Writing elegant MATLAB code

2011 6 22
5 个回答
15 次查看（30 天）

0 个投票

I have a great interest in writing my MATLAB code the "right" way. I want it to be compact, readable and fast. As such, I have a few patterns that I see occasionally. Some I have found neat solutions to, some I have not.
1. Matrix mixing or tensors? If you want to do image processing, you may want to do linear color transforms. Each output channel should be a weighted sum of each input channel (channels indexed by the third matrix dimension). Now, there are probably neat ways of doing this in image processing toolbaox, but tht is expensive, does not make for portable code, and this pattern could appear in other contexts than image processing. I guess that some tensor/kron() operation would do this cleaner, but I did not figure it out.
rgb = double(imread('ngc6543a.jpg'));
M = [0.1 0.8 0.1; -0.1 1.2 -0.1; 0 0.5 0.5];
rgb2(:,:,1) = M(1,1)*rgb(:,:,1) + M(1,2)*rgb(:,:,2) + M(1,3)*rgb(:,:,3);
rgb2(:,:,2) = M(2,1)*rgb(:,:,1) + M(2,2)*rgb(:,:,2) + M(2,3)*rgb(:,:,3);
rgb2(:,:,3) = M(3,1)*rgb(:,:,1) + M(3,2)*rgb(:,:,2) + M(3,3)*rgb(:,:,3);
figure
image(rgb2);
2. Indexing I want to make an irregular index into a vector/matrix. For instance, [x1, x2, x11, x12, x21, x22]. Ideally, I'd like to type something like: x(1:(10+1:2):N) Is there a neat pattern for this? Perhaps using ind2sub?

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Andrew Newell
Andrew Newell 2011-6-22
Do you really want to mix 1D and 2D indices?
Knut
Knut 2011-6-22
Perhaps I was being unclear.
x = 1:100;
y = [x(1) x(2) x(11) x(12) x(21)...];
I want a neat way to index x so as to produce y.
Sean de Wolski
Sean de Wolski 2011-6-22
Huh?
do you mean
y = [x(1),x(11),x(12),x(13)...x(2),x(21)...x(99)]
Where would x(100) go?
Knut
Knut 2011-6-23
I want to create an index to pick out a series of elements from a 1-d vector. For simple cases (regular stride), one can type idx=1:10:100 to get something that picks out every 10th element. I want to pick out e.g. two elements every 10th element. My request was if there is some neat way of doing this using the a:b:c-method.
idx = [];
period = 10;
duty_cycle = 3;
for i=1:length(x)
if rem(i,period) < duty_cycle
idx = [idx i];
end
end
Typing this yields:
idx =
Columns 1 through 23
1 2 10 11 12 20 21 22 30 31 32 40 41 42 50 51 52 60 61 62 70 71 72
Columns 24 through 29
80 81 82 90 91 92
Knut
Knut 2011-6-23
Bummer. Repeating the question below to get a proper code formatting and fixing the flaw.
Andrei Bobrov
Andrei Bobrov 2011-6-23
more variant
idx = reshape(bsxfun(@plus,strfind(rem(1:length(x),10),[1 2]),[0;1]),1,[])

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回答（5 个）

Titus Edelhofer
Titus Edelhofer 2011-6-22

2 个投票

Hi,
for these cases I usually cange from 3D to 2D:
[n,m,~] = size(rgb);
rgb2D = reshape(rgb, n*m, 3);
rgb2D2 = rgb2D * M';
rgb2 = reshape(rgb2D2, n, m, 3);
Titus

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Andrei Bobrov
Andrei Bobrov 2011-6-22
+1, cool!
rgb2 = reshape(reshape(rgb,[],3)*M',size(rgb))
Igor
Igor 2011-6-23
your code is valid, isn't?
>> rgb=rand(5,5,3);M=ones(3,3);rgb2 = reshape(reshape(rgb,[],3)*M',size(rgb))
Igor
Igor 2011-6-23
Titus' solutions is good but more general way:
1. to present rgb-matrix as a cell array (n*m) with 3-vector at each point -- rgbX
2. to apply the rotation function:
fun1=@(r) M*r;
rgbX=cellfun(rgbX)
3. reverse to iss.1 1
Sean de Wolski
Sean de Wolski 2011-6-23
Igor, cellfun is slow and inefficient, not to mention the to and from conversions. Titus' method is the winner.
Andrei Bobrov
Andrei Bobrov 2011-6-23
Hi friends!
@Igor. Valid.
Analog for your example: repmat(sum(rgb,3),1,1,3)
Igor
Igor 2011-6-23
Yes... cells doesn't needed
Very excellent 3-columns, I see, after reshape(rgb,[],3) in bobrov's variant

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Andrew Newell
Andrew Newell 2011-6-22

1 个投票

  1. To improve on your code, you'll probably need a third-party package like TPROD. EDIT 06/24/11: Another package is MTIMESX.
  2. One way is
x(sort([1:10:21 2:10:22]))
Another is
I = [1:10:21; 2:10:22];
x(I(:)')
A more general approach is the following:
lastDigit = 0:2;
I = 0:99; I = I(ismember(mod(I,10),lastDigit));
Whether it is "neat" will be a matter for personal taste.
Sean de Wolski
Sean de Wolski 2011-6-22

0 个投票

I would do your above operation with:
rgb3 = zeros(size(rgb));
for ii = 1:3
rgb3(:,:,ii) = sum(bsxfun(@times,reshape(M(ii,:),[1 1 3]),rgb),3);
end
There's probably a way use just one call after reshaping rgb into the 4th dimension but I'm too busy right now to toy with it. Maybe later.
Andrei Bobrov
Andrei Bobrov 2011-6-22

0 个投票

1. My variant - is worse than variant of Titus Edelhofer
reshape(sum(bsxfun(@times,reshape(rgb,[],3),permute(M,[3 2 1])),2),size(rgb))
2. My variant - is worse than variant of Andrew Newell
ind = cumsum([1:2;10*ones(length(x)/10-1,2)])';
x(ind(:)')
Knut
Knut 2011-6-23

0 个投票

Regarding question#2:
idx = [];
period = 10;
duty_cycle = 3;
for i=0:length(x)
if rem(i,period) < duty_cycle
idx = [idx i];
end
end
idx
idx =
Columns 1 through 23
0 1 2 10 11 12 20 21 22 30 31 32 40 41 42 50 51 52 60 61 62 70 71
Columns 24 through 30
72 80 81 82 90 91 92
I could do like this, but that it not neat nor general:
idx = sort([0:10:100, 1:10:100, 2:10:100])

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Andrei Bobrov
Andrei Bobrov 2011-6-23
more variant
bsxfun(@plus,2:10:100,(-2:0)')
Andrew Newell
Andrew Newell 2011-6-23
Nice. You'll need to reshape it, though.

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Knut
Knut
2011-6-22

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