问题:如何在NumPy中标准化数组?
我想拥有一个NumPy数组的规范。更具体地说,我正在寻找此功能的等效版本
def normalize(v):
norm = np.linalg.norm(v)
if norm == 0:
return v
return v / norm
skearn
或中有类似的东西numpy
吗?
该函数在v
向量为0 的情况下起作用。
I would like to have the norm of one NumPy array. 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.
回答 0
如果您使用的是scikit-learn,则可以使用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
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
回答 1
我同意,如果这样的功能是随附电池的一部分,那就太好了。但据我所知不是。这是适用于任意轴并提供最佳性能的版本。
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)))
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)))
回答 2
您可以指定ord以获得L1范数。为了避免零除,我使用了eps,但这可能不是很好。
def normalize(v):
norm=np.linalg.norm(v, ord=1)
if norm==0:
norm=np.finfo(v.dtype).eps
return v/norm
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
回答 3
这可能也适合您
import numpy as np
normalized_v = v / np.sqrt(np.sum(v**2))
但在v
长度为0 时失败。
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.
回答 4
如果您具有多维数据,并且希望将每个轴归一化为其最大值或总和:
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
使用numpys 峰到峰功能。
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
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
回答 5
Christoph Gohlke unit_vector()
在流行的转换模块中还具有将向量标准化的功能:
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))
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))
回答 6
您提到了sci-kit学习,所以我想分享另一个解决方案。
科学工具学习 MinMaxScaler
在sci-kit学习中,有一个名为的API MinMaxScaler
,可以根据需要自定义值范围。
它还为我们处理了NaN问题。
将NaN视为缺失值:忽略适合度,并保留其变换值。…参见参考文献[1]
代码样例
代码很简单,只需键入
# 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)
参考
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
回答 7
没有sklearn
和使用正义numpy
。只需定义一个函数即可:
假设行是变量,列是样本(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)
输出:
X
array([[1, 2, 3],
[4, 5, 6]])
stdmtx(X)
array([[-1., 0., 1.],
[-1., 0., 1.]])
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.]])
回答 8
如果要标准化存储在3D张量中的n维特征向量,则也可以使用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()
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()
回答 9
如果您正在使用3D向量,则可以使用toolbelt vg简洁地执行此操作。它是numpy之上的一个轻层,它支持单个值和堆叠的向量。
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
我在上次启动时创建了该库,它的使用动机如下:简单的想法在NumPy中太冗长了。
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.
回答 10
如果您不需要极高的精度,则可以将函数简化为:
v_norm = v / (np.linalg.norm(v) + 1e-16)
If you don’t need utmost precision, your function can be reduced to:
v_norm = v / (np.linalg.norm(v) + 1e-16)
回答 11
如果使用多维数组,则可以快速解决。
假设我们有2D数组,我们想通过最后一个轴对其进行归一化,而有些行的范数为零。
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 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]]
声明:本站所有文章,如无特殊说明或标注,均为本站原创发布。任何个人或组织,在未征得本站同意时,禁止复制、盗用、采集、发布本站内容到任何网站、书籍等各类媒体平台。如若本站内容侵犯了原著者的合法权益,可联系我们进行处理。