blocks|key|32825|text|正常乘法，如您所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|32826|>>>+import+numpy+as+np
>>>+m+=+np.array([[1,2,3],[4,5,6],[7,8,9]])
>>>+c+=+np.array([0,1,2])
>>>+m+*+c
array([[+0,++2,++6],
+++++++[+0,++5,+12],
+++++++[+0,++8,+18]])|code-block|syntax|javascript|32827|如果添加一个轴，它将按您所需的方式进行倍增：|32828|>>>+m+*+c[:,+np.newaxis]
array([[+0,++0,++0],
+++++++[+4,++5,++6],
+++++++[14,+16,+18]])|32829|您还可以转置两次：|32830|>>>+(m.T+*+c).T
array([[+0,++0,++0],
+++++++[+4,++5,++6],
+++++++[14,+16,+18]])|32831|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|$]]

Normal multiplication like you showed:

<pre><code>&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; m = np.array([[1,2,3],[4,5,6],[7,8,9]])
&gt;&gt;&gt; c = np.array([0,1,2])
&gt;&gt;&gt; m * c
array([[ 0, 2, 6],
 [ 0, 5, 12],
 [ 0, 8, 18]])
</code></pre>

If you add an axis, it will multiply the way you want:

<pre><code>&gt;&gt;&gt; m * c[:, np.newaxis]
array([[ 0, 0, 0],
 [ 4, 5, 6],
 [14, 16, 18]])
</code></pre>

You could also transpose twice:

<pre><code>&gt;&gt;&gt; (m.T * c).T
array([[ 0, 0, 0],
 [ 4, 5, 6],
 [14, 16, 18]])
</code></pre>

blocks|key|249200|text|您还可以使用矩阵乘法(也称为点积)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|249201|a+=+[[1,2,3],[4,5,6],[7,8,9]]
b+=+[0,1,2]
c+=+numpy.diag(b)

numpy.dot(c,a)|code-block|syntax|javascript|249202|哪一个更优雅可能是一个品味问题。|249203|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You could also use matrix multiplication (aka dot product):

<pre><code>a = [[1,2,3],[4,5,6],[7,8,9]]
b = [0,1,2]
c = numpy.diag(b)

numpy.dot(c,a)
</code></pre>

Which is more elegant is probably a matter of taste.

blocks|key|249233|text|还有另一个技巧(从v1.6开始)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|249234|A=np.arange(1,10).reshape(3,3)
b=np.arange(3)

np.einsum('ij,i->ij',A,b)|code-block|syntax|javascript|249235|我精通numpy广播(newaxis)，但是我仍然在寻找使用这个新的einsum工具的方法。因此，我花了一些时间来寻找这个解决方案。|offset|length|style|CODE|249236|计时(使用Ipython+timeit)：|249237|einsum:+4.9+micro
transpose:+8.1+micro
newaxis:+8.35+micro
dot-diag:+10.5+micro|249238|顺便说一下，将i更改为j，np.einsum('ij,j->ij',A,b)会生成Alex不想要的矩阵。而np.einsum('ji,j->ji',A,b)实际上做了双重转座。|249239|entityMap^0|0|0|B|7|Y|6|0|0|0|7|1|B|1|D|P|1H|P|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|13|8|@$I|14|J|15|K|L]|$I|16|J|17|K|L]|$I|18|J|19|K|L]|$I|1A|J|1B|K|L]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|T|$]]

Yet another trick (as of v1.6)

<pre><code>A=np.arange(1,10).reshape(3,3)
b=np.arange(3)

np.einsum('ij,i-&gt;ij',A,b)
</code></pre>

I'm proficient with the numpy broadcasting (<code>newaxis</code>), but I'm still finding my way around this new <code>einsum</code> tool. So I had play around a bit to find this solution.

Timings (using Ipython timeit):

<pre><code>einsum: 4.9 micro
transpose: 8.1 micro
newaxis: 8.35 micro
dot-diag: 10.5 micro
</code></pre>

Incidentally, changing a <code>i</code> to <code>j</code>, <code>np.einsum('ij,j-&gt;ij',A,b)</code>, produces the matrix that Alex does not want. And <code>np.einsum('ji,j-&gt;ji',A,b)</code> does, in effect, the double transpose.

blocks|key|249259|text|你为什么不干脆|type|unstyled|depth|inlineStyleRanges|entityRanges|data|249260|>>>+m+=+np.array([[1,2,3],[4,5,6],[7,8,9]])
>>>+c+=+np.array([0,1,2])
>>>+(m.T+*+c).T|code-block|syntax|javascript|249261|?？|249262|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Why don't you just do

<pre><code>&gt;&gt;&gt; m = np.array([[1,2,3],[4,5,6],[7,8,9]])
&gt;&gt;&gt; c = np.array([0,1,2])
&gt;&gt;&gt; (m.T * c).T
</code></pre>

??

blocks|key|33001|text|我比较了不同选项的速度，发现--令我惊讶的是--所有选项(除了diag)都是同样快的。我个人使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|33002|A+*+b[:,+None]|code-block|syntax|javascript|33003|(或(A.T+*+b).T)，因为它很短。|33004|​|33005|📷|atomic|33006|33007|33008|重现该图的代码：|33009|import+numpy
import+perfplot


def+newaxis(data):
++++A,+b+=+data
++++return+A+*+b[:,+numpy.newaxis]


def+none(data):
++++A,+b+=+data
++++return+A+*+b[:,+None]


def+double_transpose(data):
++++A,+b+=+data
++++return+(A.T+*+b).T


def+double_transpose_contiguous(data):
++++A,+b+=+data
++++return+numpy.ascontiguousarray((A.T+*+b).T)


def+diag_dot(data):
++++A,+b+=+data
++++return+numpy.dot(numpy.diag(b),+A)


