blocks|key|1122852|text|避免使用matrix类的主要原因是:+a)它本质上是二维的，b)与“普通”numpy数组相比有额外的开销。如果你所做的一切都是线性代数，那么请随意使用矩阵类……不过，就我个人而言，我发现它的麻烦比它的价值更大。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1122853|对于数组(Python3.5之前的版本)，请使用dot而不是matrixmultiply。|1122854|例如。|1122855|import+numpy+as+np
x+=+np.arange(9).reshape((3,3))
y+=+np.arange(3)

print+np.dot(x,y)|code-block|syntax|javascript|1122856|或者在较新版本的numpy中，只需使用x.dot(y)|1122857|就我个人而言，我发现它比暗示矩阵乘法的*运算符更具可读性……|1122858|对于Python3.5中的数组，请使用x+@+y。|1122859|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.dot.html^0|4|6|0|O|3|U|E|O|3|0|0|0|0|J|8|0|J|1|0|J|5|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@$9|1A|A|1B|1|1C]]|E|$]]|$1|H|3|I|5|6|7|1D|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1E|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|1L|8|@$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

The main reason to avoid using the <code>matrix</code> class is that a) it's inherently 2-dimensional, and b) there's additional overhead compared to a "normal" numpy array. If all you're doing is linear algebra, then by all means, feel free to use the matrix class... Personally I find it more trouble than it's worth, though.

For arrays (prior to Python 3.5), use <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.dot.html" rel="noreferrer"><code>dot</code></a> instead of <code>matrixmultiply</code>.

E.g.

<pre><code>import numpy as np
x = np.arange(9).reshape((3,3))
y = np.arange(3)

print np.dot(x,y)
</code></pre>

Or in newer versions of numpy, simply use <code>x.dot(y)</code>

Personally, I find it much more readable than the <code>*</code> operator implying matrix multiplication... 

For arrays in Python 3.5, use <code>x @ y</code>.

blocks|key|1122907|text|This+trick可能就是你要找的东西。它是一种简单的操作符重载。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1122908|然后，您可以使用类似建议的Infix类，如下所示：|1122909|a+=+np.random.rand(3,4)
b+=+np.random.rand(4,3)
x+=+Infix(lambda+x,y:+np.dot(x,y))
c+=+a+%7Cx%7C+b|code-block|syntax|javascript|1122910|entityMap|0|LINK|mutability|MUTABLE|url|http://code.activestate.com/recipes/384122-infix-operators/^0|0|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<a href="http://code.activestate.com/recipes/384122-infix-operators/">This trick</a> could be what you are looking for. It is a kind of simple operator overload.

You can then use something like the suggested Infix class like this:

<pre><code>a = np.random.rand(3,4)
b = np.random.rand(4,3)
x = Infix(lambda x,y: np.dot(x,y))
c = a |x| b
</code></pre>

blocks|key|1122961|text|在处理数组和处理矩阵时，点运算符会给出不同的答案。例如，假设如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1122962|>>>+a=numpy.array([1,+2,+3])
>>>+b=numpy.array([1,+2,+3])|code-block|syntax|javascript|1122963|让我们将它们转换为矩阵：|1122964|>>>+am=numpy.mat(a)
>>>+bm=numpy.mat(b)|1122965|现在，我们可以看到这两种情况的不同输出：|1122966|>>>+print+numpy.dot(a.T,+b)
14
>>>+print+am.T*bm
[[1.++2.++3.]
+[2.++4.++6.]
+[3.++6.++9.]]|1122967|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

There is a situation where the dot operator will give different answers when dealing with arrays as with dealing with matrices. For example, suppose the following:

<pre><code>&gt;&gt;&gt; a=numpy.array([1, 2, 3])
&gt;&gt;&gt; b=numpy.array([1, 2, 3])
</code></pre>

Lets convert them into matrices:

<pre><code>&gt;&gt;&gt; am=numpy.mat(a)
&gt;&gt;&gt; bm=numpy.mat(b)
</code></pre>

Now, we can see a different output for the two cases:

<pre><code>&gt;&gt;&gt; print numpy.dot(a.T, b)
14
&gt;&gt;&gt; print am.T*bm
[[1. 2. 3.]
 [2. 4. 6.]
 [3. 6. 9.]]
</code></pre>

blocks|key|908038|text|在3.5中，Python最终实现了got+a+matrix+multiplication+operator。语法为a+@+b。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|908039|entityMap|0|LINK|mutability|MUTABLE|url|http://legacy.python.org/dev/peps/pep-0465/^0|1L|5|H|10|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

