blocks|key|1189780|text|如果你想要更具可读性的东西，你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1189781|A+=+np.squeeze(np.asarray(M))|code-block|syntax|javascript|1189782|等同的，你也可以这样做：A+=+np.asarray(M).reshape(-1)，但是读起来不太容易。|offset|length|style|CODE|1189783|entityMap^0|0|0|C|T|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If you'd like something a bit more readable, you can do this:

<pre><code>A = np.squeeze(np.asarray(M))
</code></pre>

Equivalently, you could also do: <code>A = np.asarray(M).reshape(-1)</code>, but that's a bit less easy to read.

blocks|key|975568|text|或者你可以试着避免一些临时工|type|unstyled|depth|inlineStyleRanges|entityRanges|data|975569|A+=+M.view(np.ndarray)
A.shape+=+-1|code-block|syntax|javascript|975570|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|$]]

Or you could try to avoid some temps with

<pre><code>A = M.view(np.ndarray)
A.shape = -1
</code></pre>

blocks|key|1189852|text|A,+=+np.array(M.T)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1189853|这要看你说的优雅是什么意思了，我想这就是我要做的。|unstyled|1189854|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>A, = np.array(M.T)
</code></pre>

depends what you mean by elegance i suppose but thats what i would do

blocks|key|975615|text|您可以尝试以下变体：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|975616|result=np.array(M).flatten()|code-block|syntax|javascript|975617|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 try the following variant:

<pre><code>result=np.array(M).flatten()
</code></pre>

blocks|key|975630|text|result+=+M.A1|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|975631|https://numpy.org/doc/stable/reference/generated/numpy.matrix.A1.html|unstyled|offset|length|975632|matrix.A1
1-d+base+array|975633|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1X|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|S|8|@]|9|@$G|T|H|U|1|V]]|A|$]]|$1|I|3|J|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|K|3|-4|5|F|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|E]]]]

<pre><code>result = M.A1
</code></pre>
<a href="https://numpy.org/doc/stable/reference/generated/numpy.matrix.A1.html" rel="nofollow noreferrer">https://numpy.org/doc/stable/reference/generated/numpy.matrix.A1.html</a>
<pre><code>matrix.A1
1-d base array
</code></pre>

blocks|key|1189935|text|np.array(M).ravel()|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1189936|如果你关心速度；但是如果你关心内存：|unstyled|1189937|np.asarray(M).ravel()|1189938|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>np.array(M).ravel()
</code></pre>

If you care for speed; But if you care for memory:

<pre><code>np.asarray(M).ravel()
</code></pre>

blocks|key|1189958|text|首先是Mv+=+numpy.asarray(M.T)，它提供了一个4x1的二维数组。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1189959|然后，执行A+=+Mv[0,:]，这将为您提供所需的内容。你可以把它们放在一起，就像numpy.asarray(M.T)[0,:]。|1189960|entityMap^0|3|N|0|5|B|16|N|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

First, <code>Mv = numpy.asarray(M.T)</code>, which gives you a 4x1 but 2D array.

Then, perform <code>A = Mv[0,:]</code>, which gives you what you want. You could put them together, as <code>numpy.asarray(M.T)[0,:]</code>.

blocks|key|1189978|text|这将把矩阵转换成数组|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1189979|A+=+np.ravel(M).T|code-block|syntax|javascript|1189980|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|$]]

This will convert the matrix into array 

<pre><code>A = np.ravel(M).T
</code></pre>

blocks|key|1190057|text|来得有点晚，希望这能帮到什么人，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1190058|np.array(M.flat)|code-block|syntax|javascript|1190059|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|$]]

Came in a little late, hope this helps someone,
<pre><code>np.array(M.flat)
</code></pre>

I am using numpy. I have a matrix with 1 column and N rows and I want to get an array from with N elements.

For example, if i have <code>M = matrix([[1], [2], [3], [4]])</code>, I want to get <code>A = array([1,2,3,4])</code>.

To achieve it, I use <code>A = np.array(M.T)[0]</code>. Does anyone know a more elegant way to get the same result?

Thanks!

Numpy matrix to array

我正在使用numpy。我有一个1列N行的矩阵，我想从得到一个N个元素的数组。例如，如果我有M = matrix([[1], [2], [3], [4]])，我想要获取A = array([1,2,3,4])。为了实现它，我使用了A = np.array(M.T)[0]。有没有人知道更优雅的方法来获得同样的结果？谢谢!

问Numpy矩阵到数组
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Numpy矩阵到数组EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Numpy矩阵到数组
EN