In Matlab, when is it optimal to use bsxfun?

半城伤御伤魂 提交于 2019-11-26 04:32:39

问题


My Question: I\'ve noticed that a lot of good answers to Matlab questions on SO frequently use the function bsxfun. Why?

Motivation: In the Matlab documentation for bsxfun, the following example is provided:

A = magic(5);
A = bsxfun(@minus, A, mean(A))

Of course we could do the same operation using:

A = A - (ones(size(A, 1), 1) * mean(A));

And in fact a simple speed test demonstrates the second method is about 20% faster. So why use the first method? I\'m guessing there are some circumstances where using bsxfun will be much faster than the \"manual\" approach. I\'d be really interested in seeing an example of such a situation and an explanation as to why it is faster.

Also, one final element to this question, again from the Matlab documentation for bsxfun: \"C = bsxfun(fun,A,B) applies the element-by-element binary operation specified by the function handle fun to arrays A and B, with singleton expansion enabled.\". What does the phrase \"with singleton expansion enabled\" mean?


回答1:


There are three reasons I use bsxfun (documentation, blog link)

  1. bsxfun is faster than repmat (see below)
  2. bsxfun requires less typing
  3. Using bsxfun, like using accumarray, makes me feel good about my understanding of Matlab.

bsxfun will replicate the input arrays along their "singleton dimensions", i.e. the dimensions along which the size of the array is 1, so that they match the size of the corresponding dimension of the other array. This is what is called "singleton expasion". As an aside, the singleton dimensions are the ones that will be dropped if you call squeeze.

It is possible that for very small problems, the repmat approach is faster - but at that array size, both operations are so fast that it likely won't make any difference in terms of overall performance. There are two important reasons bsxfun is faster: (1) the calculation happens in compiled code, which means that the actual replication of the array never happens, and (2) bsxfun is one of the multithreaded Matlab functions.

I have run a speed comparison between repmat and bsxfun with R2012b on my decently fast laptop.

For me, bsxfun is about 3 times faster than repmat. The difference becomes more pronounced if the arrays get larger

The jump in runtime of repmat happens around an array size of 1Mb, which could have something to do with the size of my processor cache - bsxfun doesn't get as bad of a jump, because it only needs to allocate the output array.

Below you find the code I used for timing:

n = 300;
k=1; %# k=100 for the second graph
a = ones(10,1);
rr = zeros(n,1);
bb=zeros(n,1);
ntt=100;
tt=zeros(ntt,1);
for i=1:n;
   r = rand(1,i*k);
   for it=1:ntt;
      tic,
      x=bsxfun(@plus,a,r);
      tt(it)=toc;
   end;
   bb(i)=median(tt);
   for it=1:ntt;
      tic,
      y=repmat(a,1,i*k)+repmat(r,10,1);
      tt(it)=toc;
   end;
   rr(i)=median(tt);
end



回答2:


In my case, I use bsxfun because it avoids me to think about the column or row issues.

In order to write your example:

A = A - (ones(size(A, 1), 1) * mean(A));

I have to solve several problems:

1) size(A,1) or size(A,2)

2) ones(sizes(A,1),1) or ones(1,sizes(A,1))

3) ones(size(A, 1), 1) * mean(A) or mean(A)*ones(size(A, 1), 1)

4) mean(A) or mean(A,2)

When I use bsxfun, I just have to solve the last one:

a) mean(A) or mean(A,2)

You might think it is lazy or something, but when I use bsxfun, I have less bugs and I program faster.

Moreover, it is shorter, which improves typing speed and readability.




回答3:


Very interesting question! I have recently stumbled upon exactly such situation while answering this question. Consider the following code that computes indices of a sliding window of size 3 through a vector a:

a = rand(1e7,1);

tic;
idx = bsxfun(@plus, [0:2]', 1:numel(a)-2);
toc

% equivalent code from im2col function in MATLAB
tic;
idx0 = repmat([0:2]', 1, numel(a)-2);
idx1 = repmat(1:numel(a)-2, 3, 1);
idx2 = idx0+idx1;
toc;

isequal(idx, idx2)

Elapsed time is 0.297987 seconds.
Elapsed time is 0.501047 seconds.

ans =

 1

In this case bsxfun is almost twice faster! It is useful and fast because it avoids explicit allocation of memory for matrices idx0 and idx1, saving them to the memory, and then reading them again just to add them. Since memory bandwidth is a valuable asset and often the bottleneck on todays architectures, you want to use it wisely and decrease the memory requirements of your code to improve the performance.

bsxfun allows you to do just that: create a matrix based on applying an arbitrary operator to all pairs of elements of two vectors, instead of operating explicitly on two matrices obtained by replicating the vectors. That is singleton expansion. You can also think about it as the outer product from BLAS:

v1=[0:2]';
v2 = 1:numel(a)-2;
tic;
vout = v1*v2;
toc
Elapsed time is 0.309763 seconds.

You multiply two vectors to obtain a matrix. Just that the outer product only performs multiplication, and bsxfun can apply arbitrary operators. As a side note, it is very interesting to see that bsxfun is as fast as the BLAS outer product. And BLAS is usually considered to deliver the performance..

Edit Thanks to Dan's comment, here is a great article by Loren discussing exactly that.




回答4:


As of R2016b, Matlab supports Implicit Expansion for a wide variety of operators, so in most cases it is no longer necessary to use bsxfun:

Previously, this functionality was available via the bsxfun function. It is now recommended that you replace most uses of bsxfun with direct calls to the functions and operators that support implicit expansion. Compared to using bsxfun, implicit expansion offers faster speed, better memory usage, and improved readability of code.

There's a detailed discussion of Implicit Expansion and its performance on Loren's blog. To quote Steve Eddins from MathWorks:

In R2016b, implicit expansion works as fast or faster than bsxfun in most cases. The best performance gains for implicit expansion are with small matrix and array sizes. For large matrix sizes, implicit expansion tends to be roughly the same speed as bsxfun.




回答5:


Things are not always consistent with the 3 common methods: repmat, expension by ones indexing, and bsxfun. It gets rather more interesting when you increase the vector size even further. See plot:

bsxfun actually becomes slightly slower than the other two at some point, but what surprised me is if you increase the vector size even more (>13E6 output elements), bsxfun suddenly becomes faster again by about 3x. Their speeds seem to jump in steps and the order are not always consistent. My guess is it could be processor/memory size dependent too, but generally I think I'd stick with bsxfun whenever possible.



来源:https://stackoverflow.com/questions/12951453/in-matlab-when-is-it-optimal-to-use-bsxfun

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