blocks|key|1250792|text|按a的第二列排序|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1250793|a[a[:,+1].argsort()]|code-block|syntax|javascript|1250794|entityMap^0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

To sort by the second column of <code>a</code>:
<pre><code>a[a[:, 1].argsort()]
</code></pre>

blocks|key|1250838|text|@steve的answer实际上是最优雅的方式。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1250839|有关“正确”的方法，请参见numpy.ndarray.sort的order关键字参数|1250840|但是，您需要将数组视为具有字段的数组(结构化数组)。|1250841|如果你最初没有用字段定义你的数组，“正确”的方法是相当丑陋的……|1250842|作为一个快速示例，要对其进行排序并返回一个副本：|1250843|In+[1]:+import+numpy+as+np

In+[2]:+a+=+np.array([[1,2,3],[4,5,6],[0,0,1]])

In+[3]:+np.sort(a.view('i8,i8,i8'),+order=['f1'],+axis=0).view(np.int)
Out[3]:+
array([[0,+0,+1],
+++++++[1,+2,+3],
+++++++[4,+5,+6]])|code-block|syntax|javascript|1250844|要就地排序，请执行以下操作：|1250845|In+[6]:+a.view('i8,i8,i8').sort(order=['f1'],+axis=0)+#<--+returns+None

In+[7]:+a
Out[7]:+
array([[0,+0,+1],
+++++++[1,+2,+3],
+++++++[4,+5,+6]])|1250846|据我所知，@Steve's确实是最优雅的方式。|1250847|这种方法的唯一优点是，"+order“参数是对搜索进行排序所依据的字段列表。例如，通过提供order='f1'，'f2'，'f0‘，可以按第二列排序，然后按第三列排序，然后按第一列排序。|1250848|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/208339/steve-tjoa|1|https://stackoverflow.com/questions/2828059/sorting-arrays-in-numpy-by-column/2828121#2828121|2|http://docs.scipy.org/doc/numpy/reference/generated/numpy.ndarray.sort.html#numpy.ndarray.sort^0|0|6|0|7|6|1|0|D|I|2|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]|$A|1E|B|1F|1|1G]]|C|$]]|$1|D|3|E|5|6|7|1H|8|@]|9|@$A|1I|B|1J|1|1K]]|C|$]]|$1|F|3|G|5|6|7|1L|8|@]|9|@]|C|$]]|$1|H|3|I|5|6|7|1M|8|@]|9|@]|C|$]]|$1|J|3|K|5|6|7|1N|8|@]|9|@]|C|$]]|$1|L|3|M|5|N|7|1O|8|@]|9|@]|C|$O|P]]|$1|Q|3|R|5|6|7|1P|8|@]|9|@]|C|$]]|$1|S|3|T|5|N|7|1Q|8|@]|9|@]|C|$O|P]]|$1|U|3|V|5|6|7|1R|8|@]|9|@]|C|$]]|$1|W|3|X|5|6|7|1S|8|@]|9|@]|C|$]]|$1|Y|3|-4|5|6|7|1T|8|@]|9|@]|C|$]]]|Z|$10|$5|11|12|13|C|$14|15]]|16|$5|11|12|13|C|$14|17]]|18|$5|11|12|13|C|$14|19]]]]

<a href="https://stackoverflow.com/users/208339/steve-tjoa">@steve</a>'s <a href="https://stackoverflow.com/questions/2828059/sorting-arrays-in-numpy-by-column/2828121#2828121">answer</a> is actually the most elegant way of doing it.

For the "correct" way see the order keyword argument of <a href="http://docs.scipy.org/doc/numpy/reference/generated/numpy.ndarray.sort.html#numpy.ndarray.sort" rel="noreferrer">numpy.ndarray.sort</a> 

However, you'll need to view your array as an array with fields (a structured array).

The "correct" way is quite ugly if you didn't initially define your array with fields...

