blocks|key|4266609|text|将矩阵转换为数组，执行(广播)数组乘法，然后转换回来：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4266610|np.matrix(b+*+np.asarray(C))|code-block|syntax|javascript|4266611|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|$]]

Convert the matrix to an array, perform (broadcast) array multiplication, and convert back:

<pre><code>np.matrix(b * np.asarray(C))
</code></pre>

blocks|key|3786012|text|使用*，这是具有适当广播功能的元素产品：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3786013|>>>+b+=+array([1,2,3])
>>>+C+=+arange(9).reshape(3,3)
>>>+dot(diag(b),+C)
array([[+0,++1,++2],
+++++++[+6,++8,+10],
+++++++[18,+21,+24]])
>>>+atleast_2d(b).T+*+C
array([[+0,++1,++2],
+++++++[+6,++8,+10],
+++++++[18,+21,+24]])|code-block|syntax|javascript|3786014|atleast_2d(b).T+(或b.reshape(-1,1))将向量b重塑为列向量。|3786015|entityMap^0|2|1|0|0|0|F|I|F|11|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

Use <code>*</code>, the element-wise product with the appropriate broadcasting:

<pre><code>&gt;&gt;&gt; b = array([1,2,3])
&gt;&gt;&gt; C = arange(9).reshape(3,3)
&gt;&gt;&gt; dot(diag(b), C)
array([[ 0, 1, 2],
 [ 6, 8, 10],
 [18, 21, 24]])
&gt;&gt;&gt; atleast_2d(b).T * C
array([[ 0, 1, 2],
 [ 6, 8, 10],
 [18, 21, 24]])
</code></pre>

<code>atleast_2d(b).T</code> (or <code>b.reshape(-1,1)</code>) reshapes the vector <code>b</code> to a column vector.

blocks|key|5995205|text|做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5995206|b[:,np.newaxis]+*+C|code-block|syntax|javascript|5995207|或|5995208|b[:,None]+*+C|5995209|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Do

<pre><code>b[:,np.newaxis] * C
</code></pre>

or

<pre><code>b[:,None] * C
</code></pre>

blocks|key|3786067|text|您也可以使用einsum|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3786068|>>>+b+=+np.array([1,2,3])
>>>+C+=+np.arange(9).reshape(3,3)
>>>+np.einsum('i,ij->ij',+b,+C)
array([[+0,++1,++2],
+++++++[+6,++8,+10],
+++++++[18,+21,+24]])|code-block|syntax|javascript|3786069|在我的机器上，这种方法比两种方法都快|3786070|np.dot(np.diag(b),+C)|3786071|和|3786072|np.atleast_2d(b).T+*+C|3786073|对于100,000个循环，einsum耗时1.16秒，dot耗时1.61秒，atleast_2d耗时2.03秒。|3786074|如果您熟悉Einstein+summation+notation，那么einsum是一个非常有用的函数。使用索引表示法可以更容易地表示许多简单的线性代数运算，并且增加的执行速度只是锦上添花。|3786075|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy/reference/generated/numpy.einsum.html|1|http://en.wikipedia.org/wiki/Einstein_notation^0|6|6|0|0|0|0|0|0|0|0|5|R|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|19|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1B|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1D|8|@]|9|@]|C|$]]|$1|S|3|T|5|6|7|1E|8|@]|9|@$A|1F|B|1G|1|1H]]|C|$]]|$1|U|3|-4|5|6|7|1I|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]|12|$5|X|Y|Z|C|$10|13]]]]

You can also use <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.einsum.html" rel="nofollow">einsum</a>:

<pre><code>&gt;&gt;&gt; b = np.array([1,2,3])
&gt;&gt;&gt; C = np.arange(9).reshape(3,3)
&gt;&gt;&gt; np.einsum('i,ij-&gt;ij', b, C)
array([[ 0, 1, 2],
 [ 6, 8, 10],
 [18, 21, 24]])
</code></pre>

On my machine, this method is faster than both

<pre><code>np.dot(np.diag(b), C)
</code></pre>

and

<pre><code>np.atleast_2d(b).T * C
</code></pre>

For 100,000 loops, einsum takes 1.16 seconds, dot takes 1.61 seconds and atleast_2d takes 2.03 seconds.

If you are familiar with <a href="http://en.wikipedia.org/wiki/Einstein_notation" rel="nofollow">Einstein summation notation</a>, then einsum is an incredibly useful function. It is easier to formulate many simple linear algebra operations using index notation, and the added speed of execution is just icing on the cake.

