gha - 1 year ago 109

Python Question

I have a very large 2D array which looks something like this:

`a=`

[[a1, b1, c1],

[a2, b2, c2],

...,

[an, bn, cn]]

Using numpy, is there an easy way to get a new 2D array with e.g. 2 random rows from the initial array a (without replacement)?

e.g.

`b=`

[[a4, b4, c4],

[a99, b99, c99]]

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Answer Source

```
>>> A = np.random.randint(5, size=(10,3))
>>> A
array([[1, 3, 0],
[3, 2, 0],
[0, 2, 1],
[1, 1, 4],
[3, 2, 2],
[0, 1, 0],
[1, 3, 1],
[0, 4, 1],
[2, 4, 2],
[3, 3, 1]])
>>> idx = np.random.randint(10, size=2)
>>> idx
array([7, 6])
>>> A[idx,:]
array([[0, 4, 1],
[1, 3, 1]])
```

Putting it together for a general case:

```
A[np.random.randint(A.shape[0], size=2), :]
```

For non replacement (numpy 1.7.0+):

```
A[np.random.choice(A.shape[0], 2, replace=False), :]
```

I do not believe there is a good way to generate random list without replacement before 1.7. Perhaps you can setup a small definition that ensures the two values are not the same.

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