I would like to convert a NumPy array to a unit vector. More specifically, I am looking for an equivalent version of this normalisation function:
def normalize(v):
norm = np.linalg.norm(v)
if norm == 0:
return v
return v / norm
This function handles the situation where vector v has the norm value of 0.
Is there any similar functions provided in sklearn or numpy?
Answers:
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Method 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
Method 2
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)))
Method 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.
Method 4
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
Method 5
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
Method 6
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)
Reference
Method 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))
Method 8
If you don’t need utmost precision, your function can be reduced to:
v_norm = v / (np.linalg.norm(v) + 1e-16)
Method 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]]
Method 10
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.
Method 11
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.]])
Method 12
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()
Method 13
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
Method 14
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)
Method 15
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.
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0