How to normalize a 2-dimensional numpy array in python less verbose?

Given a 3 times 3 numpy array

a = numpy.arange(0,27,3).reshape(3,3) # array([[ 0, 3, 6], # [ 9, 12, 15], # [18, 21, 24]]) 

To normalize the rows of the 2-dimensional array I thought of

row_sums = a.sum(axis=1) # array([ 9, 36, 63]) new_matrix = numpy.zeros((3,3)) for i, (row, row_sum) in enumerate(zip(a, row_sums)): new_matrix[i,:] = row / row_sum 

There must be a better way, isn't there?

Perhaps to clearify: By normalizing I mean, the sum of the entrys per row must be one. But I think that will be clear to most people.

11 Answers

Broadcasting is really good for this:

row_sums = a.sum(axis=1) new_matrix = a / row_sums[:, numpy.newaxis] 

row_sums[:, numpy.newaxis] reshapes row_sums from being (3,) to being (3, 1). When you do a / b, a and b are broadcast against each other.

You can learn more about broadcasting here or even better here.

Scikit-learn offers a function normalize() that lets you apply various normalizations. The "make it sum to 1" is called L1-norm. Therefore:

from sklearn.preprocessing import normalize matrix = numpy.arange(0,27,3).reshape(3,3).astype(numpy.float64) # array([[ 0., 3., 6.], # [ 9., 12., 15.], # [ 18., 21., 24.]]) normed_matrix = normalize(matrix, axis=1, norm='l1') # [[ 0. 0.33333333 0.66666667] # [ 0.25 0.33333333 0.41666667] # [ 0.28571429 0.33333333 0.38095238]] 

Now your rows will sum to 1.

I think this should work,

a = numpy.arange(0,27.,3).reshape(3,3) a /= a.sum(axis=1)[:,numpy.newaxis] 

In case you are trying to normalize each row such that its magnitude is one (i.e. a row's unit length is one or the sum of the square of each element in a row is one):

import numpy as np a = np.arange(0,27,3).reshape(3,3) result = a / np.linalg.norm(a, axis=-1)[:, np.newaxis] # array([[ 0. , 0.4472136 , 0.89442719], # [ 0.42426407, 0.56568542, 0.70710678], # [ 0.49153915, 0.57346234, 0.65538554]]) 

Verifying:

np.sum( result**2, axis=-1 ) # array([ 1., 1., 1.]) 

I think you can normalize the row elements sum to 1 by this: new_matrix = a / a.sum(axis=1, keepdims=1). And the column normalization can be done with new_matrix = a / a.sum(axis=0, keepdims=1). Hope this can hep.

it appears that this also works

def normalizeRows(M): row_sums = M.sum(axis=1) return M / row_sums 

You could use built-in numpy function: np.linalg.norm(a, axis = 1, keepdims = True)

You could also use matrix transposition:

(a.T / row_sums).T 

Here is one more possible way using reshape:

a_norm = (a/a.sum(axis=1).reshape(-1,1)).round(3) print(a_norm) 

Or using None works too:

a_norm = (a/a.sum(axis=1)[:,None]).round(3) print(a_norm) 

Output:

array([[0. , 0.333, 0.667], [0.25 , 0.333, 0.417], [0.286, 0.333, 0.381]]) 

Or using lambda function, like

>>> vec = np.arange(0,27,3).reshape(3,3) >>> import numpy as np >>> norm_vec = map(lambda row: row/np.linalg.norm(row), vec) 

each vector of vec will have a unit norm.

We can achieve the same effect by premultiplying with the diagonal matrix whose main diagonal is the reciprocal of the row sums.

A = np.diag(A.sum(1)**-1) @ A