In 3.5, Python finally <a href="http://legacy.python.org/dev/peps/pep-0465/" rel="noreferrer">got a matrix multiplication operator</a>. The syntax is <code>a @ b</code>.

blocks|key|1123073|text|函数matmul+(从numpy+1.10.1开始)可以很好地处理类型和以numpy矩阵类的形式返回结果：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1123074|import+numpy+as+np

A+=+np.mat('1+2+3;+4+5+6;+7+8+9;+10+11+12')
B+=+np.array(np.mat('1+1+1+1;+1+1+1+1;+1+1+1+1'))
print+(A,+type(A))
print+(B,+type(B))

C+=+np.matmul(A,+B)
print+(C,+type(C))|code-block|syntax|javascript|1123075|输出：|1123076|(matrix([[+1,++2,++3],
++++++++[+4,++5,++6],
++++++++[+7,++8,++9],
++++++++[10,+11,+12]]),+<class+'numpy.matrixlib.defmatrix.matrix'>)
(array([[1,+1,+1,+1],
+++++++[1,+1,+1,+1],
+++++++[1,+1,+1,+1]]),+<type+'numpy.ndarray'>)
(matrix([[+6,++6,++6,++6],
++++++++[15,+15,+15,+15],
++++++++[24,+24,+24,+24],
++++++++[33,+33,+33,+33]]),+<class+'numpy.matrixlib.defmatrix.matrix'>)|1123077|由于Python3.5作为mentioned+early，您还可以使用新的矩阵乘法运算符@，例如|style|CODE|1123078|C+=+A+@+B|1123079|并得到与上面相同的结果。|1123080|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.scipy.org/doc/numpy/reference/generated/numpy.matmul.html|1|https://stackoverflow.com/a/25476540/2666859|2|https://docs.python.org/3/whatsnew/3.5.html#whatsnew-pep-465^0|2|6|0|0|0|0|0|18|1|D|F|1|18|1|2|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@$A|17|B|18|1|19]]|C|$]]|$1|D|3|E|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1B|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1D|8|@$A|1E|B|1F|O|P]]|9|@$A|1G|B|1H|1|1I]|$A|1J|B|1K|1|1L]]|C|$]]|$1|Q|3|R|5|F|7|1M|8|@]|9|@]|C|$G|H]]|$1|S|3|T|5|6|7|1N|8|@]|9|@]|C|$]]|$1|U|3|-4|5|6|7|1O|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]|12|$5|X|Y|Z|C|$10|13]]|14|$5|X|Y|Z|C|$10|15]]]]

Function <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.matmul.html" rel="nofollow noreferrer">matmul</a> (since numpy 1.10.1) works fine for both types and return result as a numpy matrix class:

<pre><code>import numpy as np

A = np.mat('1 2 3; 4 5 6; 7 8 9; 10 11 12')
B = np.array(np.mat('1 1 1 1; 1 1 1 1; 1 1 1 1'))
print (A, type(A))
print (B, type(B))

C = np.matmul(A, B)
print (C, type(C))
</code></pre>

Output:

<pre><code>(matrix([[ 1, 2, 3],
 [ 4, 5, 6],
 [ 7, 8, 9],
 [10, 11, 12]]), &lt;class 'numpy.matrixlib.defmatrix.matrix'&gt;)
(array([[1, 1, 1, 1],
 [1, 1, 1, 1],
 [1, 1, 1, 1]]), &lt;type 'numpy.ndarray'&gt;)
(matrix([[ 6, 6, 6, 6],
 [15, 15, 15, 15],
 [24, 24, 24, 24],
 [33, 33, 33, 33]]), &lt;class 'numpy.matrixlib.defmatrix.matrix'&gt;)
</code></pre>

Since python 3.5 as <a href="https://stackoverflow.com/a/25476540/2666859">mentioned early</a> you also can use a new matrix multiplication operator <a href="https://docs.python.org/3/whatsnew/3.5.html#whatsnew-pep-465" rel="nofollow noreferrer"><code>@</code></a> like

<pre><code>C = A @ B
</code></pre>

and get the same result as above.

The numpy docs recommend using array instead of matrix for working with matrices. However, unlike octave (which I was using till recently), * doesn't perform matrix multiplication, you need to use the function matrixmultipy(). I feel this makes the code very unreadable.

Does anybody share my views, and has found a solution?

how does multiplication differ for NumPy Matrix vs Array classes?

Python

numpy文档建议使用数组而不是矩阵来处理矩阵。但是，与octave (我最近还在使用它)不同，*不执行矩阵乘法，您需要使用matrixmultipy()函数。我觉得这使得代码很难读。有没有人同意我的观点，并找到了解决方案？

问NumPy矩阵与数组类的乘法有何不同？
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