How to normalize a numpy array to a unit vector

I agree that it would be nice if such a function were part of the included libraries. But it isn't, as far as I know. So here is a version for arbitrary axes that gives optimal performance.

import numpy as np

def normalized(a, axis=-1, order=2):
    l2 = np.atleast_1d(np.linalg.norm(a, order, axis))
    l2[l2==0] = 1
    return a / np.expand_dims(l2, axis)

A = np.random.randn(3,3,3)
print(normalized(A,0))
print(normalized(A,1))
print(normalized(A,2))

print(normalized(np.arange(3)[:,None]))
print(normalized(np.arange(3)))

This might also work for you

import numpy as np
normalized_v = v / np.sqrt(np.sum(v**2))

but fails when v has length 0.

In that case, introducing a small constant to prevent the zero division solves this.

As proposed in the comments one could also use

v/np.linalg.norm(v)

To avoid zero division I use eps, but that's maybe not great.

def normalize(v):
    norm=np.linalg.norm(v)
    if norm==0:
        norm=np.finfo(v.dtype).eps
    return v/norm

If you don't need utmost precision, your function can be reduced to:

v_norm = v / (np.linalg.norm(v) + 1e-16)

If you have multidimensional data and want each axis normalized to its max or its sum:

def normalize(_d, to_sum=True, copy=True):
    # d is a (n x dimension) np array
    d = _d if not copy else np.copy(_d)
    d -= np.min(d, axis=0)
    d /= (np.sum(d, axis=0) if to_sum else np.ptp(d, axis=0))
    return d

Uses numpys peak to peak function.

a = np.random.random((5, 3))

b = normalize(a, copy=False)
b.sum(axis=0) # array([1., 1., 1.]), the rows sum to 1

c = normalize(a, to_sum=False, copy=False)
c.max(axis=0) # array([1., 1., 1.]), the max of each row is 1

There is also the function unit_vector() to normalize vectors in the popular transformations module by Christoph Gohlke:

import transformations as trafo
import numpy as np

data = np.array([[1.0, 1.0, 0.0],
                 [1.0, 1.0, 1.0],
                 [1.0, 2.0, 3.0]])

print(trafo.unit_vector(data, axis=1))

You mentioned sci-kit learn, so I want to share another solution.

sci-kit learn MinMaxScaler

In sci-kit learn, there is a API called MinMaxScaler which can customize the the value range as you like.

It also deal with NaN issues for us.

NaNs are treated as missing values: disregarded in fit, and maintained in transform. ... see reference [1]

Code sample

The code is simple, just type

# Let's say X_train is your input dataframe
from sklearn.preprocessing import MinMaxScaler
# call MinMaxScaler object
min_max_scaler = MinMaxScaler()
# feed in a numpy array
X_train_norm = min_max_scaler.fit_transform(X_train.values)
# wrap it up if you need a dataframe
df = pd.DataFrame(X_train_norm)

• [1] sklearn.preprocessing.MinMaxScaler

If you work with multidimensional array following fast solution is possible.

Say we have 2D array, which we want to normalize by last axis, while some rows have zero norm.

import numpy as np
arr = np.array([
    [1, 2, 3], 
    [0, 0, 0],
    [5, 6, 7]
], dtype=np.float)

lengths = np.linalg.norm(arr, axis=-1)
print(lengths)  # [ 3.74165739  0.         10.48808848]
arr[lengths > 0] = arr[lengths > 0] / lengths[lengths > 0][:, np.newaxis]
print(arr)
# [[0.26726124 0.53452248 0.80178373]
# [0.         0.         0.        ]
# [0.47673129 0.57207755 0.66742381]]

If you want to normalize n dimensional feature vectors stored in a 3D tensor, you could also use PyTorch:

import numpy as np
from torch import from_numpy
from torch.nn.functional import normalize

vecs = np.random.rand(3, 16, 16, 16)
norm_vecs = normalize(from_numpy(vecs), dim=0, eps=1e-16).numpy()

If you're working with 3D vectors, you can do this concisely using the toolbelt vg. It's a light layer on top of numpy and it supports single values and stacked vectors.

import numpy as np
import vg

x = np.random.rand(1000)*10
norm1 = x / np.linalg.norm(x)
norm2 = vg.normalize(x)
print np.all(norm1 == norm2)
# True

I created the library at my last startup, where it was motivated by uses like this: simple ideas which are way too verbose in NumPy.

Without sklearn and using just numpy. Just define a function:.

Assuming that the rows are the variables and the columns the samples (axis= 1):

import numpy as np

# Example array
X = np.array([[1,2,3],[4,5,6]])

def stdmtx(X):
    means = X.mean(axis =1)
    stds = X.std(axis= 1, ddof=1)
    X= X - means[:, np.newaxis]
    X= X / stds[:, np.newaxis]
    return np.nan_to_num(X)

output:

X
array([[1, 2, 3],
       [4, 5, 6]])

stdmtx(X)
array([[-1.,  0.,  1.],
       [-1.,  0.,  1.]])

For a 2D array, you can use the following one-liner to normalize across rows. To normalize across columns, simply set axis=0.

a / np.linalg.norm(a, axis=1, keepdims=True)

If you want all values in [0; 1] for 1d-array then just use

(a - a.min(axis=0)) / (a.max(axis=0) - a.min(axis=0))

Where a is your 1d-array.

An example:

>>> a = np.array([0, 1, 2, 4, 5, 2])
>>> (a - a.min(axis=0)) / (a.max(axis=0) - a.min(axis=0))
array([0. , 0.2, 0.4, 0.8, 1. , 0.4])

Note for the method. For saving proportions between values there is a restriction: 1d-array must have at least one 0 and consists of 0 and positive numbers.

A simple dot product would do the job. No need for any extra package.

x = x/np.sqrt(x.dot(x))

By the way, if the norm of x is zero, it is inherently a zero vector, and cannot be converted to a unit vector (which has norm 1). If you want to catch the case of np.array([0,0,...0]), then use

norm = np.sqrt(x.dot(x))
x = x/norm if norm != 0 else x

Unfortunately the simple solution x/numpy.linalg.norm(x) doesn't work if x is an array of vectors. But with a simple reshape() you can force it into a flat list, use a list comprehension, and use reshape() again to get back the original shape.

s=x.shape
np.array([ v/np.linalg.norm(v)  for v in x.reshape(-1, s[-1])]).reshape(s)

First we store the shape of the array

s=x.shape

Then we reshape it into a simple (one-dimensional) array of vectors

x.reshape(-1, s[-1])

by making use of the '-1' argument of reshape() which essentially means "take as many as it needs", e.g . if x was a (4,5,3) array, x.reshape(-1,3) would be of shape (20,3). The use of s[-1] allows for an arbitrary dimension of the vectors.

Then we use a list comprehension to step through the array and calculate the unit vector one vector at a time

[ v/np.linalg.norm(v)  for v in x.reshape(-1, s[-1])]

and finally we turn it back into an numpy array and give it back its original shape

np.array([ v/np.linalg.norm(v)  for v in x.reshape(-1, s[-1])]).reshape(s)