# Acess all last array elements withtin cells

2 ビュー (過去 30 日間)
Marc Laub 2022 年 10 月 25 日
コメント済み: Chunru 2022 年 10 月 25 日
Hey,
I need to store somae data and later acess them again. For further calculation that is done repetedly I would prefer to acces them as fast as possible so that I can use them in a cevtor function. Right know I dont know how to acess them.
They would be either stored in a cell or a class, not sure yet, but the structutre would be similar.
For the cells I would have something like:
precipitate{1}=[1,2,3,4;1,2,3,4];
precipitate{2}=[1,2,3,4,5;1,2,3,4,5];
recipitate{2}=[1,2,3;1,2,3];
in the class, ech element of the class would have a property that keeps that array.
I know need to acess all last colums of the arrays at once, process them and attach new data at the end.
So I would need to acces them so that the result would be
res=[4,5,3;4,5,3];
process them to like
[5,6,4;5,6,4]
and attach them all at once so that it would look like
precipitate{1}=[1,2,3,4,5;1,2,3,4,5];
precipitate{2}=[1,2,3,4,5,6;1,2,3,4,5,6];
precipitate{2}=[1,2,3,4;1,2,3,4];
Right now I am struggling to acces them all at once without a loop.
Does anybody know how to manage that?

サインインしてコメントする。

### 採用された回答

Chunru 2022 年 10 月 25 日

You can use cellfun:
precipitate{1}=[1,2,3,4;1,2,3,4];
precipitate{2}=[1,2,3,4,5;1,2,3,4,5];
precipitate{3}=[1,2,3;1,2,3];
% get the last column
res = cellfun(@(x) x(:, end), precipitate, 'UniformOutput', false)
res = 1×3 cell array
{2×1 double} {2×1 double} {2×1 double}
% process it
res=cellfun(@(x) x+1, res, 'UniformOutput', false)
res = 1×3 cell array
{2×1 double} {2×1 double} {2×1 double}
% attach it
precipitate = cellfun(@(x, y) [x y], precipitate, res, 'UniformOutput', false);
celldisp(precipitate)
precipitate{1} = 1 2 3 4 5 1 2 3 4 5 precipitate{2} = 1 2 3 4 5 6 1 2 3 4 5 6 precipitate{3} = 1 2 3 4 1 2 3 4
##### 4 件のコメント2 件の古いコメントを表示2 件の古いコメントを非表示
Marc Laub 2022 年 10 月 25 日
sry to annoy, not comfotable with cell indexing..
what if in that case
res=cellfun(@(x) x+1, res, 'UniformOutput',
the added 1 is not a contant but a vector, for examle a 1 for the first case, 0.5 for the second...
I only get as res something like {[5,5,5]} {(5.5, 5.5 , 5.5)}... instead of {5} {5.5}...
Chunru 2022 年 10 月 25 日
You can perform whatsoever processing you want. Add 1 is just an example you have given.

サインインしてコメントする。

### その他の回答 (1 件)

Rik 2022 年 10 月 25 日
Loops are faster than cellfun (excluding the legacy syntax). The reason is that cellfun will hide the loop internally and you have the additional overhead of an anonymous function.
So there is no reason in terms of performance to avoid loops.
The main exception to this is when there exists a vectorized function to do something. Using sum is obviously faster than using plus in a loop.

サインインしてコメントする。

### カテゴリ

Help Center および File ExchangeMatrices and Arrays についてさらに検索

R2020b

### Community Treasure Hunt

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

Start Hunting!

Translated by