blocks|key|963571|text|corrcoef返回归一化协方差矩阵。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|963572|协方差矩阵就是矩阵|963573|Cov(+X,+X+)++++Cov(+X,+Y+)

Cov(+Y,+X+)++++Cov(+Y,+Y+)|code-block|syntax|javascript|963574|归一化后，这将产生矩阵：|963575|Corr(+X,+X+)++++Corr(+X,+Y+)

Corr(+Y,+X+)++++Corr(+Y,+Y+)|963576|correlation1[0,+0+]是Strategy1Returns与自身的关系，必须为1，您只需要correlation1[+0,+1+]。|963577|entityMap|0|LINK|mutability|MUTABLE|url|http://www.scipy.org/Numpy_Example_List_With_Doc#corrcoef^0|0|8|0|8|0|0|0|0|0|0|0|J|K|G|1G|K|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|17|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|18|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|19|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

<a href="http://www.scipy.org/Numpy_Example_List_With_Doc#corrcoef" rel="noreferrer"><code>corrcoef</code></a> returns the normalised covariance matrix.

The covariance matrix is the matrix

<pre><code>Cov( X, X ) Cov( X, Y )

Cov( Y, X ) Cov( Y, Y )
</code></pre>

Normalised, this will yield the matrix:

<pre><code>Corr( X, X ) Corr( X, Y )

Corr( Y, X ) Corr( Y, Y )
</code></pre>

<code>correlation1[0, 0 ]</code> is the correlation between <code>Strategy1Returns</code> and itself, which must be 1. You just want <code>correlation1[ 0, 1 ]</code>.

blocks|key|1177847|text|它允许您计算>2个数据集的相关系数，例如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177848|>>>+from+numpy+import+*
>>>+a+=+array([1,2,3,4,6,7,8,9])
>>>+b+=+array([2,4,6,8,10,12,13,15])
>>>+c+=+array([-1,-2,-2,-3,-4,-6,-7,-8])
>>>+corrcoef([a,b,c])
array([[+1.++++++++,++0.99535001,+-0.9805214+],
+++++++[+0.99535001,++1.++++++++,+-0.97172394],
+++++++[-0.9805214+,+-0.97172394,++1.++++++++]])|code-block|syntax|javascript|1177849|这里我们可以一次得到a，b+(0.995)，a，c+(-0.981)和b，c+(-0.972)的相关系数。两个数据集的情况只是N数据集类的一个特例。而且可能保持相同的返回类型会更好。由于"one+value“可以简单地通过以下命令获得|1177850|>>>+corrcoef(a,b)[1,0]
0.99535001355530017|1177851|没有什么重要的理由去创造这种特殊的情况。|1177852|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

It allows you to compute correlation coefficients of >2 data sets, e.g.

<pre><code>&gt;&gt;&gt; from numpy import *
&gt;&gt;&gt; a = array([1,2,3,4,6,7,8,9])
&gt;&gt;&gt; b = array([2,4,6,8,10,12,13,15])
&gt;&gt;&gt; c = array([-1,-2,-2,-3,-4,-6,-7,-8])
&gt;&gt;&gt; corrcoef([a,b,c])
array([[ 1. , 0.99535001, -0.9805214 ],
 [ 0.99535001, 1. , -0.97172394],
 [-0.9805214 , -0.97172394, 1. ]])
</code></pre>

Here we can get the correlation coefficient of a,b (0.995), a,c (-0.981) and b,c (-0.972) at once. The two-data-set case is just a special case of N-data-set class. And probably it's better to keep the same return type. Since the "one value" can be obtained simply with

<pre><code>&gt;&gt;&gt; corrcoef(a,b)[1,0]
0.99535001355530017
</code></pre>

there's no big reason to create the special case.

blocks|key|1177881|text|考虑使用matplotlib.cbook片段|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1177882|例如：|1177883|import+matplotlib.cbook+as+cbook
segments+=+cbook.pieces(np.arange(20),+3)
for+s+in+segments:
+++++print+s|code-block|syntax|javascript|1177884|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Consider using matplotlib.cbook pieces

for example:

<pre><code>import matplotlib.cbook as cbook
segments = cbook.pieces(np.arange(20), 3)
for s in segments:
 print s
</code></pre>

blocks|key|963680|text|numpy的Correlate函数处理2个要关联的1D数组，并返回一个相关值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|963681|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The function Correlate of numpy works with 2 1D arrays that you want to correlate and returns one correlation value.

blocks|key|963713|text|您可以使用以下函数仅返回相关系数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|963714|def+pearson_r(x,+y):
"""Compute+Pearson+correlation+coefficient+between+two+arrays."""

+++#+Compute+correlation+matrix
+++corr_mat+=+np.corrcoef(x,+y)

+++#+Return+entry+[0,1]
+++return+corr_mat[0,1]|code-block|syntax|javascript|963715|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can use the following function to return only the correlation coefficient:

<pre><code>def pearson_r(x, y):
"""Compute Pearson correlation coefficient between two arrays."""

 # Compute correlation matrix
 corr_mat = np.corrcoef(x, y)

 # Return entry [0,1]
 return corr_mat[0,1]
</code></pre>

It seems strange to me that np.corrcoef returns a matrix.

<pre><code> correlation1 = corrcoef(Strategy1Returns,Strategy2Returns)

[[ 1. -0.99598935]
 [-0.99598935 1. ]]
</code></pre>

Does anyone know why this is the case and whether it is possible to return just one value in the classical sense?

Why does corrcoef return a matrix?

对我来说，np.corrcoef返回一个矩阵似乎很奇怪。 correlation1 = corrcoef(Strategy1Returns,Strategy2Returns)[[ 1.         -0.99598935] [-0.99598935  1.        ]]有没有人知道为什么会这样，是否有可能只返...

问为什么corrcoef返回一个矩阵？
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