wwl - 10 months ago 71

Python Question

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.

Answer Source

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
```