In Python 3, how to swap two sub-arrays in a 2-dimensional numpy array?
<p>Suppose there is an 2-dimensional numpy array (or matrix in maths) named "A", I want to swap its 1st row with another row "n". n is can be any natural number. The following codes do not work</p>

<pre><code>A = np.eye(3)
n = 2
A[0], A[n] = A[n], A[0]
print(A)
</code></pre>

<p>you can see it gives (I show the matrix form A for simplicity)</p>

<pre><code>A = 
   0, 0, 1
   0, 1, 0
   0, 0, 1
</code></pre>

<p>But what I want is</p>

<pre><code>A =
   0, 0, 1
   0, 1, 0
   1, 0, 0
</code></pre>

<p>I thought about one solution is introducing another matrix "B" which is equal to "A", but "A" and "B" are different objects. Then do this</p>

<pre><code>A = np.eye(3)
B = np.eye(3)
A[0], A[n] = B[n], B[0]
</code></pre>

<p>This can give the correct swap on A. But it needs an additional matrix "B", I don't know if it is computational efficient. Or maybe you have better idea? Thanks :)</p>

<p>List swapping swaps the variable through pass by reference. So, it does not work to do the inner element swapping by traditional swap by value which is <code>a,b=b,a</code></p>

<p>Here you can do the inner list modifications. That performs one after another which eliminates the overwriting problem.</p>

<p><strong>Code:</strong></p>

<pre><code>import numpy as np
A = np.eye(3)
n = 2
A[[0,n]] = A[[n,0]]
print(A)
</code></pre>

<p><strong>Output:</strong></p>

<pre><code>[[ 0.  0.  1.]
 [ 0.  1.  0.]
 [ 1.  0.  0.]]
</code></pre>

python arrays numpy matrix