As a quick example, to sort it and return a copy:

<pre><code>In [1]: import numpy as np

In [2]: a = np.array([[1,2,3],[4,5,6],[0,0,1]])

In [3]: np.sort(a.view('i8,i8,i8'), order=['f1'], axis=0).view(np.int)
Out[3]: 
array([[0, 0, 1],
 [1, 2, 3],
 [4, 5, 6]])
</code></pre>

To sort it in-place:

<pre><code>In [6]: a.view('i8,i8,i8').sort(order=['f1'], axis=0) #&lt;-- returns None

In [7]: a
Out[7]: 
array([[0, 0, 1],
 [1, 2, 3],
 [4, 5, 6]])
</code></pre>

@Steve's really is the most elegant way to do it, as far as I know... 

The only advantage to this method is that the "order" argument is a list of the fields to order the search by. For example, you can sort by the second column, then the third column, then the first column by supplying order=['f1','f2','f0'].

blocks|key|1250917|text|在the+Python+documentation+wiki中，我认为你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1250918|a+=+([[1,+2,+3],+[4,+5,+6],+[0,+0,+1]]);+
a+=+sorted(a,+key=lambda+a_entry:+a_entry[1])+
print+a|code-block|syntax|javascript|1250919|输出为：|1250920|[[[0,+0,+1],+[1,+2,+3],+[4,+5,+6]]]|1250921|entityMap|0|LINK|mutability|MUTABLE|url|http://wiki.python.org/moin/HowTo/Sorting^0|1|T|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

From <a href="http://wiki.python.org/moin/HowTo/Sorting" rel="noreferrer">the Python documentation wiki</a>, I think you can do:

<pre><code>a = ([[1, 2, 3], [4, 5, 6], [0, 0, 1]]); 
a = sorted(a, key=lambda a_entry: a_entry[1]) 
print a
</code></pre>

The output is:

<pre><code>[[[0, 0, 1], [1, 2, 3], [4, 5, 6]]]
</code></pre>

blocks|key|1035903|text|来自the+NumPy+mailing+list的另一个解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1035904|>>>+a
array([[1,+2],
+++++++[0,+0],
+++++++[1,+0],
+++++++[0,+2],
+++++++[2,+1],
+++++++[1,+0],
+++++++[1,+0],
+++++++[0,+0],
+++++++[1,+0],
++++++[2,+2]])
>>>+a[np.lexsort(np.fliplr(a).T)]
array([[0,+0],
+++++++[0,+0],
+++++++[0,+2],
+++++++[1,+0],
+++++++[1,+0],
+++++++[1,+0],
+++++++[1,+0],
+++++++[1,+2],
+++++++[2,+1],
+++++++[2,+2]])|code-block|syntax|javascript|1035905|entityMap|0|LINK|mutability|MUTABLE|url|http://mail.scipy.org/pipermail/numpy-discussion/2008-December/039332.html^0|2|M|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

From <a href="http://mail.scipy.org/pipermail/numpy-discussion/2008-December/039332.html" rel="noreferrer">the NumPy mailing list</a>, here's another solution:

<pre><code>&gt;&gt;&gt; a
array([[1, 2],
 [0, 0],
 [1, 0],
 [0, 2],
 [2, 1],
 [1, 0],
 [1, 0],
 [0, 0],
 [1, 0],
 [2, 2]])
&gt;&gt;&gt; a[np.lexsort(np.fliplr(a).T)]
array([[0, 0],
 [0, 0],
 [0, 2],
 [1, 0],
 [1, 0],
 [1, 0],
 [1, 0],
 [1, 2],
 [2, 1],
 [2, 2]])
</code></pre>

blocks|key|1251004|text|如果有人想要在程序的关键部分使用排序，下面是不同建议的性能比较：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1251005|import+numpy+as+np
table+=+np.random.rand(5000,+10)

%25timeit+table.view('f8,f8,f8,f8,f8,f8,f8,f8,f8,f8').sort(order=['f9'],+axis=0)
1000+loops,+best+of+3:+1.88+ms+per+loop

%25timeit+table[table[:,9].argsort()]
10000+loops,+best+of+3:+180+µs+per+loop

import+pandas+as+pd
df+=+pd.DataFrame(table)
%25timeit+df.sort_values(9,+ascending=True)
1000+loops,+best+of+3:+400+µs+per+loop|code-block|syntax|javascript|1251006|因此，看起来使用argsort进行索引是目前为止最快的方法…|offset|length|1251007|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.argsort.html^0|0|0|8|7|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

In case someone wants to make use of sorting at a critical part of their programs here's a performance comparison for the different proposals:

