 aura -4 years ago 231
Python Question

# In Python 3, how to swap two sub-arrays in a 2-dimensional numpy array?

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

``````A = np.eye(3)
n = 2
A, A[n] = A[n], A
print(A)
``````

you can see it gives (I show the matrix form A for simplicity)

``````A =
0, 0, 1
0, 1, 0
0, 0, 1
``````

But what I want is

``````A =
0, 0, 1
0, 1, 0
1, 0, 0
``````

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

``````A = np.eye(3)
B = np.eye(3)
A, A[n] = B[n], B
``````

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 :) arsho
Answer Source

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 `a,b=b,a`

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

Code:

``````import numpy as np
A = np.eye(3)
n = 2
A[[0,n]] = A[[n,0]]
print(A)
``````

Output:

``````[[ 0.  0.  1.]
[ 0.  1.  0.]
[ 1.  0.  0.]]
``````
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