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]])
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]])
axis=1. Without the axis argument, sum() returns the sum of all the values in the array.
keepdims argument that lets you do e/e.sum(axis=1, keepdims=True)
1 part, I said you could drop the axis= part.
[:,None] notation? I see the change it makes, but don't get the coding convention.You can do it mathematically as 
.
Here, E is your original matrix and D is a diagonal matrix where each entry is the sum of the corresponding row in E. If you're lucky enough to have an invertible D, this is a pretty mathematically convenient way to do things.
In numpy:
import numpy as np
diagonal_entries = [sum(e[row]) for row in range(e.shape[0])]
D = np.diag(diagonal_entries)
D_inv = np.linalg.inv(D)
e = np.dot(e, D_inv)
You can also use reshape method of numpy as follows:
e = np.array([[0., 1.],[2., 4.],[1., 5.]])
e/=np.sum(e, axis=1).reshape(-1,1)
e
array([[0. , 1. ],
[0.33333333, 0.66666667],
[0.16666667, 0.83333333]])
