Normalizing a numpy array

问题:

Given an array, I want to normalize it such that each row sums to 1.

I currently have the following code:

import numpy
w = numpy.array([[0, 1, 0, 1, 0, 0], 
                 [1, 0, 0, 0, 0, 1], 
                 [0, 0, 0, 0, 0, 1], 
                 [1, 0, 0, 0, 1, 0], 
                 [0, 0, 0, 1, 0, 1], 
                 [0, 1, 1, 0, 1, 0]], dtype = float)


def rownormalize(array):
    i = 0
    for row in array:
        array[i,:] = array[i,:]/sum(row)
        i += 1

I've two questions:

1) The code works, but I'm wondering if there's a more elegant way.

2) How can I convert the data type into a float if it's int? I tried

if array.dtype == int:
    array.dtype = float

But it doesn't work.

回答1:

You can do 1) like that:

array /= array.sum(axis=1, keepdims=True)

and 2) like that:

array = array.astype(float)

回答2:

Divisions even though broadcasted across all elements could be expensive. An alternative with focus on performance, would be to pre-compute the reciprocal of row-summations and use those to perform broadcasted multiplications instead, like so -

w *= 1.0/w.sum(1,keepdims=1)

Runtime test -

In [588]: w = np.random.rand(3000,3000)

In [589]: out1 = w/w.sum(axis=1, keepdims=True) #@Julien Bernu's soln

In [590]: out2 = w*(1.0/w.sum(1,keepdims=1))

In [591]: np.allclose(out1,out2)
Out[591]: True

In [592]: %timeit w/w.sum(axis=1, keepdims=True) #@Julien Bernu's soln
10 loops, best of 3: 66.7 ms per loop

In [593]: %timeit w*(1.0/w.sum(1,keepdims=1))
10 loops, best of 3: 40 ms per loop