Selecting Value in matrix already indexed

1 次查看（过去 30 天）
Raphaël
Raphaël 2018-5-15
评论： Raphaël 2018-5-15
Hi, I'm trying to something relatively simple. Is there a way to select certain value in an already indexed matrix. Here is a simplified exemple of what I want to do.
A = rand(3);
B = @(x) A + x.^2; %Where x is a vector 3x1. Result is thus a 3x3 matrix.
C = @(x) B(x)(:,1:2); %I want C to be the first two columns of the result of B(x)
Obsiously, B(x)(:,1:2) isn't a valid way to write things. Is there a way to do this without finding a work around?
Thanks for the help.
RMT
  1 个评论
Stephen23
Stephen23 2018-5-15
@Raphael: is there a particular requirement to use anonymous functions? You could get around this quite easily by writing a normal function in an Mfile.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Stephen23
Stephen23 2018-5-15
编辑：Stephen23 2018-5-15
Some options:
1. avoid the situation by defining a function in an Mfile. This is the simplest and probably most efficient solution.
2. call subsref directly to do the indexing:
>> A = rand(3);
>> Bf = @(x) bsxfun(@plus,A,x.^2);
>> Cf = @(x) subsref(Bf(x),substruct('()',{':',1:2}));
>> Bf(1:3)
ans =
1.1974 4.4016 9.1932
1.6292 4.8217 9.0481
1.6670 4.1590 9.4776
>> Cf(1:3)
ans =
1.1974 4.4016
1.6292 4.8217
1.6670 4.1590
3. Define an intermediate function to do the indexing:
>> A = rand(3);
>> Bf = @(x) bsxfun(@plus,A,x.^2);
>> If = @(M) M(:,1:2);
>> Cf = @(x) If(Bf(x));
>> Bf(1:3)
ans =
1.5517 4.7931 9.7715
1.0754 4.6154 9.4393
1.6711 4.5820 9.2449
>> Cf(1:3)
ans =
1.5517 4.7931
1.0754 4.6154
1.6711 4.5820
  1 个评论
Raphaël
Raphaël 2018-5-15
I like option 3, thanks.
But there's really no way to ask twice in a row an indexing. A(5)(3). I find it weird you don't have that option.

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 Matrix Indexing 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