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

## Question:

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.

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.

I think this should work,

``````a = numpy.arange(0,27.,3).reshape(3,3)

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

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.

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.])
``````

it appears that this also works

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

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.

You could also use matrix transposition:

``````(a.T / row_sums).T
``````

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.

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

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]])
``````

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
``````

Use

``````a = a / np.linalg.norm(a, ord = 2, axis = 0, keepdims = True)
``````

Due to the broadcasting, it will work as intended.