How do I divide matrix elements by column sums in

问题:

Is there an easy way to divide each matrix element by the column sum? For example:

input:

1  4

4  10

output:

1/5  4/14

4/5  10/14

回答1:

Here's a list of the different ways to do this ...

• ... using bsxfun:

  B = bsxfun(@rdivide,A,sum(A));
  

• ... using repmat:

  B = A./repmat(sum(A),size(A,1),1);
  

• ... using an outer product (as suggested by Amro):

  B = A./(ones(size(A,1),1)*sum(A));
  

• ... and using a for loop (as suggested by mtrw):

  B = A;
  columnSums = sum(B);
  for i = 1:numel(columnSums)
    B(:,i) = B(:,i)./columnSums(i);
  end
  

Update:

As of MATLAB R2016b and later, most built-in binary functions (list can be found here) support implicit expansion, meaning they have the behavior of bsxfun by default. So, in the newest MATLAB versions, all you have to do is:

B = A./sum(A);

回答2:

a=[1 4;4 10]
a =
     1     4
     4    10

a*diag(1./sum(a,1))
ans =
    0.2000    0.2857
    0.8000    0.7143

回答3:

Couldn't resist trying a list comprehension. If this matrix was represented in a row-major list of lists, try this:

>>> A = [[1,4],[4,10]]
>>> [[float(i)/j for i,j in zip(a,map(sum,zip(*A)))] for a in A]
[[0.20000000000000001, 0.2857142857142857], [0.80000000000000004, 0.7142857142857143]]

Yes, I know that this is not super-efficient, as we compute the column sums once per row. Saving this in a variable named colsums looks like:

>>> colsums = map(sum,zip(*A))
>>> [[float(i)/j for i,j in zip(a,colsums)] for a in A]
[[0.20000000000000001, 0.2857142857142857], [0.80000000000000004, 0.7142857142857143]]

Note that zip(*A) gives transpose(A).