规范化矩阵 python 的行
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/36267936/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Normalizing rows of a matrix python
提问by Yas
Given a 2-dimensional array in python, I would like to normalize each row with the following norms:
给定 python 中的二维数组，我想使用以下规范对每一行进行标准化：
- Norm 1: L_1
- Norm 2: L_2
- Norm Inf: L_Inf
- 范数 1：L_1
- 范数 2：L_2
- 规范信息：L_Inf
I have started this code:
我已经开始这个代码：
from numpy import linalg as LA
X = np.array([[1, 2, 3, 6],
[4, 5, 6, 5],
[1, 2, 5, 5],
[4, 5,10,25],
[5, 2,10,25]])
print X.shape
x = np.array([LA.norm(v,ord=1) for v in X])
print x
Output:
输出：
(5, 4) # array dimension
[12 20 13 44 42] # L1 on each Row
How can I modify the code such that WITHOUT using LOOP, I can directly have the rows of the matrix normalized? (Given the norm values above)
如何修改代码，以便不使用 LOOP，我可以直接对矩阵的行进行归一化？（鉴于上面的标准值）
I tried :
我试过 ：
l1 = X.sum(axis=1)
print l1
print X/l1.reshape(5,1)
[12 20 13 44 42]
[[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]]
but the output is zero.
但输出为零。
回答by wim
This is the L? norm:
这是L？规范：
>>> np.abs(X).sum(axis=1)
array([12, 20, 13, 44, 42])
This is the L? norm:
这是L？规范：
>>> np.sqrt((X * X).sum(axis=1))
array([ 7.07106781, 10.09950494, 7.41619849, 27.67670501, 27.45906044])
This is the L∞ norm:
这是 L∞ 范数：
>>> np.abs(X).max(axis=1)
array([ 6, 6, 5, 25, 25])
To normalise rows, just divide by the norm. For example, using L? normalisation:
要规范化行，只需除以规范。例如，使用 L? 正常化：
>>> l2norm = np.sqrt((X * X).sum(axis=1))
>>> X / l2norm.reshape(5,1)
array([[ 0.14142136, 0.28284271, 0.42426407, 0.84852814],
[ 0.39605902, 0.49507377, 0.59408853, 0.49507377],
[ 0.13483997, 0.26967994, 0.67419986, 0.67419986],
[ 0.14452587, 0.18065734, 0.36131469, 0.90328672],
[ 0.18208926, 0.0728357 , 0.36417852, 0.9104463 ]])
>>> np.sqrt((_ * _).sum(axis=1))
array([ 1., 1., 1., 1., 1.])
More direct is the normmethod in numpy.linalg, if you have it available:
更直接的是 中的norm方法numpy.linalg，如果你有的话：
>>> from numpy.linalg import norm
>>> norm(X, axis=1, ord=1) # L-1 norm
array([12, 20, 13, 44, 42])
>>> norm(X, axis=1, ord=2) # L-2 norm
array([ 7.07106781, 10.09950494, 7.41619849, 27.67670501, 27.45906044])
>>> norm(X, axis=1, ord=np.inf) # L-∞ norm
array([ 6, 6, 5, 25, 25])
(after OP edit):You saw zero values because /is an integer division in Python 2.x. Either upgrade to Python 3, or change dtype to float to avoid that integer division:
（在 OP 编辑之后）：您看到零值，因为/是 Python 2.x 中的整数除法。要么升级到 Python 3，要么将 dtype 更改为 float 以避免整数除法：
>>> linfnorm = norm(X, axis=1, ord=np.inf)
>>> X.astype(np.float) / linfnorm[:,None]
array([[ 0.16666667, 0.33333333, 0.5 , 1. ],
[ 0.66666667, 0.83333333, 1. , 0.83333333],
[ 0.2 , 0.4 , 1. , 1. ],
[ 0.16 , 0.2 , 0.4 , 1. ],
[ 0.2 , 0.08 , 0.4 , 1. ]])
回答by ayhan
You can pass axis=1parameter:
您可以传递axis=1参数：
In [58]: LA.norm(X, axis=1, ord=1)
Out[58]: array([12, 20, 13, 44, 42])
In [59]: LA.norm(X, axis=1, ord=2)
Out[59]: array([ 7.07106781, 10.09950494, 7.41619849, 27.67670501, 27.45906044])
