×5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 NaN NaN NaN
  4 个评论
显示 2更早的评论
tinkyminky93
tinkyminky93 2022-6-21
Yes, I know. Is it necessary to write NaN to those elements? What if I don't write anything?
Voss
Voss 2022-6-21
It is necessary to have something there. It doesn't have to be a NaN, but it's a good idea to use something so that you can distinguish that part from the rest of the data, i.e., discern where the data stops and the filled-in section begins.
Something must be there because a matrix must be rectangular in shape. It can't be some weird polygon shape with a corner missing. In other words, every row must have the same number of elements (or, put another way, every column must have the same number of elements). This is the defining characteristic of a matrix.
If you want something where the number of elements can vary by row, use a cell array.

请先登录，再进行评论。

更多回答（1 个）

Jan
Jan 2022-6-21
Ran in:
x = rand(897880, 1);
nx = numel(x);
len = 300000;
tile = [repelem(len, floor(nx / len)), rem(nx, len)]
tile = 1×3
300000 300000 297880
xC = mat2cell(x, tile, 1);
  2 个评论
tinkyminky93
tinkyminky93 2022-6-21
Suppose that I have 7 groups which have 30000 elements inside of it. And I want like 7x30000 array not a cell sir. From your code, I got 3x1 cell and every element of the cell has 30000 elements in it.
Jan
Jan 2022-6-21
编辑：Jan 2022-6-21
Ran in:
In your question you have 897880 elements and want 3 groups. Because the last group is smaller than the others, you cannot simply use reshape. Therefore I create 3 different arrays and store them in a cell.
Of course you can store the first groups with the matching sizes in a matrix also:
x = rand(897880, 1);
len = 300000;
nx = numel(x);
match = nx - mod(nx, len);
Groups = reshape(x(1:match), len, []).';
Remainder = x(match:nx).';
size(Groups)
ans = 1×2
2 300000
size(Remainder)
ans = 1×2
1 297881

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

产品


版本

R2019b

Community Treasure Hunt

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

Start Hunting!

Translated by