def+einsum(data):
++++A,+b+=+data
++++return+numpy.einsum("ij,i->ij",+A,+b)


perfplot.save(
++++"p.png",
++++setup=lambda+n:+(numpy.random.rand(n,+n),+numpy.random.rand(n)),
++++kernels=[
++++++++newaxis,
++++++++none,
++++++++double_transpose,
++++++++double_transpose_contiguous,
++++++++diag_dot,
++++++++einsum,
++++],
++++n_range=[2+**+k+for+k+in+range(13)],
++++xlabel="len(A),+len(b)",
)|33010|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/222276239b1c1c739007c497fda8e1ef.png|imageAlt^0|V|4|0|0|2|B|0|0|0|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1D|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|R|3|N|5|6|7|1I|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1K|8|@]|D|@]|E|$]]|$1|V|3|W|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|X|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14|15|-4]]]]

I've compared the different options for speed and found that – much to my surprise – all options (except <code>diag</code>) are equally fast. I personally use
<pre class="lang-py prettyprint-override"><code>A * b[:, None]
</code></pre>
(or <code>(A.T * b).T</code>) because it's short.
<a href="https://i.stack.imgur.com/dGicS.png" rel="noreferrer"><img src="https://i.stack.imgur.com/dGicS.png" alt="enter image description here" /></a>
<hr />
Code to reproduce the plot:
<pre class="lang-py prettyprint-override"><code>import numpy
import perfplot


def newaxis(data):
 A, b = data
 return A * b[:, numpy.newaxis]


def none(data):
 A, b = data
 return A * b[:, None]


def double_transpose(data):
 A, b = data
 return (A.T * b).T


def double_transpose_contiguous(data):
 A, b = data
 return numpy.ascontiguousarray((A.T * b).T)


def diag_dot(data):
 A, b = data
 return numpy.dot(numpy.diag(b), A)


def einsum(data):
 A, b = data
 return numpy.einsum(&quot;ij,i-&gt;ij&quot;, A, b)


perfplot.save(
 &quot;p.png&quot;,
 setup=lambda n: (numpy.random.rand(n, n), numpy.random.rand(n)),
 kernels=[
 newaxis,
 none,
 double_transpose,
 double_transpose_contiguous,
 diag_dot,
 einsum,
 ],
 n_range=[2 ** k for k in range(13)],
 xlabel=&quot;len(A), len(b)&quot;,
)
</code></pre>

blocks|key|33062|text|对于那些在谷歌上迷失的灵魂，使用numpy.expand_dims然后numpy.repeat可以工作，并且在更高维度的情况下也可以工作(例如，将形状(10，12，3)乘以(10，12))。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|33063|>>>+import+numpy
>>>+a+=+numpy.array([[1,2,3],[4,5,6],[7,8,9]])
>>>+b+=+numpy.array([0,1,2])
>>>+b0+=+numpy.expand_dims(b,+axis+=+0)
>>>+b0+=+numpy.repeat(b0,+a.shape[0],+axis+=+0)
>>>+b1+=+numpy.expand_dims(b,+axis+=+1)
>>>+b1+=+numpy.repeat(b1,+a.shape[1],+axis+=+1)
>>>+a*b0
array([[+0,++2,++6],
+++[+0,++5,+12],
+++[+0,++8,+18]])
>>>+a*b1
array([[+0,++0,++0],
+++[+4,++5,++6],
+++[14,+16,+18]])|code-block|syntax|javascript|33064|entityMap^0|G|H|Z|C|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

For those lost souls on google, using <code>numpy.expand_dims</code> then <code>numpy.repeat</code> will work, and will also work in higher dimensional cases (i.e. multiplying a shape (10, 12, 3) by a (10, 12)).

<pre><code>&gt;&gt;&gt; import numpy
&gt;&gt;&gt; a = numpy.array([[1,2,3],[4,5,6],[7,8,9]])
&gt;&gt;&gt; b = numpy.array([0,1,2])
&gt;&gt;&gt; b0 = numpy.expand_dims(b, axis = 0)
&gt;&gt;&gt; b0 = numpy.repeat(b0, a.shape[0], axis = 0)
&gt;&gt;&gt; b1 = numpy.expand_dims(b, axis = 1)
&gt;&gt;&gt; b1 = numpy.repeat(b1, a.shape[1], axis = 1)
&gt;&gt;&gt; a*b0
array([[ 0, 2, 6],
 [ 0, 5, 12],
 [ 0, 8, 18]])
&gt;&gt;&gt; a*b1
array([[ 0, 0, 0],
 [ 4, 5, 6],
 [14, 16, 18]])
</code></pre>

I'm trying to multiply each of the terms in a 2D array by the corresponding terms in a 1D array. This is very easy if I want to multiply every column by the 1D array, as shown in the <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.multiply.html">numpy.multiply</a> function. But I want to do the opposite, multiply each term in the row.
In other words I want to multiply:

<pre><code>[1,2,3] [0]
[4,5,6] * [1]
[7,8,9] [2]
</code></pre>

and get

<pre><code>[0,0,0]
[4,5,6]
[14,16,18]
</code></pre>

but instead I get

<pre><code>[0,2,6]
[0,5,12]
[0,8,18]
</code></pre>

Does anyone know if there's an elegant way to do that with numpy?
Thanks a lot,
Alex

Multiplying across in a numpy array

我尝试将二维数组中的每个项与一维数组中的相应项相乘。如果我想将每一列乘以一维数组，这是非常简单的，如函数所示。但我想做相反的事情，将每一项相乘。换句话说，我想做乘法：[1,2,3]   [0][4,5,6] * [1][7,8,9]   [2]并获取[0,0,0][4,5,6][14,16,18]但我却得到了[0,2,...

问在numpy数组中乘以
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