<pre><code>import numpy as np
table = np.random.rand(5000, 10)

%timeit table.view('f8,f8,f8,f8,f8,f8,f8,f8,f8,f8').sort(order=['f9'], axis=0)
1000 loops, best of 3: 1.88 ms per loop

%timeit table[table[:,9].argsort()]
10000 loops, best of 3: 180 µs per loop

import pandas as pd
df = pd.DataFrame(table)
%timeit df.sort_values(9, ascending=True)
1000 loops, best of 3: 400 µs per loop
</code></pre>

So, it looks like indexing with <a href="http://docs.scipy.org/doc/numpy-1.10.0/reference/generated/numpy.argsort.html" rel="noreferrer">argsort</a> is the quickest method so far...

blocks|key|1251097|text|您可以按照Steve+Tjoa的方法对多个列进行排序，方法是使用像mergesort这样的稳定排序，并将索引从最不重要的列排序到最重要的列：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1251098|a+=+a[a[:,2].argsort()]+#+First+sort+doesn't+need+to+be+stable.
a+=+a[a[:,1].argsort(kind='mergesort')]
a+=+a[a[:,0].argsort(kind='mergesort')]|code-block|syntax|javascript|1251099|这将按列0排序，然后是1，然后是2。|1251100|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 can sort on multiple columns as per Steve Tjoa's method by using a stable sort like mergesort and sorting the indices from the least significant to the most significant columns:

<pre><code>a = a[a[:,2].argsort()] # First sort doesn't need to be stable.
a = a[a[:,1].argsort(kind='mergesort')]
a = a[a[:,0].argsort(kind='mergesort')]
</code></pre>

This sorts by column 0, then 1, then 2.

blocks|key|1036110|text|一个稍微复杂一点的lexsort示例-在第一列递减，在第二列递增。lexsort的诀窍在于它对行进行排序(因此是.T)，并将优先级给予最后一行。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1036111|In+[120]:+b=np.array([[1,2,1],[3,1,2],[1,1,3],[2,3,4],[3,2,5],[2,1,6]])
In+[121]:+b
Out[121]:+
array([[1,+2,+1],
+++++++[3,+1,+2],
+++++++[1,+1,+3],
+++++++[2,+3,+4],
+++++++[3,+2,+5],
+++++++[2,+1,+6]])
In+[122]:+b[np.lexsort(([1,-1]*b[:,[1,0]]).T)]
Out[122]:+
array([[3,+1,+2],
+++++++[3,+2,+5],
+++++++[2,+1,+6],
+++++++[2,+3,+4],
+++++++[1,+1,+3],
+++++++[1,+2,+1]])|code-block|syntax|javascript|1036112|entityMap^0|9|7|X|7|1K|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

A little more complicated <code>lexsort</code> example - descending on the 1st column, secondarily ascending on the 2nd. The tricks with <code>lexsort</code> are that it sorts on rows (hence the <code>.T</code>), and gives priority to the last.

<pre><code>In [120]: b=np.array([[1,2,1],[3,1,2],[1,1,3],[2,3,4],[3,2,5],[2,1,6]])
In [121]: b
Out[121]: 
array([[1, 2, 1],
 [3, 1, 2],
 [1, 1, 3],
 [2, 3, 4],
 [3, 2, 5],
 [2, 1, 6]])
In [122]: b[np.lexsort(([1,-1]*b[:,[1,0]]).T)]
Out[122]: 
array([[3, 1, 2],
 [3, 2, 5],
 [2, 1, 6],
 [2, 3, 4],
 [1, 1, 3],
 [1, 2, 1]])
</code></pre>

blocks|key|1036156|text|简单地使用排序，使用你想要排序的列编号。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1036157|a+=+np.array([1,1],+[1,-1],+[-1,1],+[-1,-1]])
print+(a)
a=a.tolist()+
a+=+np.array(sorted(a,+key=lambda+a_entry:+a_entry[0]))
print+(a)|code-block|syntax|javascript|1036158|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|$]]

Simply using sort, use coloumn number based on which you want to sort.

