blocks|key|675265|text|Scikit-learn提供了一个函数normalize()，可以让您应用各种标准化。“使其和为1”称为L1范数。因此：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|675266|from+sklearn.preprocessing+import+normalize

matrix+=+numpy.arange(0,27,3).reshape(3,3).astype(numpy.float64)
#+array([[++0.,+++3.,+++6.],
#++++++++[++9.,++12.,++15.],
#++++++++[+18.,++21.,++24.]])

normed_matrix+=+normalize(matrix,+axis=1,+norm='l1')
#+[[+0.++++++++++0.33333333++0.66666667]
#++[+0.25++++++++0.33333333++0.41666667]
#++[+0.28571429++0.33333333++0.38095238]]|code-block|syntax|javascript|675267|现在，您的行加起来将为1。|675268|entityMap|0|LINK|mutability|MUTABLE|url|https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.normalize.html^0|J|B|J|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Scikit-learn offers a function <a href="https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.normalize.html" rel="noreferrer"><code>normalize()</code></a> that lets you apply various normalizations. The &quot;make it sum to 1&quot; is called L1-norm. Therefore:
<pre><code>from sklearn.preprocessing import normalize

matrix = numpy.arange(0,27,3).reshape(3,3).astype(numpy.float64)
# array([[ 0., 3., 6.],
# [ 9., 12., 15.],
# [ 18., 21., 24.]])

normed_matrix = normalize(matrix, axis=1, norm='l1')
# [[ 0. 0.33333333 0.66666667]
# [ 0.25 0.33333333 0.41666667]
# [ 0.28571429 0.33333333 0.38095238]]
</code></pre>
Now your rows will sum to 1.

blocks|key|460873|text|我想这应该行得通，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|460874|a+=+numpy.arange(0,27.,3).reshape(3,3)

a+/=++a.sum(axis=1)[:,numpy.newaxis]|code-block|syntax|javascript|460875|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|$]]

I think this should work,

<pre><code>a = numpy.arange(0,27.,3).reshape(3,3)

a /= a.sum(axis=1)[:,numpy.newaxis]
</code></pre>

blocks|key|460977|text|如果您试图将每行归一化，使其大小为1(即，行的单位长度为1或行中每个元素的平方和为1)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|460978|import+numpy+as+np

a+=+np.arange(0,27,3).reshape(3,3)

result+=+a+/+np.linalg.norm(a,+axis=-1)[:,+np.newaxis]
#+array([[+0.++++++++,++0.4472136+,++0.89442719],
#++++++++[+0.42426407,++0.56568542,++0.70710678],
#++++++++[+0.49153915,++0.57346234,++0.65538554]])|code-block|syntax|javascript|460979|验证：|460980|np.sum(+result**2,+axis=-1+)
#+array([+1.,++1.,++1.])+|460981|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|$]]

In case you are trying to normalize each row such that its magnitude is one (i.e. a row's unit length is one or the sum of the square of each element in a row is one):

<pre><code>import numpy as np

a = np.arange(0,27,3).reshape(3,3)

result = a / np.linalg.norm(a, axis=-1)[:, np.newaxis]
# array([[ 0. , 0.4472136 , 0.89442719],
# [ 0.42426407, 0.56568542, 0.70710678],
# [ 0.49153915, 0.57346234, 0.65538554]])
</code></pre>

Verifying:

<pre><code>np.sum( result**2, axis=-1 )
# array([ 1., 1., 1.]) 
</code></pre>

blocks|key|461104|text|我认为您可以通过以下命令将行元素sum规范化为1：new_matrix+=+a+/+a.sum(axis=1,+keepdims=1)。列的规范化可以用new_matrix+=+a+/+a.sum(axis=0,+keepdims=1)来完成。希望这能有所帮助。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|461105|entityMap^0|P|16|24|16|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

I think you can normalize the row elements sum to 1 by this:
 <code>new_matrix = a / a.sum(axis=1, keepdims=1)</code>.
And the column normalization can be done with <code>new_matrix = a / a.sum(axis=0, keepdims=1)</code>. Hope this can hep.

blocks|key|461158|text|您可以使用内置的numpy函数：np.linalg.norm(a,+axis+=+1,+keepdims+=+True)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|461159|entityMap^0|G|18|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

You could use built-in numpy function: 
<code>np.linalg.norm(a, axis = 1, keepdims = True)</code>

blocks|key|461010|text|这似乎也是可行的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|461011|def+normalizeRows(M):
++++row_sums+=+M.sum(axis=1)
++++return+M+/+row_sums|code-block|syntax|javascript|461012|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|$]]

it appears that this also works

<pre><code>def normalizeRows(M):
 row_sums = M.sum(axis=1)
 return M / row_sums
</code></pre>

blocks|key|461072|text|您还可以使用矩阵转置：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|461073|(a.T+/+row_sums).T|code-block|syntax|javascript|461074|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 could also use matrix transposition:

<pre><code>(a.T / row_sums).T
</code></pre>

blocks|key|461032|text|或者使用lambda函数，比如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|461033|>>>+vec+=+np.arange(0,27,3).reshape(3,3)
>>>+import+numpy+as+np
>>>+norm_vec+=+map(lambda+row:+row/np.linalg.norm(row),+vec)|code-block|syntax|javascript|461034|vec的每个向量都有一个单位范数。|461035|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|$]]

Or using lambda function, like

<pre><code>&gt;&gt;&gt; vec = np.arange(0,27,3).reshape(3,3)
&gt;&gt;&gt; import numpy as np
&gt;&gt;&gt; norm_vec = map(lambda row: row/np.linalg.norm(row), vec)
</code></pre>

each vector of vec will have a unit norm.

blocks|key|676427|text|我们可以通过预乘对角矩阵来达到同样的效果，对角线的主对角线是行和的倒数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|676428|A+=+np.diag(A.sum(1)**-1)+@+A|code-block|syntax|javascript|676429|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|$]]

We can achieve the same effect by premultiplying with the diagonal matrix whose main diagonal is the reciprocal of the row sums.
<pre class="lang-py prettyprint-override"><code>A = np.diag(A.sum(1)**-1) @ A
</code></pre>

blocks|key|461120|text|normed_matrix+=+normalize(input_data,+axis=1,+norm='l1')
print(normed_matrix)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|461121|其中input_data是2D数组的名称|unstyled|461122|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>normed_matrix = normalize(input_data, axis=1, norm='l1')
print(normed_matrix)
</code></pre>

where input_data is the name of your 2D array

Given a 3 times 3 numpy array

<pre><code>a = numpy.arange(0,27,3).reshape(3,3)

# array([[ 0, 3, 6],
# [ 9, 12, 15],
# [18, 21, 24]])
</code></pre>

To normalize the rows of the 2-dimensional array I thought of

<pre><code>row_sums = a.sum(axis=1) # array([ 9, 36, 63])
new_matrix = numpy.zeros((3,3))
for i, (row, row_sum) in enumerate(zip(a, row_sums)):
 new_matrix[i,:] = row / row_sum
</code></pre>

There must be a better way, isn't there?

Perhaps to clearify: By normalizing I mean, the sum of the entrys per row must be one. But I think that will be clear to most people.

How to normalize a 2-dimensional numpy array in python less verbose?

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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给定一个3乘以3的数值数组a = numpy.arange(0,27,3).reshape(3,3)# array([[ 0,  3,  6],#        [ 9, 12, 15],#        [18, 21, 24]])为了标准化二维数组的行，我想到了row_sums = a.sum(axis=1) # array([ 9, 36, 63])new_matrix = numpy.ze

问如何在python中对二维numpy数组进行规范化处理？
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