python - How to normalize an array in NumPy to a unit vector?

Question

I would like to convert a NumPy array to a unit vector. More specifically, I am looking for an equivalent version of this function

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

Is there something like that in skearn or numpy?

This function works in a situation where v is the 0 vector.

Answer 1

If you're using scikit-learn you can use sklearn.preprocessing.normalize:

import numpy as np
from sklearn.preprocessing import normalize

x = np.random.rand(1000)*10
norm1 = x / np.linalg.norm(x)
norm2 = normalize(x[:,np.newaxis], axis=0).ravel()
print np.all(norm1 == norm2)
# True

Answer 2

I would agree that it were nice if such a function was part of the included batteries. But it isn't, as far as I know. Here is a version for arbitrary axes, and giving 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)))

Answer 3

You can specify ord to get the L1 norm. To avoid zero division I use eps, but that's maybe not great.

def normalize(v):
    norm=np.linalg.norm(v, ord=1)
    if norm==0:
        norm=np.finfo(v.dtype).eps
    return v/norm

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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))

Answer 8

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.

Answer 9

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]]

Answer 10

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.]])

Answer 11

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 FloatTensor
from torch.nn.functional import normalize

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

Answer 12

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

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