<pre><code>a = np.array([1,1], [1,-1], [-1,1], [-1,-1]])
print (a)
a=a.tolist() 
a = np.array(sorted(a, key=lambda a_entry: a_entry[0]))
print (a)
</code></pre>

blocks|key|1251236|text|这是一个古老的问题，但如果你需要将其推广到高于2维数组，以下是可以很容易推广的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1251237|np.einsum('ij->ij',+a[a[:,1].argsort(),:])|code-block|syntax|javascript|1251238|这对于两个维度来说是多余的，根据@steve的答案，a[a[:,1].argsort()]就足够了，但是这个答案不能推广到更高的维度。你可以找到an+example+of+3D+array+in+this+question.|offset|length|style|CODE|1251239|输出：|1251240|[[7+0+5]
+[9+2+3]
+[4+5+6]]|1251241|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/61449930/is-there-another-way-for-sorting-3d-numpy-array-by-column-values/61451923#61451923^0|0|0|Q|J|20|14|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|N|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

It is an old question but if you need to generalize this to a higher than 2 dimension arrays, here is the solution than can be easily generalized: 

<pre><code>np.einsum('ij-&gt;ij', a[a[:,1].argsort(),:])
</code></pre>

This is an overkill for two dimensions and <code>a[a[:,1].argsort()]</code> would be enough per @steve's answer, however that answer cannot be generalized to higher dimensions. You can find <a href="https://stackoverflow.com/questions/61449930/is-there-another-way-for-sorting-3d-numpy-array-by-column-values/61451923#61451923">an example of 3D array in this question.</a>

Output:

<pre><code>[[7 0 5]
 [9 2 3]
 [4 5 6]]
</code></pre>

blocks|key|1251298|text|import+numpy+as+np
a=np.array([[21,20,19,18,17],[16,15,14,13,12],[11,10,9,8,7],[6,5,4,3,2]])
y=np.argsort(a[:,2],kind='mergesort')#+a[:,2]=[19,14,9,4]
a=a[y]
print(a)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1251299|所需输出为[[6,5,4,3,2],[11,10,9,8,7],[16,15,14,13,12],[21,20,19,18,17]]|unstyled|offset|length|style|CODE|1251300|请注意，argsort(numArray)返回numArray的索引，因为它应该以排序的方式排列。|1251301|示例|1251302|x=np.array([8,1,5])+
z=np.argsort(x)+#[1,3,0]+are+the+**indices+of+the+predicted+sorted+array**
print(x[z])+#boolean+indexing+which+sorts+the+array+on+basis+of+indices+saved+in+z|1251303|答案是[1,5,8]|1251304|entityMap^0|0|5|1P|0|4|H|N|8|0|0|0|3|7|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@$G|W|H|X|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@$G|Z|H|10|I|J]|$G|11|H|12|I|J]]|9|@]|A|$]]|$1|M|3|N|5|F|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|F|7|15|8|@$G|16|H|17|I|J]]|9|@]|A|$]]|$1|S|3|-4|5|F|7|18|8|@]|9|@]|A|$]]]|T|$]]

<pre><code>import numpy as np
a=np.array([[21,20,19,18,17],[16,15,14,13,12],[11,10,9,8,7],[6,5,4,3,2]])
y=np.argsort(a[:,2],kind='mergesort')# a[:,2]=[19,14,9,4]
a=a[y]
print(a)
</code></pre>
Desired output is <code>[[6,5,4,3,2],[11,10,9,8,7],[16,15,14,13,12],[21,20,19,18,17]]</code>
note that <code>argsort(numArray)</code> returns the indices of an <code>numArray</code> as it was supposed to be arranged in a sorted manner.
example
<pre><code>x=np.array([8,1,5]) 
z=np.argsort(x) #[1,3,0] are the **indices of the predicted sorted array**
print(x[z]) #boolean indexing which sorts the array on basis of indices saved in z
</code></pre>
answer would be <code>[1,5,8]</code>

blocks|key|1251339|text|#用于沿第1列排序|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1251340|indexofsort=np.argsort(dataset[:,0],axis=-1,kind='stable')+
dataset+++=+dataset[indexofsort,:]|code-block|syntax|javascript|1251341|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|$]]

