blocks|key|1093338|text|矩阵X的索引是行(i)和列(j)。因此，X,j表示矩阵X的列j。即将矩阵X的每一列归一化为z-scores。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1093339|您可以使用pandas来完成此操作：|1093340|df_new_zscores+=+(df_new+-+df_new.mean())+/+df_new.std()|code-block|syntax|javascript|1093341|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Standard_score^0|K|3|R|1|U|1|19|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

The indexes for matrix X are row (i) and column (j). Hence, <code>X,j</code> means column <code>j</code> of matrix <code>X</code>. I.e. normalize each column of matrix X to <a href="https://en.wikipedia.org/wiki/Standard_score" rel="nofollow noreferrer">z-scores</a>.

You can do that using pandas:

<pre><code>df_new_zscores = (df_new - df_new.mean()) / df_new.std()
</code></pre>

blocks|key|997822|text|实际上，您不需要自己编写规范化代码-它已经为sklearn.preprocessing.scale准备好了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|997823|下面是一个from+the+docs示例|997824|>>>+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...]])|code-block|syntax|javascript|997825|当与默认设置axis=0一起使用时，正规化按列发生(即，对于每个列j，就像您的查询一样)。因此，很容易确认缩放数据的均值和单位方差为零：|997826|>>>+X_scaled.mean(axis=0)
array([+0.,++0.,++0.])

>>>+X_scaled.std(axis=0)
array([+1.,++1.,++1.])|997827|entityMap|0|LINK|mutability|MUTABLE|url|http://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.scale.html|1|http://scikit-learn.org/stable/modules/preprocessing.html#preprocessing-scaler^0|M|R|M|R|0|0|5|D|1|0|0|6|6|X|1|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@$9|17|A|18|1|19]]|E|$]]|$1|H|3|I|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1G|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]]]

You don't actually need to write code for the normalization yourself - it comes ready with <a href="http://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.scale.html" rel="nofollow noreferrer"><code>sklearn.preprocessing.scale</code></a>.

Here is an example <a href="http://scikit-learn.org/stable/modules/preprocessing.html#preprocessing-scaler" rel="nofollow noreferrer">from the docs</a>:

<pre><code>&gt;&gt;&gt; from sklearn import preprocessing
&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; X_train = np.array([[ 1., -1., 2.],
... [ 2., 0., 0.],
... [ 0., 1., -1.]])
&gt;&gt;&gt; X_scaled = preprocessing.scale(X_train)

&gt;&gt;&gt; X_scaled 
array([[ 0. ..., -1.22..., 1.33...],
 [ 1.22..., 0. ..., -0.26...],
 [-1.22..., 1.22..., -1.06...]])
</code></pre>

When used with the default setting <code>axis=0</code>, the mormalization happens column-wise (i.e. for each column <code>j</code>, as in your equestion). As a result, it is easy to confirm that scaled data has zero mean and unit variance:

<pre><code>&gt;&gt;&gt; X_scaled.mean(axis=0)
array([ 0., 0., 0.])

&gt;&gt;&gt; X_scaled.std(axis=0)
array([ 1., 1., 1.])
</code></pre>

blocks|key|997829|text|我不知道熊猫，但我认为这个等式意味着归一化矩阵由下式给出|type|unstyled|depth|inlineStyleRanges|entityRanges|data|997830|?|atomic|offset|length|997831|你减去经验平均值，然后除以每列的经验标准差。|997832|有时，您可以将其用于主成分分析。|997833|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/f17252a3b39ac7c9d1f04eead92282d5.png|imageAlt^0|0|0|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@$E|V|F|W|1|X]]|A|$]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R|S|-4]]]]

I do not know pandas but I think that the equation means that the normalized matrix is given by
<a href="https://i.stack.imgur.com/TQOzP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TQOzP.png" alt="enter image description here"></a>
You subtract the empirical mean and devide by the empirical standard deviation per column.

You sometimes use this for Principal Component Analysis.

I have a 150x4 matrix <code>X</code> which I created from a pandas dataframe using the following code:

<pre><code>X = df_new.as_matrix()
</code></pre>

I have to normalize it using this function:

<a href="https://i.stack.imgur.com/3Ziqh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Ziqh.png" alt="this"></a>

I know that <code>Uj</code> is the mean val of <code>j</code>, and that <code>σ j</code> is the standard deviation of <code>j</code>, but I don't understand what <code>j</code> is. I'm having a little trouble understanding what the bar on <code>X</code> is, and I'm confused by the commas in the equation (I don't know if they have any significance or not). 

Can anyone help me understand what this equation means so I can then write the normalization using sklearn?

Normalization of a matrix

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个150x4的矩阵X，它是我使用以下代码从熊猫数据帧创建的：X = df_new.as_matrix()我必须使用这个函数对其进行规范化：​​我知道Uj是j的平均值，σ j是j的标准差，但我不明白j是什么。我在理解X上的栏是什么时遇到了一点麻烦，我被方程式中的逗号搞糊涂了(我不知道它们是否有任何意义)。有人能帮我...

问矩阵的正规化
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