numpy divide row by row sum

Stefan Profanter picture Stefan Profanter · Apr 24, 2013 · Viewed 37.8k times · Source

How can I divide a numpy array row by the sum of all values in this row?

This is one example. But I'm pretty sure there is a fancy and much more efficient way of doing this:

import numpy as np
e = np.array([[0., 1.],[2., 4.],[1., 5.]])
for row in xrange(e.shape[0]):
    e[row] /= np.sum(e[row])

Result:

array([[ 0.        ,  1.        ],
       [ 0.33333333,  0.66666667],
       [ 0.16666667,  0.83333333]])

Answer

DSM picture DSM · Apr 24, 2013

Method #1: use None (or np.newaxis) to add an extra dimension so that broadcasting will behave:

>>> e
array([[ 0.,  1.],
       [ 2.,  4.],
       [ 1.,  5.]])
>>> e/e.sum(axis=1)[:,None]
array([[ 0.        ,  1.        ],
       [ 0.33333333,  0.66666667],
       [ 0.16666667,  0.83333333]])

Method #2: go transpose-happy:

>>> (e.T/e.sum(axis=1)).T
array([[ 0.        ,  1.        ],
       [ 0.33333333,  0.66666667],
       [ 0.16666667,  0.83333333]])

(You can drop the axis= part for conciseness, if you want.)

Method #3: (promoted from Jaime's comment)

Use the keepdims argument on sum to preserve the dimension:

>>> e/e.sum(axis=1, keepdims=True)
array([[ 0.        ,  1.        ],
       [ 0.33333333,  0.66666667],
       [ 0.16666667,  0.83333333]])