blocks|key|4266723|text|结合已经给出的答案，并尝试一种新的(但效率低下的)方法，我提出了以下时间安排。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4266724|4266725|einsum:+0.54s(百里香)|unordered-list-item|4266726|广播:0.44s(幼虫)|4266727|点对角线:15.6s(亨利，TS)+|4266728|稀疏媒体矩阵:16.8s(自己尝试，惨败：-p)|4266729|<代码>F29|4266730|如果我重复einsum并广播每十次(更精确的计时)，我发现：|4266731|4266732|einsum:+5.2+s|4266733|broadcasting:+4.5+s|4266734|4266735|下面是它的测试代码：|4266736|import+timeit
from+scipy.sparse.dia+import+dia_matrix
import+numpy+as+np

def+get_bc():
++++N+=+10000
++++M+=+10000
++++b+=+np.random.random(size=N)
++++c+=+np.random.random(size=[N,+M])
++++return+b,+c

def+multiply_dot_diag():
++++global+b,+c
++++return+np.dot(np.diag(b),+c)

def+multiply_broadcast():
++++global+b,+c
++++return+np.atleast_2d(b).T+*+c

def+mulitply_sparse():
++++global+b,+c
++++N+=+b.size
++++return+dia_matrix((b,+0),+shape=[N,+N]).todense().dot(c)

def+multiply_einsum():
++++global+b,+c
++++return+np.einsum('i,ij->ij',+b,+c)

if+__name__+==+'__main__':
++++global+b,+c
++++b,+c+=+get_bc()
++++print+(b,+c)

++++print+(timeit.timeit(multiply_einsum,+number=1))
++++print+(timeit.timeit(multiply_broadcast,+number=1))
++++print+(timeit.timeit(multiply_dot_diag,+number=1))
++++print+(timeit.timeit(mulitply_sparse,+number=1))

++++np.testing.assert_array_equal(multiply_dot_diag(),+multiply_broadcast())
++++np.testing.assert_array_equal(multiply_dot_diag(),+mulitply_sparse())
++++np.testing.assert_array_equal(multiply_dot_diag(),+multiply_einsum())|code-block|syntax|javascript|4266737|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|16|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|17|8|@]|9|@]|A|$]]|$1|H|3|I|5|E|7|18|8|@]|9|@]|A|$]]|$1|J|3|K|5|E|7|19|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1A|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1B|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|Q|3|R|5|E|7|1D|8|@]|9|@]|A|$]]|$1|S|3|T|5|E|7|1E|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1G|8|@]|9|@]|A|$]]|$1|X|3|Y|5|Z|7|1H|8|@]|9|@]|A|$10|11]]|$1|12|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|13|$]]

Combining the answers already given, and trying out a new (but inefficient method), I come with the following timings.

<ul>
<li>einsum: 0.54 s (Thycydides) </li>
<li>broadcasting: 0.44 s (larsmans) </li>
<li>dot diagonal: 15.6 s (henrikr, TS) </li>
<li>sparse dia matrix: 16.8 s (own try, failed miserably :-p)</li>
</ul>

If I repeat einsum and broadcasting each ten times (more timing accuracy), I find:

<ul>
<li>einsum: 5.2 s</li>
<li>broadcasting: 4.5 s</li>
</ul>

Here is the testing code for it:

<pre><code>import timeit
from scipy.sparse.dia import dia_matrix
import numpy as np

def get_bc():
 N = 10000
 M = 10000
 b = np.random.random(size=N)
 c = np.random.random(size=[N, M])
 return b, c

def multiply_dot_diag():
 global b, c
 return np.dot(np.diag(b), c)

def multiply_broadcast():
 global b, c
 return np.atleast_2d(b).T * c

def mulitply_sparse():
 global b, c
 N = b.size
 return dia_matrix((b, 0), shape=[N, N]).todense().dot(c)

def multiply_einsum():
 global b, c
 return np.einsum('i,ij-&gt;ij', b, c)

if __name__ == '__main__':
 global b, c
 b, c = get_bc()
 print (b, c)

 print (timeit.timeit(multiply_einsum, number=1))
 print (timeit.timeit(multiply_broadcast, number=1))
 print (timeit.timeit(multiply_dot_diag, number=1))
 print (timeit.timeit(mulitply_sparse, number=1))

 np.testing.assert_array_equal(multiply_dot_diag(), multiply_broadcast())
 np.testing.assert_array_equal(multiply_dot_diag(), mulitply_sparse())
 np.testing.assert_array_equal(multiply_dot_diag(), multiply_einsum())
</code></pre>

I'm looking for a fast way of creating a new matrix <code>A_{ij}</code> from a given vector <code>b_i</code> and another
matrix <code>C_{ij}</code>. The components of the new matrix should have the form

<pre><code>A_{ij} = b_i * C_{ij}.
</code></pre>

So far I am using <code>dot(diag(b), C)</code>, but the dot product naturally has a lot of multiplications
with zero, which is quite inefficient. Is there a better way?

numpy: fast way of creating a matrix with entries c_i * C_{ij}

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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我正在寻找一种快速的方法来从给定的向量b_i和另一个矩阵C_{ij}创建一个新的矩阵A_{ij}。新矩阵的组件应具有以下形式A_{ij} = b_i * C_{ij}.到目前为止，我使用的是dot(diag(b), C)，但是点积自然有很多与零的乘法，这是相当低效的。有没有更好的方法？

问c_i :创建包含条目的矩阵的快速方法* C_{ij}
EN

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问c_i :创建包含条目的矩阵的快速方法* C_{ij}EN