How do I break an array into individual element without using loops

2 次查看(过去 30 天)
x=[3,4,5,6,7,8,9,10]
x = 1×8
3 4 5 6 7 8 9 10
for i=1:length(x)
y=x(i)
end
y = 3
y = 4
y = 5
y = 6
y = 7
y = 8
y = 9
y = 10
I want a function to give me this result without using a loop
  5 个评论
DGM
DGM 2023-12-17
编辑:DGM 2023-12-17
Repetition doesn't answer the question. The example you gave is ostensibly the thing you want, but in that example, y is a scalar which changes value with each loop iteration. So if there is no loop, what is the equivalent? I posed two interpretations:
  1. A single scalar variable which is multivalued (which is nonsense)
  2. A pile of numbered scalar variables (which are not practically addressable)
If we assume that you mean to create a pile of numbered variables, then there's another question you need to answer. Once you have a hundred numbered variables in memory, how do you intend on addressing them? Spoiler: the solution to the unnecessary problem you're creating is another problem.
As to how you'd create them all those variables, well that would still involve a loop. If you don't want a loop, then you're going to have to write it all out like @Image Analyst pointed out. At that point, if you just intend on writing out every single operation on every single element of every array in your entire workflow, then what's the point in using MATLAB to do it?
Olayinka
Olayinka 2023-12-20
@madhan ravi @Image Analyst @DGM Thank you all for your contribution.
Madhan ravi gave me the closest to what I thought I want but I have learnt that it doesn't make sense. I'll keep learning and improving.
Thank you.

请先登录,再进行评论。

回答(1 个)

madhan ravi
madhan ravi 2023-12-16
编辑:madhan ravi 2023-12-16
y = reshape(x, 1, 1, [])
y(:, :, 1)
This makes it a 3 D array
If you want it as a cell array then
y = num2cell(x)
celldisp(y)
The loop you use is just looping through x vector.

类别

Help CenterFile Exchange 中查找有关 Matrix Indexing 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by