wwl - 1 year ago 111
Python Question

# Normalizing a numpy array

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.

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
``````
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