How to make numpy array column sum up to 1

Question

I am working on building a transition matrix for implementing the PageRank algorithm. How could I use numpy to make sure that the columns add up to one.

For example:

1 1 1   
1 1 1  
1 1 1

should be normalized to be

.33 .33 .33  
.33 .33 .33  
.33 .33 .33

Answer 1

Divide the elements of each column by their column-summations -

a/a.sum(axis=0,keepdims=1) # or simply : a/a.sum(0)

For making the row-summations unity, change the axis input -

a/a.sum(axis=1,keepdims=1)

Sample run -

In [78]: a = np.random.rand(4,5)

In [79]: a
Out[79]: 
array([[ 0.37,  0.74,  0.36,  0.41,  0.44],
       [ 0.51,  0.86,  0.91,  0.03,  0.76],
       [ 0.56,  0.46,  0.01,  0.86,  0.38],
       [ 0.72,  0.66,  0.56,  0.84,  0.69]])

In [80]: b = a/a.sum(axis=0,keepdims=1)

In [81]: b.sum(0) # Verify
Out[81]: array([ 1.,  1.,  1.,  1.,  1.])

To make sure it works on int arrays as well for Python 2.x, use from __future__ import division or use np.true_divide.

---

For columns adding upto 0

For columns that add upto 0, assuming that we are okay with keeping them as they are, we can set the summations to 1, rather than divide by 0, like so -

sums = a.sum(axis=0,keepdims=1); 
sums[sums==0] = 1
out = a/sums

Answer 2

for i in range(len(A[0])):
    col_sum = A[:, i].sum()
    if col_sum != 0:
        A[:, i] = A[:, i]/col_sum
    else: 
        pass

The for loop is a bit sloppy and I'm sure there's a much more elegant way but it works.
Replace pass with A[:, i] = 1/len(A[0]) to eliminate dangling nodes and make the matrix column stocastic.