divide matrix by a vector of the same number of column

I have a matrix A[mXn] for example,

   A=[ 0     2     0     0     0
       0     1     1     0     1
       0     1     0     0     2
       0     0     0     0     1
       0     0     0     0     0]

I have a vector that computes the colume-wise sum of A, B=sum(A,1),

   B=[0 4 1 0 4]

How can I compute a matrix C, which each element is the element of A divided by the element of B that is in the same colume? Therefore, my C should be

   C=[0   0.5   0   0  0
      0   0.25  1   0  0.25
      0   0.25  0   0  0.5
      0   0     0   0  0.25 
      0   0     0   0  0   ]

Answer 1

[Star Strider]

Using bsxfun:

A=[ 0     2     0     0     0
    0     1     1     0     1
    0     1     0     0     2
    0     0     0     0     1
    0     0     0     0     0];

B=[0 4 1 0 4];

C = bsxfun(@rdivide, A, B);
C(isnan(C)) = 0;

Answer 2

[Image Analyst]

One way of probably many ways:

A=[ 0     2     0     0     0
  0     1     1     0     1
  0     1     0     0     2
  0     0     0     0     1
  0     0     0     0     0]
[rows, columns] = size(A)
% Get sum of columns and replicate vertically.
denom = repmat(sum(A, 1), [rows, 1])
% Do the division.
C = A ./ denom
% Set infinities (where denom == 0) to 0
C(denom==0) = 0

Answer 3

[Amirali Kamalian]

You can also use matrix-matrix multiplication where,

B = B.^(-1);
C = A*diag(B);
C(isnan(C)) = 0;

This code is perhaps easier and computationally more efficient.

Answer 4

[per isakson]

I like this better. It avoids dividing by zero.

ispos = B>0
C = A;
C(:,ispos) = A(:,ispos) ./ B(ispos)

Requires "Implicit expansion", which was introduced in R2016b

[Jos (10584)]

Without logical indexing (and for positive values only):

C = nanmax(A ./ B, 0)

Answer 5

[Vikram Gupta]

Issue here is divide by 0 is not defined. But if the extremely small change is introduced

>>  B=[0 4 1 0 4];
>> B = B +(10^-10)
>>  A ./ B
ans =
         0    0.5000         0         0         0
         0    0.2500    1.0000         0    0.2500
         0    0.2500         0         0    0.5000
         0         0         0         0    0.2500
         0         0         0         0         0

[Jos (10584)]

Mathematically, this will give the wrong result ...