Python 如何标准化矩阵？
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How do I standardize a matrix?
提问by pnodbnda
Basically, take a matrix and change it so that its mean is equal to 0 and variance is 1. I'm using numpy's arrays so if it can already do it it's better, but I can implement it myself as long as I can find an algorithm.
基本上，取一个矩阵并更改它，使其均值等于 0，方差为 1。我正在使用 numpy 的数组，所以如果它已经可以做到它会更好，但我可以自己实现它，只要我能找到一个算法。
edit: nvm nimrodm has a better implementation
编辑：nvm nimrodm 有更好的实现
采纳答案by ja72
Take each element and subtract with the mean and then divide by the standard deviation.
取每个元素并减去均值，然后除以标准差。
Shoot me, I don't know python. In general the above is
拍我，我不懂蟒蛇。一般来说，以上是
mu = Average()
sig = StandardDeviation()
for(i=0;i<rows;i++)
{
for(j=0;j<cols;j++)
{
A[i,j] = (A[i,j]-mu)/sig;
}
}
回答by nimrodm
The following subtracts the mean of A from each element (the new mean is 0), then normalizes the result by the standard deviation.
下面从每个元素中减去 A 的平均值（新的平均值为 0），然后通过标准偏差对结果进行归一化。
from numpy import *
A = (A - mean(A)) / std(A)
The above is for standardizing the entire matrix as a whole, If A has many dimensions and you want to standardize each column individually, specify the axis:
以上是将整个矩阵作为一个整体进行标准化，如果 A 有很多维度，并且您想单独标准化每一列，请指定轴：
from numpy import *
A = (A - mean(A, axis=0)) / std(A, axis=0)
Always verify by hand what these one-liners are doing before integrating them into your code. A simple change in orientation or dimension can drastically change (silently) what operations numpy performs on them.
在将它们集成到您的代码中之前，请始终手动验证这些单行代码在做什么。方向或维度的简单变化可以（悄悄地）彻底改变 numpy 对它们执行的操作。
回答by AmanRaj
import scipy.stats as ss
A = np.array(ss.zscore(A))
回答by DoesData
from sklearn.preprocessing import StandardScaler
standardized_data = StandardScaler().fit_transform(your_data)
Example:
例子：
>>> import numpy as np
>>> from sklearn.preprocessing import StandardScaler
>>> data = np.random.randint(25, size=(4, 4))
>>> data
array([[17, 12, 4, 17],
[ 1, 16, 19, 1],
[ 7, 8, 10, 4],
[22, 4, 2, 8]])
>>> standardized_data = StandardScaler().fit_transform(data)
>>> standardized_data
array([[ 0.63812398, 0.4472136 , -0.718646 , 1.57786412],
[-1.30663482, 1.34164079, 1.55076242, -1.07959124],
[-0.57735027, -0.4472136 , 0.18911737, -0.58131836],
[ 1.24586111, -1.34164079, -1.02123379, 0.08304548]])
Works well on large datasets.
在大型数据集上运行良好。
回答by Yuya Takashina
Use sklearn.preprocessing.scale.
使用sklearn.preprocessing.scale.
http://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.scale.html
http://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.scale.html
Here is an example.
这是一个例子。
>>> from sklearn import preprocessing
>>> import numpy as np
>>> X_train = np.array([[ 1., -1., 2.],
... [ 2., 0., 0.],
... [ 0., 1., -1.]])
>>> X_scaled = preprocessing.scale(X_train)
>>> X_scaled
array([[ 0. ..., -1.22..., 1.33...],
[ 1.22..., 0. ..., -0.26...],
[-1.22..., 1.22..., -1.06...]])
回答by Alexander Drobyshevsky
import numpy as np
A = np.array([[1,2,6], [3000,1000,2000]]).T
A_means = np.mean(A, axis=0)
A_centr = A - A_means
A_norms = np.linalg.norm(A_centr, axis=0)
A_std = A_centr / A_norms