#for sorting along column 1
<pre><code>indexofsort=np.argsort(dataset[:,0],axis=-1,kind='stable') 
dataset = dataset[indexofsort,:]
</code></pre>

blocks|key|1251372|text|def+sort_np_array(x,+column=None,+flip=False):
++++x+=+x[np.argsort(x[:,+column])]
++++if+flip:
++++++++x+=+np.flip(x,+axis=0)
++++return+x|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1251373|原问题中的数组：|unstyled|1251374|a+=+np.array([[9,+2,+3],
++++++++++++++[4,+5,+6],
++++++++++++++[7,+0,+5]])|1251375|问题作者所期望的sort_np_array函数的结果：|offset|length|style|CODE|1251376|sort_np_array(a,+column=1,+flip=False)|1251377|[2]:+array([[7,+0,+5],
++++++++++++[9,+2,+3],
++++++++++++[4,+5,+6]])|1251378|entityMap^0|0|0|0|8|D|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|X|8|@$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$B|C]]|$1|S|3|-4|5|F|7|12|8|@]|9|@]|A|$]]]|T|$]]

<pre class="lang-py prettyprint-override"><code>def sort_np_array(x, column=None, flip=False):
 x = x[np.argsort(x[:, column])]
 if flip:
 x = np.flip(x, axis=0)
 return x
</code></pre>
Array in the original question:
<pre class="lang-py prettyprint-override"><code>a = np.array([[9, 2, 3],
 [4, 5, 6],
 [7, 0, 5]])
</code></pre>
The result of the <code>sort_np_array</code> function as expected by the author of the question:
<pre class="lang-py prettyprint-override"><code>sort_np_array(a, column=1, flip=False)
</code></pre>
<pre><code>[2]: array([[7, 0, 5],
 [9, 2, 3],
 [4, 5, 6]])
</code></pre>

blocks|key|1251424|text|感谢这篇文章：https://stackoverflow.com/a/5204280/13890678|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1251425|我用structured+array找到了一个更“通用”的答案。我认为这种方法的一个优点是代码更容易阅读。|1251426|import+numpy+as+np
a+=+np.array([[9,+2,+3],
+++++++++++[4,+5,+6],
+++++++++++[7,+0,+5]])

struct_a+=+np.core.records.fromarrays(
++++a.transpose(),+names="col1,+col2,+col3",+formats="i8,+i8,+i8"
)
struct_a.sort(order="col2")

print(struct_a)|code-block|syntax|javascript|1251427|[(7,+0,+5)+(9,+2,+3)+(4,+5,+6)]|1251428|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/5204280/13890678|1|https://numpy.org/doc/stable/user/basics.rec.html^0|7|18|0|0|2|G|1|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|F|3|G|5|H|7|14|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|H|7|15|8|@]|9|@]|C|$I|J]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

Thanks to this post: <a href="https://stackoverflow.com/a/5204280/13890678">https://stackoverflow.com/a/5204280/13890678</a>
I found a more &quot;generic&quot; answer using <a href="https://numpy.org/doc/stable/user/basics.rec.html" rel="nofollow noreferrer">structured array</a>.
I think one advantage of this method is that the code is easier to read.
<pre class="lang-py prettyprint-override"><code>import numpy as np
a = np.array([[9, 2, 3],
 [4, 5, 6],
 [7, 0, 5]])

struct_a = np.core.records.fromarrays(
 a.transpose(), names=&quot;col1, col2, col3&quot;, formats=&quot;i8, i8, i8&quot;
)
struct_a.sort(order=&quot;col2&quot;)

print(struct_a)
</code></pre>
<pre><code>[(7, 0, 5) (9, 2, 3) (4, 5, 6)]
</code></pre>

How can I sort an array in NumPy by the nth column?

For example,

<pre><code>a = array([[9, 2, 3],
 [4, 5, 6],
 [7, 0, 5]])
</code></pre>

I'd like to sort rows by the second column, such that I get back:

<pre><code>array([[7, 0, 5],
 [9, 2, 3],
 [4, 5, 6]])
</code></pre>

Sorting arrays in NumPy by column

如何在NumPy中按第n列对数组进行排序？例如,a = array([[9, 2, 3],           [4, 5, 6],           [7, 0, 5]])我想按第二列对行进行排序，这样我就可以返回：array([[7, 0, 5],       [9, 2, 3],       [4, 5, 6]...

问按列对NumPy中的数组进行排序
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