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

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Stephen23
Stephen23 2018년 5월 15일
@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.

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Stephen23
Stephen23 2018년 5월 15일
편집: Stephen23 2018년 5월 15일

1 개 추천

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

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Raphaël
Raphaël 2018년 5월 15일
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.

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